DP (Grade 11 & 12) · Maths AA SL
Functions
300 questions across 6 sub-topics
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Composite & Inverse Functions
Linear and Quadratic Functions
Rational Functions
Exponential and Logarithmic Functions
Transformations
Function Sketching
Composite & Inverse Functions 50 questions
QUESTION 1
3 marks
Easy
Let \(f(x) = \sqrt{x-3}\).
(a) Write down the largest possible domain of \(f\).
(b) Write down the range of \(f\) for this domain.
Show complete worked solution
**(a)**
For a square root to be defined (over the reals), the expression under it must be non-negative:
$$x - 3 \geq 0$$
Solving:
$$x \geq 3$$
$$\boxed{\text{Domain: } x \in \mathbb{R}, \ x \geq 3}$$
**(b)**
A square root always gives a value greater than or equal to 0, and as $x$ increases from 3, $f(x)$ increases from 0 without bound, so every non-negative value is attained.
$$\boxed{\text{Range: } f(x) \geq 0}$$
QUESTION 2
4 marks
Easy
Let \(g(x) = \dfrac{2x+1}{x-4}\), \(x \neq 4\).
Find \(g^{-1}(x)\), and write down its domain.
Show complete worked solution
Let $y = g(x)$ and interchange the roles of $x$ and $y$, then solve for $x$ in terms of $y$:
$$y = \frac{2x+1}{x-4}$$
Multiplying both sides by $(x-4)$:
$$y(x-4) = 2x+1$$
$$yx - 4y = 2x + 1$$
Collecting the terms containing $x$ on one side:
$$yx - 2x = 4y + 1$$
$$x(y-2) = 4y+1$$
Solving for $x$:
$$x = \frac{4y+1}{y-2}$$
Renaming $y$ as $x$ to express the inverse function:
$$\boxed{g^{-1}(x) = \frac{4x+1}{x-2}}$$
The domain of $g^{-1}$ excludes the value that makes the denominator zero:
$$\boxed{\text{Domain: } x \neq 2}$$
Checking: $g(5) = \dfrac{11}{1} = 11$, and $g^{-1}(11) = \dfrac{45}{9} = 5$, which agrees.
QUESTION 3
5 marks
Medium
Let \(f(x) = x^2 - 4x + 1\) for \(x \geq 2\).
(a) By completing the square, find the range of \(f\).
(b) Find \(f^{-1}(x)\), and state its domain and range.
Show complete worked solution
**(a)**
Completing the square:
$$f(x) = x^2 - 4x + 1 = (x-2)^2 - 4 + 1 = (x-2)^2 - 3$$
For $x \geq 2$, $(x-2)^2 \geq 0$ and $f$ is increasing, so the minimum value occurs at $x=2$:
$$f(2) = (2-2)^2-3=-3$$
$$\boxed{\text{Range of } f: \ f(x) \geq -3}$$
**(b)**
Let $y = (x-2)^2 - 3$ and solve for $x$:
$$y + 3 = (x-2)^2$$
$$x - 2 = \pm\sqrt{y+3}$$
Since $x \geq 2$, take the positive root:
$$x = 2 + \sqrt{y+3}$$
Renaming $y$ as $x$:
$$\boxed{f^{-1}(x) = 2 + \sqrt{x+3}}$$
The domain of $f^{-1}$ equals the range of $f$, and the range of $f^{-1}$ equals the domain of $f$:
$$\boxed{\text{Domain: } x \geq -3, \quad \text{Range: } f^{-1}(x) \geq 2}$$
QUESTION 4
5 marks
Medium
Let \(h(x) = 5 - 2^{x-1}\) for \(x \in \mathbb{R}\).
(a) Find the range of \(h\).
(b) Find \(h^{-1}(x)\), stating its domain and range.
Show complete worked solution
**(a)**
Since $2^{x-1} > 0$ for all real $x$, and can be made arbitrarily close to 0 (as $x \to -\infty$) or arbitrarily large (as $x \to \infty$):
$$h(x) = 5 - 2^{x-1} < 5$$
$$\boxed{\text{Range of } h: \ h(x) < 5}$$
**(b)**
Let $y = 5 - 2^{x-1}$ and solve for $x$:
$$2^{x-1} = 5 - y$$
Taking $\log_2$ of both sides:
$$x - 1 = \log_2(5-y)$$
$$x = 1 + \log_2(5-y)$$
Renaming $y$ as $x$:
$$\boxed{h^{-1}(x) = 1 + \log_2(5-x)}$$
The domain of $h^{-1}$ equals the range of $h$, and the range of $h^{-1}$ equals the domain of $h$:
$$\boxed{\text{Domain: } x < 5, \quad \text{Range: } h^{-1}(x) \in \mathbb{R}}$$
QUESTION 5
7 marks
Hard
Consider \(f(x) = x^2 + 6x + 5\), defined for \(x \geq k\), where \(k\) is the smallest possible value such that \(f\) is a one-to-one function.
(a) Find the value of \(k\).
(b) For this domain, find \(f^{-1}(x)\), stating its domain and range.
Show complete worked solution
**(a)**
Completing the square to find the vertex:
$$f(x) = x^2+6x+5 = (x+3)^2 - 9 + 5 = (x+3)^2 - 4$$
The vertex is at $x=-3$. A quadratic is one-to-one only on the branch either side of its vertex, so restricting to $x \geq -3$ (the increasing branch) gives the smallest such $k$:
$$\boxed{k = -3}$$
**(b)**
Let $y = (x+3)^2 - 4$, $x \geq -3$, and solve for $x$:
$$y + 4 = (x+3)^2$$
$$x + 3 = \pm\sqrt{y+4}$$
Since $x \geq -3$, take the positive root:
$$x = -3 + \sqrt{y+4}$$
Renaming $y$ as $x$:
$$\boxed{f^{-1}(x) = -3 + \sqrt{x+4}}$$
The domain of $f^{-1}$ equals the range of $f$; since the minimum of $f$ on $x \geq -3$ is $f(-3)=-4$, the range of $f$ is $f(x) \geq -4$. The range of $f^{-1}$ equals the domain of $f$:
$$\boxed{\text{Domain: } x \geq -4, \quad \text{Range: } f^{-1}(x) \geq -3}$$
QUESTION 6
4 marks
Easy
Let \(f(x) = x+3\) and \(g(x) = 2x^2\).
(a) Find \((g \circ f)(x)\), giving your answer in expanded form.
(b) Find \((f \circ g)(2)\).
Show complete worked solution
**(a)**
Since $(g\circ f)(x) = g(f(x))$, substitute $f(x)=x+3$ into $g$:
$$g(f(x)) = 2(x+3)^2$$
Expanding:
$$2(x^2+6x+9) = 2x^2+12x+18$$
$$\boxed{(g\circ f)(x) = 2x^2+12x+18}$$
**(b)**
Since $(f\circ g)(2) = f(g(2))$, first find $g(2)$:
$$g(2) = 2(2)^2 = 8$$
Then substitute into $f$:
$$f(8) = 8+3=11$$
$$\boxed{(f\circ g)(2) = 11}$$
QUESTION 7
4 marks
Easy
Let \(f(x) = 4x - 1\).
(a) Find \(f^{-1}(x)\).
(b) Hence find \(f(f^{-1}(5))\), showing your working.
Show complete worked solution
**(a)**
Let $y=4x-1$ and solve for $x$:
$$y+1 = 4x \implies x = \frac{y+1}{4}$$
Renaming $y$ as $x$:
$$\boxed{f^{-1}(x) = \frac{x+1}{4}}$$
**(b)**
Finding $f^{-1}(5)$ using the formula from part (a):
$$f^{-1}(5) = \frac{5+1}{4} = \frac{6}{4} = \frac{3}{2}$$
Substituting into $f$:
$$f\left(\frac{3}{2}\right) = 4\left(\frac{3}{2}\right) - 1 = 6-1 = 5$$
$$\boxed{f(f^{-1}(5)) = 5}$$
(This confirms the general property $f(f^{-1}(x)) = x$.)
QUESTION 8
4 marks
Medium
Let \(f(x) = 2x-3\) and \(g(x) = x^2+1\).
(a) Find \((f \circ g)(x)\).
(b) Solve the equation \((f\circ g)(x) = 15\).
Show complete worked solution
**(a)**
Since $(f\circ g)(x) = f(g(x))$, substitute $g(x) = x^2+1$ into $f$:
$$(f\circ g)(x) = 2(x^2+1) - 3$$
Expanding and simplifying:
$$= 2x^2+2-3 = 2x^2-1$$
$$\boxed{(f\circ g)(x) = 2x^2-1}$$
**(b)**
Setting the expression from part (a) equal to 15:
$$2x^2 - 1 = 15$$
Solving for $x^2$:
$$2x^2 = 16$$
$$x^2 = 8$$
Taking square roots of both sides:
$$x = \pm\sqrt{8} = \pm 2\sqrt{2}$$
$$\boxed{x = 2\sqrt2 \text{ or } x=-2\sqrt2}$$
QUESTION 9
5 marks
Medium
Let \(f(x) = \sqrt{x-1}\) and \(g(x) = x^2-4\), for \(x \geq 0\).
(a) Find \((f\circ g)(x)\) and state its domain.
(b) Solve \((f\circ g)(x) = 3\), giving your answer correct to three significant figures.
Show complete worked solution
**(a)**
Since $(f\circ g)(x) = f(g(x))$, substitute $g(x) = x^2-4$ into $f$:
$$(f\circ g)(x) = \sqrt{(x^2-4)-1} = \sqrt{x^2-5}$$
For the square root to be defined we need $x^2-5 \geq 0$, i.e. $x^2 \geq 5$; combined with the given domain $x\geq 0$:
$$x \geq \sqrt{5}$$
$$\boxed{(f\circ g)(x) = \sqrt{x^2-5}, \quad \text{domain } x \geq \sqrt{5} \ (\approx 2.24)}$$
**(b)**
Setting the composite function equal to 3:
$$\sqrt{x^2-5} = 3$$
Squaring both sides:
$$x^2 - 5 = 9$$
$$x^2 = 14$$
Taking the positive square root, since the domain requires $x \geq \sqrt5 > 0$:
$$x = \sqrt{14} \approx 3.74$$
$$\boxed{x \approx 3.74}$$
QUESTION 10
5 marks
Hard
Let \(f(x) = 2x+1\).
Given that \((f \circ g)(x) = 6x^2 - 4x + 3\), find \(g(x)\).
Show complete worked solution
Writing the composite explicitly using the definition of $f$:
$$(f\circ g)(x) = f(g(x)) = 2g(x) + 1$$
This must equal the given expression:
$$2g(x) + 1 = 6x^2 - 4x + 3$$
Solving for $g(x)$:
$$2g(x) = 6x^2-4x+2$$
$$g(x) = 3x^2 - 2x + 1$$
$$\boxed{g(x) = 3x^2-2x+1}$$
Checking: $f(g(x)) = 2(3x^2-2x+1)+1 = 6x^2-4x+2+1 = 6x^2-4x+3$, which agrees.
QUESTION 11
5 marks
Easy
Let \(f(x) = 3x - 2\) and \(g(x) = x + 5\).
(a) Find \((f \circ g)(x)\). [2]
(b) Find \((g \circ f)(x)\). [2]
(c) Hence state whether \(f \circ g = g \circ f\). [1]
Show complete worked solution
**(a)**
Since $(f\circ g)(x) = f(g(x))$, substitute $g(x) = x+5$ into $f$:
$$f(g(x)) = 3(x+5) - 2$$
Expanding:
$$= 3x+15-2 = 3x+13$$
$$\boxed{(f\circ g)(x) = 3x+13}$$
**(b)**
Since $(g\circ f)(x) = g(f(x))$, substitute $f(x) = 3x-2$ into $g$:
$$g(f(x)) = (3x-2) + 5 = 3x+3$$
$$\boxed{(g\circ f)(x) = 3x+3}$$
**(c)**
Comparing the two results, $3x+13 \neq 3x+3$ for any value of $x$, so:
$$\boxed{f\circ g \neq g\circ f}$$
Composition of functions is not commutative in general.
QUESTION 12
4 marks
Easy
Let \(f(x) = \dfrac{x}{2} + 1\).
(a) Find \(f^{-1}(x)\). [2]
(b) Verify that \(f(f^{-1}(6)) = 6\). [2]
Show complete worked solution
**(a)**
Let $y = \dfrac{x}{2}+1$ and solve for $x$:
$$y - 1 = \frac{x}{2}$$
$$x = 2(y-1) = 2y-2$$
Renaming $y$ as $x$:
$$\boxed{f^{-1}(x) = 2x-2}$$
**(b)**
Using the formula from part (a):
$$f^{-1}(6) = 2(6)-2 = 10$$
Substituting into $f$:
$$f(10) = \frac{10}{2}+1 = 5+1 = 6$$
$$\boxed{f(f^{-1}(6)) = 6}$$
This confirms the general property $f(f^{-1}(x)) = x$.
QUESTION 13
3 marks
Easy
Let \(f(x) = 2x^2\) for \(x \geq 0\), and \(g(x) = x-1\).
Find \((f \circ g)(3)\).
Show complete worked solution
Since $(f\circ g)(3) = f(g(3))$, first find $g(3)$:
$$g(3) = 3-1 = 2$$
Then substitute into $f$:
$$f(2) = 2(2)^2 = 2(4) = 8$$
$$\boxed{(f\circ g)(3) = 8}$$
QUESTION 14
3 marks
Easy
Let \(h(x) = \dfrac{1}{x+2}\), \(x \neq -2\).
(a) State the largest possible domain of \(h\). [1]
(b) State the range of \(h\) for this domain. [2]
Show complete worked solution
**(a)**
The expression $\dfrac{1}{x+2}$ is undefined only when the denominator is zero, i.e. when $x=-2$. Every other real value of $x$ is allowed:
$$\boxed{\text{Domain: } x \in \mathbb{R}, \ x \neq -2}$$
**(b)**
As $x$ takes every real value except $-2$, the quantity $x+2$ takes every real value except $0$, so $\dfrac{1}{x+2}$ takes every real value except $0$ (it can never equal zero, since a fraction with a fixed non-zero numerator cannot be zero):
$$\boxed{\text{Range: } h(x) \neq 0}$$
QUESTION 15
3 marks
Easy
Let \(f(x) = \sqrt{10-3x}\).
(a) Write down the largest possible domain of \(f\). [1]
(b) Write down the range of \(f\) for this domain. [2]
Show complete worked solution
**(a)**
For a square root to be defined (over the reals), the expression under it must be non-negative:
$$10-3x \geq 0$$
Solving:
$$10 \geq 3x \implies x \leq \frac{10}{3}$$
$$\boxed{\text{Domain: } x \in \mathbb{R}, \ x \leq \tfrac{10}{3}}$$
**(b)**
A square root always gives a value greater than or equal to 0. At $x = \tfrac{10}{3}$, $f(x)=0$, and as $x$ decreases without bound, $10-3x \to \infty$ so $f(x)$ increases without bound. Every non-negative value is therefore attained:
$$\boxed{\text{Range: } f(x) \geq 0}$$
QUESTION 16
5 marks
Easy
A conversion formula between Celsius \(C\) and Fahrenheit \(F\) is given by \(F(C) = \dfrac{9C}{5} + 32\).
(a) Find \(F(20)\). [2]
(b) Find the inverse function \(F^{-1}\), which converts a Fahrenheit temperature back to Celsius. [2]
(c) Hence find \(F^{-1}(68)\). [1]
Show complete worked solution
**(a)**
Substituting $C=20$ into the formula:
$$F(20) = \frac{9(20)}{5} + 32 = \frac{180}{5}+32 = 36+32$$
$$\boxed{F(20) = 68}$$
**(b)**
Let $y = \dfrac{9C}{5}+32$ and solve for $C$:
$$y - 32 = \frac{9C}{5}$$
$$C = \frac{5(y-32)}{9}$$
Renaming $y$ as $x$:
$$\boxed{F^{-1}(x) = \frac{5(x-32)}{9}}$$
**(c)**
Using the formula from part (b):
$$F^{-1}(68) = \frac{5(68-32)}{9} = \frac{5(36)}{9} = \frac{180}{9}$$
$$\boxed{F^{-1}(68) = 20}$$
This agrees with part (a), as expected since $F^{-1}$ undoes $F$.
QUESTION 17
4 marks
Easy
Let \(f(x) = 5x+2\).
(a) Find \(f^{-1}(x)\). [2]
(b) Hence evaluate \(f^{-1}(17)\). [2]
Show complete worked solution
**(a)**
Let $y=5x+2$ and solve for $x$:
$$y-2 = 5x \implies x = \frac{y-2}{5}$$
Renaming $y$ as $x$:
$$\boxed{f^{-1}(x) = \frac{x-2}{5}}$$
**(b)**
Using the formula from part (a):
$$f^{-1}(17) = \frac{17-2}{5} = \frac{15}{5}$$
$$\boxed{f^{-1}(17) = 3}$$
QUESTION 18
3 marks
Easy
Let \(g(x) = x^3 - 1\), \(x \in \mathbb{R}\).
Find \(g^{-1}(x)\).
Show complete worked solution
Let $y = x^3-1$ and solve for $x$:
$$y+1 = x^3$$
Taking the cube root of both sides (valid for all real numbers, since cube roots of negative numbers are defined):
$$x = \sqrt[3]{y+1}$$
Renaming $y$ as $x$:
$$\boxed{g^{-1}(x) = \sqrt[3]{x+1}}$$
QUESTION 19
5 marks
Easy
Let \(f(x) = 2x-1\) and \(g(x) = 3x+4\).
(a) Find \((f \circ g)(x)\). [2]
(b) Solve the equation \((f\circ g)(x) = 25\). [3]
Show complete worked solution
**(a)**
Since $(f\circ g)(x) = f(g(x))$, substitute $g(x) = 3x+4$ into $f$:
$$(f\circ g)(x) = 2(3x+4) - 1$$
Expanding and simplifying:
$$= 6x+8-1 = 6x+7$$
$$\boxed{(f\circ g)(x) = 6x+7}$$
**(b)**
Setting the expression from part (a) equal to 25:
$$6x+7 = 25$$
Solving for $x$:
$$6x = 18$$
$$\boxed{x = 3}$$
QUESTION 20
4 marks
Easy
Consider \(f(x) = x^2\), \(x \in \mathbb{R}\).
(a) Explain, giving a specific numerical example, why \(f\) is not a one-to-one function. [2]
(b) State the largest domain containing \(x=0\) for which \(f\) becomes one-to-one. [2]
Show complete worked solution
**(a)**
A function is one-to-one only if no two different inputs give the same output. For $f(x)=x^2$:
$$f(-2) = (-2)^2 = 4 \qquad \text{and} \qquad f(2) = (2)^2 = 4$$
Since $f(-2) = f(2) = 4$ with $-2 \neq 2$, two different inputs produce the same output.
$$\boxed{\text{$f$ is not one-to-one, since } f(-2)=f(2)=4}$$
(Equivalently, a horizontal line such as $y=4$ crosses the graph of $f$ more than once.)
**(b)**
Restricting the domain to one side of the vertex $x=0$ removes the repetition seen in part (a). The largest such domain containing $x=0$ is:
$$\boxed{x \geq 0}$$
(On this restricted domain, $f$ is increasing, so it is one-to-one and an inverse function exists.)
QUESTION 21
4 marks
Easy
Let \(f(x) = \dfrac{x-4}{3}\).
(a) Find \(f^{-1}(x)\). [2]
(b) Verify algebraically that \(f(f^{-1}(x)) = x\) for all \(x\). [2]
Show complete worked solution
**(a)**
Let $y = \dfrac{x-4}{3}$ and solve for $x$:
$$3y = x-4$$
$$x = 3y+4$$
Renaming $y$ as $x$:
$$\boxed{f^{-1}(x) = 3x+4}$$
**(b)**
Substituting $f^{-1}(x) = 3x+4$ into $f$:
$$f(f^{-1}(x)) = f(3x+4) = \frac{(3x+4)-4}{3} = \frac{3x}{3}$$
$$\boxed{f(f^{-1}(x)) = x} \quad \text{for all } x$$
This confirms $f^{-1}$ correctly undoes $f$.
QUESTION 22
4 marks
Easy
The diagram shows the graph of \(y=f(x)\), a straight line, together with the line \(y=x\). The point \(P(1, 4)\) lies on the graph of \(f\).
(a) Write down the coordinates of the point on the graph of \(y = f^{-1}(x)\) that corresponds to \(P\). [2]
(b) Given that \(f(x) = 3x+1\), find \(f^{-1}(x)\), and use it to verify your answer to part (a). [2]
Show complete worked solution
**(a)**
The graph of $y=f^{-1}(x)$ is the reflection of the graph of $y=f(x)$ in the line $y=x$. Reflecting a point in $y=x$ swaps its coordinates, so the point on $f$ with coordinates $(1,4)$ corresponds to a point on $f^{-1}$ with coordinates:
$$\boxed{(4, 1)}$$
**(b)**
Let $y = 3x+1$ and solve for $x$:
$$y-1 = 3x \implies x = \frac{y-1}{3}$$
Renaming $y$ as $x$:
$$\boxed{f^{-1}(x) = \frac{x-1}{3}}$$
Checking against part (a):
$$f^{-1}(4) = \frac{4-1}{3} = \frac{3}{3} = 1$$
This gives the point $(4,1)$ on $f^{-1}$, which agrees with part (a).
QUESTION 23
3 marks
Easy
Let \(f(x) = 4-x\), \(x \in \mathbb{R}\).
Show that \(f\) is a self-inverse function, i.e. that \(f^{-1}(x) = f(x)\).
Show complete worked solution
Let $y=4-x$ and solve for $x$:
$$x = 4-y$$
Renaming $y$ as $x$:
$$f^{-1}(x) = 4-x$$
Comparing this with the original function $f(x) = 4-x$:
$$\boxed{f^{-1}(x) = f(x) = 4-x}$$
Since $f^{-1}(x)$ is identical to $f(x)$, $f$ is self-inverse.
QUESTION 24
5 marks
Easy
A company's total cost, in dollars, for producing \(n\) items is modelled by \(C(n) = 15n+20\), for \(n \geq 0\).
(a) Find the cost of producing 12 items. [2]
(b) Find the inverse function \(C^{-1}(x)\), and interpret what it represents in this context. [3]
Show complete worked solution
**(a)**
Substituting $n=12$:
$$C(12) = 15(12)+20 = 180+20$$
$$\boxed{C(12) = \$200}$$
**(b)**
Let $y = 15n+20$ and solve for $n$:
$$y-20 = 15n$$
$$n = \frac{y-20}{15}$$
Renaming $y$ as $x$:
$$\boxed{C^{-1}(x) = \frac{x-20}{15}}$$
Since $C$ takes a number of items and gives a total cost, $C^{-1}$ takes a total cost (in dollars) and gives the number of items that could be produced for that cost.
QUESTION 25
4 marks
Easy
Let \(f(x) = -2x+7\).
(a) Find \(f^{-1}(x)\). [2]
(b) By first finding \(f(3)\), verify that \(f^{-1}(f(3)) = 3\). [2]
Show complete worked solution
**(a)**
Let $y=-2x+7$ and solve for $x$:
$$y-7 = -2x$$
$$x = \frac{7-y}{2}$$
Renaming $y$ as $x$:
$$\boxed{f^{-1}(x) = \frac{7-x}{2}}$$
**(b)**
Finding $f(3)$:
$$f(3) = -2(3)+7 = -6+7 = 1$$
Substituting into $f^{-1}$:
$$f^{-1}(1) = \frac{7-1}{2} = \frac{6}{2} = 3$$
$$\boxed{f^{-1}(f(3)) = 3}$$
QUESTION 26
4 marks
Easy
The diagram shows part of the graph of \(y=f(x)\) and the line \(y=x\). The points \(A(0,-2)\), \(B(2,0)\) and \(C(3,2)\) lie on the graph of \(f\).
(a) Write down the coordinates of the three corresponding points on the graph of \(y=f^{-1}(x)\). [3]
(b) State the single geometric transformation that maps the graph of \(f\) onto the graph of \(f^{-1}\). [1]
Show complete worked solution
**(a)**
The graph of $f^{-1}$ is obtained by reflecting the graph of $f$ in the line $y=x$, which swaps the $x$- and $y$-coordinates of every point. Applying this to $A$, $B$ and $C$:
| Point on $f$ | Corresponding point on $f^{-1}$ |
|---|---|
| $A(0,-2)$ | $(-2, 0)$ |
| $B(2,0)$ | $(0, 2)$ |
| $C(3,2)$ | $(2, 3)$ |
$$\boxed{(-2,0), \ (0,2), \ (2,3)}$$
**(b)**
$$\boxed{\text{Reflection in the line } y=x}$$
QUESTION 27
7 marks
Medium
Let \(f(x) = x^2 - 2x + 5\), defined for \(x \geq 1\).
(a) Write \(f(x)\) in the form \((x-h)^2+k\). [1]
(b) Hence find the range of \(f\). [3]
(c) Find \(f^{-1}(x)\), stating its domain and range. [3]
Show complete worked solution
**(a)**
Completing the square:
$$f(x) = x^2-2x+5 = (x-1)^2 - 1 + 5$$
$$\boxed{f(x) = (x-1)^2+4}$$
**(b)**
The vertex of $f(x) = (x-1)^2+4$ is at $x=1$, where $(x-1)^2=0$ gives the minimum value $f(1)=4$. For $x \geq 1$, $(x-1)^2$ is increasing, so $f$ is increasing on this domain:
$$\boxed{\text{Range of } f: \ f(x) \geq 4}$$
**(c)**
Let $y = (x-1)^2+4$, $x\geq1$, and solve for $x$:
$$y-4 = (x-1)^2$$
$$x-1 = \pm\sqrt{y-4}$$
Since $x \geq 1$, take the positive root:
$$x = 1+\sqrt{y-4}$$
Renaming $y$ as $x$:
$$\boxed{f^{-1}(x) = 1+\sqrt{x-4}}$$
The domain of $f^{-1}$ equals the range of $f$, and the range of $f^{-1}$ equals the domain of $f$:
$$\boxed{\text{Domain: } x \geq 4, \quad \text{Range: } f^{-1}(x) \geq 1}$$
QUESTION 28
6 marks
Medium
Let \(f(x) = \dfrac{3x-1}{x+2}\), \(x \neq -2\).
(a) Find \(f^{-1}(x)\), stating the value excluded from its domain. [3]
(b) Verify that \(f^{-1}(f(4)) = 4\). [2]
(c) State the value excluded from the domain of \(f^{-1}\), and explain how it relates to the horizontal asymptote of the graph of \(f\). [1]
Show complete worked solution
**(a)**
Let $y = \dfrac{3x-1}{x+2}$ and interchange the roles of $x$ and $y$, then solve for $x$ in terms of $y$:
$$y(x+2) = 3x-1$$
$$yx + 2y = 3x-1$$
Collecting the terms containing $x$ on one side:
$$yx - 3x = -1-2y$$
$$x(y-3) = -(1+2y)$$
Solving for $x$:
$$x = \frac{2y+1}{3-y}$$
Renaming $y$ as $x$:
$$\boxed{f^{-1}(x) = \frac{2x+1}{3-x}, \quad x \neq 3}$$
**(b)**
Finding $f(4)$:
$$f(4) = \frac{3(4)-1}{4+2} = \frac{11}{6}$$
Substituting into $f^{-1}$:
$$f^{-1}\left(\frac{11}{6}\right) = \frac{2\left(\frac{11}{6}\right)+1}{3-\frac{11}{6}} = \frac{\frac{11}{3}+1}{\frac{7}{6}} = \frac{\frac{14}{3}}{\frac{7}{6}} = \frac{14}{3} \times \frac{6}{7} = 4$$
$$\boxed{f^{-1}(f(4)) = 4}$$
**(c)**
The value $x=3$ is excluded from the domain of $f^{-1}$. This is consistent with the fact that the graph of $f$ has a horizontal asymptote at $y=3$ (found by considering $x \to \infty$), which means $y=3$ is never actually attained by $f$, so $3$ is not in the range of $f$ and hence not in the domain of $f^{-1}$:
$$\boxed{x \neq 3}$$
QUESTION 29
6 marks
Medium
Let \(f(x) = x+1\) and \(g(x) = x^2-3\).
(a) Find \((f\circ g)(x)\). [2]
(b) Find \((g\circ f)(x)\). [2]
(c) Evaluate both composite functions at \(x=2\), and hence state whether composition of functions is commutative. [2]
Show complete worked solution
**(a)**
Since $(f\circ g)(x) = f(g(x))$, substitute $g(x)=x^2-3$ into $f$:
$$(f\circ g)(x) = (x^2-3)+1$$
$$\boxed{(f\circ g)(x) = x^2-2}$$
**(b)**
Since $(g\circ f)(x) = g(f(x))$, substitute $f(x)=x+1$ into $g$:
$$(g\circ f)(x) = (x+1)^2 - 3$$
Expanding:
$$= x^2+2x+1-3$$
$$\boxed{(g\circ f)(x) = x^2+2x-2}$$
**(c)**
Substituting $x=2$ into each result:
$$(f\circ g)(2) = 2^2-2 = 2$$
$$(g\circ f)(2) = 2^2+2(2)-2 = 4+4-2 = 6$$
Since $(f\circ g)(2) = 2$ and $(g\circ f)(2) = 6$ are different:
$$\boxed{f\circ g \neq g\circ f \text{, so composition is not commutative}}$$
QUESTION 30
8 marks
Medium
A store applies a discount of \$15 to the marked price \(p\) dollars (for \(p \geq 15\)), giving a discounted price \(d(p) = p-15\). An 8\% sales tax is then added to the discounted price, using \(t(x) = 1.08x\).
(a) Find the composite function \(T(p) = (t \circ d)(p)\) that gives the total amount paid in terms of the marked price \(p\). [2]
(b) Find the total amount paid for an item marked \$100. [2]
(c) Find the inverse function \(T^{-1}(x)\), and use it to find the marked price of an item for which the total amount paid was \$140.40. [4]
Show complete worked solution
**(a)**
Since $T(p) = (t\circ d)(p) = t(d(p))$, substitute $d(p) = p-15$ into $t$:
$$T(p) = 1.08(p-15)$$
Expanding:
$$\boxed{T(p) = 1.08p - 16.2}$$
**(b)**
Substituting $p=100$:
$$T(100) = 1.08(100) - 16.2 = 108-16.2$$
$$\boxed{T(100) = \$91.80}$$
**(c)**
Let $y = 1.08p - 16.2$ and solve for $p$:
$$y+16.2 = 1.08p$$
$$p = \frac{y+16.2}{1.08}$$
Renaming $y$ as $x$:
$$\boxed{T^{-1}(x) = \frac{x+16.2}{1.08}}$$
Substituting $x=140.40$:
$$T^{-1}(140.40) = \frac{140.40+16.2}{1.08} = \frac{156.6}{1.08}$$
$$\boxed{T^{-1}(140.40) = \$145 \text{ (marked price)}}$$
Check: $T(145) = 1.08(145)-16.2 = 156.6-16.2 = 140.4$, which agrees.
QUESTION 31
7 marks
Medium
Let \(f(x) = 2^x\) and \(g(x) = x+3\).
(a) Find \((f\circ g)(x)\), simplifying your answer as far as possible. [3]
(b) Find \((g\circ f)(x)\). [2]
(c) Solve the equation \((f\circ g)(x) = 64\). [2]
Show complete worked solution
**(a)**
Since $(f\circ g)(x) = f(g(x))$, substitute $g(x)=x+3$ into $f$:
$$(f\circ g)(x) = 2^{x+3}$$
Using the index law $2^{a+b} = 2^a \times 2^b$:
$$2^{x+3} = 2^x \times 2^3 = 8 \times 2^x$$
$$\boxed{(f\circ g)(x) = 8\times 2^x}$$
**(b)**
Since $(g\circ f)(x) = g(f(x))$, substitute $f(x)=2^x$ into $g$:
$$\boxed{(g\circ f)(x) = 2^x+3}$$
**(c)**
Using the unsimplified form $(f\circ g)(x) = 2^{x+3}$:
$$2^{x+3} = 64$$
Writing $64$ as a power of $2$:
$$2^{x+3} = 2^6$$
Since the bases are equal, the exponents must be equal:
$$x+3 = 6$$
$$\boxed{x=3}$$
QUESTION 32
6 marks
Medium
Let \(f(x) = \sqrt{x+2}\) for \(x \geq -2\), and \(g(x) = x^2-2\) for \(x \geq 0\).
(a) Find \((f\circ g)(x)\) and simplify, showing that it equals \(x\) for \(x \geq 0\). [3]
(b) Find \((g\circ f)(x)\) and simplify, showing that it equals \(x\) for \(x \geq -2\). [3]
Show complete worked solution
**(a)**
Since $(f\circ g)(x) = f(g(x))$, substitute $g(x)=x^2-2$ into $f$:
$$(f\circ g)(x) = \sqrt{(x^2-2)+2} = \sqrt{x^2}$$
Since $x \geq 0$, $\sqrt{x^2} = x$ (not $-x$):
$$\boxed{(f\circ g)(x) = x, \quad x\geq 0}$$
**(b)**
Since $(g\circ f)(x) = g(f(x))$, substitute $f(x)=\sqrt{x+2}$ into $g$:
$$(g\circ f)(x) = \left(\sqrt{x+2}\right)^2 - 2$$
Since $x \geq -2$, $\sqrt{x+2}$ is defined, and squaring undoes the square root:
$$= (x+2) - 2$$
$$\boxed{(g\circ f)(x) = x, \quad x\geq -2}$$
Since both compositions return the input unchanged (on the appropriate domain), this confirms that $g$ is indeed the inverse of $f$, i.e. $g = f^{-1}$.
QUESTION 33
6 marks
Medium
Let \(f(x) = \dfrac{x+3}{2x-1}\), \(x \neq \dfrac{1}{2}\).
(a) Find \(f^{-1}(x)\). [3]
(b) Compare \(f^{-1}(x)\) with \(f(x)\). What do you notice? [2]
(c) State the value that must be excluded from the domain of \(f^{-1}\). [1]
Show complete worked solution
**(a)**
Let $y = \dfrac{x+3}{2x-1}$ and interchange the roles of $x$ and $y$, then solve for $x$:
$$y(2x-1) = x+3$$
$$2xy - y = x+3$$
Collecting the terms containing $x$ on one side:
$$2xy - x = y+3$$
$$x(2y-1) = y+3$$
Solving for $x$:
$$x = \frac{y+3}{2y-1}$$
Renaming $y$ as $x$:
$$\boxed{f^{-1}(x) = \frac{x+3}{2x-1}}$$
**(b)**
Comparing this with the original function $f(x) = \dfrac{x+3}{2x-1}$, the two expressions are identical:
$$\boxed{f^{-1}(x) = f(x) \text{, so } f \text{ is self-inverse}}$$
**(c)**
The denominator $2x-1$ is zero when $x = \dfrac12$, so:
$$\boxed{x \neq \tfrac{1}{2}}$$
(This is the same restriction as on $f$ itself, consistent with $f$ being self-inverse.)
QUESTION 34
7 marks
Medium
The diagram shows part of the graph of \(y=f(x)\), where \(f(x) = (x-1)^2+2\) for \(x \geq 1\), together with the line \(y=x\).
(a) Write down the range of \(f\). [2]
(b) Find \(f^{-1}(x)\), stating its domain and range. [3]
(c) The point \((5, 18)\) lies on the graph of \(y=f(x)\). State the coordinates of the corresponding point on the graph of \(y=f^{-1}(x)\), and state the geometric transformation relating the two graphs. [2]
Show complete worked solution
**(a)**
The graph has vertex $(1,2)$, and $f$ is increasing for $x\geq 1$:
$$\boxed{\text{Range of } f: \ f(x) \geq 2}$$
**(b)**
Let $y=(x-1)^2+2$, $x\geq 1$, and solve for $x$:
$$y-2 = (x-1)^2$$
$$x-1 = \pm\sqrt{y-2}$$
Since $x\geq 1$, take the positive root:
$$x = 1+\sqrt{y-2}$$
Renaming $y$ as $x$:
$$\boxed{f^{-1}(x) = 1+\sqrt{x-2}}$$
The domain of $f^{-1}$ equals the range of $f$, and the range of $f^{-1}$ equals the domain of $f$:
$$\boxed{\text{Domain: } x\geq 2, \quad \text{Range: } f^{-1}(x)\geq 1}$$
**(c)**
Reflecting a point in the line $y=x$ swaps its coordinates, so the point $(5,18)$ on $f$ corresponds to the point $(18,5)$ on $f^{-1}$:
$$\boxed{(18,5) \text{, obtained by reflection in the line } y=x}$$
QUESTION 35
6 marks
Medium
An online currency exchange converts \(x\) US dollars into euros using \(e(x) = 0.92x\), and then subtracts a fixed service fee of \$3 (in euros) using \(s(y) = y-3\).
(a) Find \((s\circ e)(x)\), the total number of euros received for \(x\) dollars exchanged. [2]
(b) Find the number of dollars that must be exchanged to receive exactly 50 euros, correct to 3 significant figures. [2]
(c) Find \((s\circ e)^{-1}(x)\) directly, and verify it gives the same value as part (b) when \(x=50\). [2]
Show complete worked solution
**(a)**
Since $(s\circ e)(x) = s(e(x))$, substitute $e(x)=0.92x$ into $s$:
$$\boxed{(s\circ e)(x) = 0.92x-3}$$
**(b)**
Setting the expression from part (a) equal to 50:
$$0.92x-3 = 50$$
Solving for $x$:
$$0.92x = 53$$
$$x = \frac{53}{0.92} = 57.6087\ldots$$
$$\boxed{x \approx 57.6 \text{ dollars (3 s.f.)}}$$
**(c)**
Let $y=0.92x-3$ and solve for $x$:
$$y+3 = 0.92x$$
$$x = \frac{y+3}{0.92}$$
Renaming $y$ as $x$:
$$\boxed{(s\circ e)^{-1}(x) = \frac{x+3}{0.92}}$$
Substituting $x=50$:
$$(s\circ e)^{-1}(50) = \frac{53}{0.92} \approx 57.6$$
This agrees with part (b).
QUESTION 36
6 marks
Medium
Let \(f(x) = 3x-2\).
(a) Given that \((f\circ g)(x) = 12x^2+6x-5\), find \(g(x)\). [4]
(b) Verify your answer to part (a) by direct substitution. [2]
Show complete worked solution
**(a)**
Writing the composite explicitly using the definition of $f$:
$$(f\circ g)(x) = f(g(x)) = 3g(x) - 2$$
This must equal the given expression:
$$3g(x) - 2 = 12x^2+6x-5$$
Solving for $g(x)$:
$$3g(x) = 12x^2+6x-3$$
$$\boxed{g(x) = 4x^2+2x-1}$$
**(b)**
Substituting $g(x) = 4x^2+2x-1$ into $f$:
$$f(g(x)) = 3(4x^2+2x-1) - 2$$
Expanding:
$$= 12x^2+6x-3-2 = 12x^2+6x-5$$
$$\boxed{f(g(x)) = 12x^2+6x-5}$$
This matches the given expression, confirming the answer to part (a).
QUESTION 37
6 marks
Medium
Let \(f(x) = \sqrt{x}\) for \(x \geq 0\), and \(g(x) = x-5\) for \(x \in \mathbb{R}\).
(a) Find \((f\circ g)(x)\) and state its largest possible domain. [3]
(b) Find \((g\circ f)(x)\) and state its range. [3]
Show complete worked solution
**(a)**
Since $(f\circ g)(x) = f(g(x))$, substitute $g(x)=x-5$ into $f$:
$$(f\circ g)(x) = \sqrt{x-5}$$
For this square root to be defined, we need $x-5\geq 0$:
$$\boxed{(f\circ g)(x) = \sqrt{x-5}, \quad \text{domain } x\geq 5}$$
**(b)**
Since $(g\circ f)(x) = g(f(x))$, substitute $f(x)=\sqrt{x}$ into $g$:
$$\boxed{(g\circ f)(x) = \sqrt{x}-5}$$
The domain of this composite is $x\geq 0$ (inherited from $f$). As $x$ ranges over $[0,\infty)$, $\sqrt{x}$ ranges over $[0,\infty)$, so $\sqrt{x}-5$ ranges over $[-5,\infty)$:
$$\boxed{\text{Range: } (g\circ f)(x) \geq -5}$$
QUESTION 38
7 marks
Medium
Let \(f(x) = 2x^3+1\), \(x \in \mathbb{R}\).
(a) Find \(f^{-1}(x)\). [3]
(b) Find \(f(2)\), and hence verify that \(f^{-1}(f(2)) = 2\). [2]
(c) State the domain and range of \(f^{-1}\). [2]
Show complete worked solution
**(a)**
Let $y=2x^3+1$ and solve for $x$:
$$y-1 = 2x^3$$
$$x^3 = \frac{y-1}{2}$$
Taking the cube root of both sides:
$$x = \sqrt[3]{\frac{y-1}{2}}$$
Renaming $y$ as $x$:
$$\boxed{f^{-1}(x) = \sqrt[3]{\frac{x-1}{2}}}$$
**(b)**
$$f(2) = 2(2)^3+1 = 2(8)+1 = 17$$
Substituting into $f^{-1}$:
$$f^{-1}(17) = \sqrt[3]{\frac{17-1}{2}} = \sqrt[3]{8} = 2$$
$$\boxed{f^{-1}(f(2)) = 2}$$
**(c)**
Since $f(x) = 2x^3+1$ is defined for all real $x$ and, being a strictly increasing cubic, takes every real value exactly once:
$$\boxed{\text{Domain of } f^{-1}: x\in\mathbb{R}, \quad \text{Range of } f^{-1}: f^{-1}(x)\in\mathbb{R}}$$
QUESTION 39
7 marks
Medium
Let \(f(x) = \ln(x-2)\), \(x > 2\).
(a) Find \(f^{-1}(x)\). [3]
(b) State the domain and range of \(f^{-1}\). [2]
(c) Find the value of \(x\) for which \(f(x) = 0\). [2]
Show complete worked solution
**(a)**
Let $y=\ln(x-2)$ and solve for $x$:
$$x-2 = e^y$$
$$x = e^y+2$$
Renaming $y$ as $x$:
$$\boxed{f^{-1}(x) = e^x+2}$$
**(b)**
Since $\ln(x-2)$ can take any real value as $x-2$ ranges over $(0,\infty)$, the range of $f$ (i.e. the domain of $f^{-1}$) is all real numbers, and the range of $f^{-1}$ equals the domain of $f$:
$$\boxed{\text{Domain of } f^{-1}: x\in\mathbb{R}, \quad \text{Range of } f^{-1}: f^{-1}(x) > 2}$$
**(c)**
Setting $f(x)=0$:
$$\ln(x-2) = 0$$
Since $\ln(1)=0$:
$$x-2 = 1$$
$$\boxed{x=3}$$
QUESTION 40
6 marks
Medium
The diagram shows a mapping diagram for functions \(f: A \to B\) and \(g: B \to C\), where \(A=\{1,2,3,4\}\). The arrows show that \(f(1)=a\), \(f(2)=b\), \(f(3)=b\), \(f(4)=c\), and \(g(a)=10\), \(g(b)=20\), \(g(c)=30\).
(a) Find \((g\circ f)(1)\) and \((g\circ f)(3)\). [2]
(b) Explain, with reference to the diagram, why \(f\) does not have an inverse function. [2]
(c) State the complete set of ordered pairs \((x, (g\circ f)(x))\) for \(x \in A\), and hence determine whether \((g\circ f)\) is one-to-one. [2]
Show complete worked solution
**(a)**
$$(g\circ f)(1) = g(f(1)) = g(a) = 10$$
$$(g\circ f)(3) = g(f(3)) = g(b) = 20$$
$$\boxed{(g\circ f)(1)=10, \quad (g\circ f)(3)=20}$$
**(b)**
From the diagram, both $2$ and $3$ in set $A$ map to the same element $b$ in set $B$, i.e. $f(2)=f(3)=b$ with $2\neq 3$. Since two different inputs give the same output, $f$ is not one-to-one:
$$\boxed{\text{$f$ is not one-to-one, so it has no inverse function}}$$
(A function only has an inverse function if it is one-to-one, since reversing the arrows would otherwise send $b$ to two different values, $2$ and $3$, which is not a function.)
**(c)**
Computing $(g\circ f)(x)$ for each $x\in A$:
$$(g\circ f)(1)=10,\quad (g\circ f)(2)=20,\quad (g\circ f)(3)=20,\quad (g\circ f)(4)=30$$
$$\boxed{\{(1,10),\ (2,20),\ (3,20),\ (4,30)\}}$$
Since $(g\circ f)(2) = (g\circ f)(3) = 20$ with $2\neq 3$:
$$\boxed{(g\circ f) \text{ is not one-to-one}}$$
QUESTION 41
8 marks
Medium
Let \(f(x) = x^2+4x+7\) for \(x \geq k\), where \(k\) is the smallest value such that \(f\) is one-to-one. Let \(g(x) = f^{-1}(x)\).
(a) By completing the square, find the value of \(k\) and the range of \(f\). [3]
(b) Find \(g(x)\), stating its domain and range. [3]
(c) Find \(g(12)\). [2]
Show complete worked solution
**(a)**
Completing the square:
$$f(x) = x^2+4x+7 = (x+2)^2 - 4 + 7 = (x+2)^2+3$$
The vertex occurs at $x=-2$. A quadratic is one-to-one only on one branch either side of its vertex, so the smallest value of $k$ for which $f$ is one-to-one (increasing) is:
$$\boxed{k=-2}$$
Since $f$ is increasing for $x\geq -2$, with minimum $f(-2)=3$:
$$\boxed{\text{Range of } f: \ f(x)\geq 3}$$
**(b)**
Let $y=(x+2)^2+3$, $x\geq -2$, and solve for $x$:
$$y-3=(x+2)^2$$
$$x+2 = \pm\sqrt{y-3}$$
Since $x\geq -2$, take the positive root:
$$x = -2+\sqrt{y-3}$$
Renaming $y$ as $x$:
$$\boxed{g(x) = -2+\sqrt{x-3}}$$
The domain of $g$ equals the range of $f$, and the range of $g$ equals the domain of $f$:
$$\boxed{\text{Domain: } x\geq 3, \quad \text{Range: } g(x)\geq -2}$$
**(c)**
$$g(12) = -2+\sqrt{12-3} = -2+\sqrt{9} = -2+3$$
$$\boxed{g(12) = 1}$$
Check: $f(1) = 1+4+7 = 12$, and $1\geq -2$, confirming this is consistent.
QUESTION 42
7 marks
Medium
Let \(f(x) = \dfrac{1}{2}x+3\).
(a) Find \(f^{-1}(x)\). [3]
(b) Any point of intersection between the graphs of \(y=f(x)\) and \(y=f^{-1}(x)\) must lie on the line \(y=x\). Using this fact, find the coordinates of the point where the graphs of \(f\) and \(f^{-1}\) intersect. [4]
Show complete worked solution
**(a)**
Let $y=\dfrac{1}{2}x+3$ and solve for $x$:
$$y-3 = \frac{1}{2}x$$
$$x = 2(y-3) = 2y-6$$
Renaming $y$ as $x$:
$$\boxed{f^{-1}(x) = 2x-6}$$
**(b)**
Since $f$ is an increasing function, any point where the graphs of $f$ and $f^{-1}$ intersect must also lie on $y=x$. So we solve $f(x)=x$:
$$\frac{1}{2}x+3 = x$$
$$3 = x - \frac{1}{2}x = \frac{1}{2}x$$
$$x=6$$
Since the intersection point lies on $y=x$, the $y$-coordinate also equals 6:
$$\boxed{(6,6)}$$
Check: $f(6) = \dfrac{1}{2}(6)+3 = 3+3=6$, and $f^{-1}(6) = 2(6)-6=6$, both confirming the point $(6,6)$.
QUESTION 43
10 marks
Hard
Let \(f(x) = x^2-6x+13\) for \(x \geq 3\), and \(g(x) = 2x-1\) for \(x \in \mathbb{R}\).
(a) By completing the square, find \(f^{-1}(x)\), stating its domain and range. [4]
(b) Find \((f\circ g)(x)\) in expanded form. [3]
(c) Solve \((f\circ g)(x) = 8\), given that the domain of \(f\) requires its input to satisfy \(g(x)\geq 3\). [3]
Show complete worked solution
**(a)**
Completing the square:
$$f(x) = x^2-6x+13 = (x-3)^2-9+13 = (x-3)^2+4$$
Let $y=(x-3)^2+4$, $x\geq3$, and solve for $x$:
$$y-4=(x-3)^2$$
$$x-3 = \pm\sqrt{y-4}$$
Since $x\geq3$, take the positive root:
$$x=3+\sqrt{y-4}$$
Renaming $y$ as $x$:
$$\boxed{f^{-1}(x) = 3+\sqrt{x-4}}$$
Since the range of $f$ (minimum $4$ at $x=3$, increasing thereafter) is $f(x)\geq4$:
$$\boxed{\text{Domain: } x\geq4, \quad \text{Range: } f^{-1}(x)\geq3}$$
**(b)**
Since $(f\circ g)(x) = f(g(x))$, substitute $g(x)=2x-1$ into $f(x) = x^2-6x+13$:
$$(f\circ g)(x) = (2x-1)^2 - 6(2x-1)+13$$
Expanding $(2x-1)^2 = 4x^2-4x+1$ and $-6(2x-1) = -12x+6$:
$$= 4x^2-4x+1-12x+6+13$$
Collecting like terms:
$$\boxed{(f\circ g)(x) = 4x^2-16x+20}$$
**(c)**
Since $f$ is only defined for inputs $\geq 3$, we require $g(x)\geq3$, i.e. $2x-1\geq3$, i.e. $x\geq2$. Solving the equation for $x\geq2$:
$$4x^2-16x+20 = 8$$
$$4x^2-16x+12=0$$
Dividing by 4:
$$x^2-4x+3=0$$
Factorising:
$$(x-1)(x-3)=0$$
$$x=1 \text{ or } x=3$$
Since the domain requires $x\geq2$, the value $x=1$ is rejected:
$$\boxed{x=3}$$
Check: $g(3)=5$, and $f(5)=25-30+13=8$, which agrees.
QUESTION 44
9 marks
Hard
Let \(f(x) = \dfrac{2x+5}{x-3}\), \(x \neq 3\).
(a) Find \(f^{-1}(x)\), and state the value excluded from its domain. [4]
(b) Show that the excluded value found in part (a) is consistent with the horizontal asymptote of the graph of \(y=f(x)\). [3]
(c) Verify that \(f^{-1}(f(1)) = 1\). [2]
Show complete worked solution
**(a)**
Let $y=\dfrac{2x+5}{x-3}$ and interchange the roles of $x$ and $y$, then solve for $x$:
$$y(x-3) = 2x+5$$
$$yx-3y = 2x+5$$
Collecting the terms containing $x$ on one side:
$$yx-2x = 5+3y$$
$$x(y-2) = 5+3y$$
Solving for $x$:
$$x = \frac{3y+5}{y-2}$$
Renaming $y$ as $x$:
$$\boxed{f^{-1}(x) = \frac{3x+5}{x-2}, \quad x\neq 2}$$
**(b)**
Performing polynomial division on $f(x)$:
$$f(x) = \frac{2x+5}{x-3} = \frac{2(x-3)+11}{x-3} = 2+\frac{11}{x-3}$$
As $x\to\infty$, $\dfrac{11}{x-3}\to0$, so $f(x)\to2$; this means $y=2$ is a horizontal asymptote. Moreover, since $\dfrac{11}{x-3}$ can never equal exactly $0$ (as $11\neq0$), $f(x)$ never actually equals $2$:
$$\boxed{2 \text{ is excluded from the range of } f \text{, matching the exclusion } x\neq2 \text{ found for } f^{-1} \text{ in part (a)}}$$
**(c)**
$$f(1) = \frac{2(1)+5}{1-3} = \frac{7}{-2} = -3.5$$
Substituting into $f^{-1}$:
$$f^{-1}(-3.5) = \frac{3(-3.5)+5}{-3.5-2} = \frac{-10.5+5}{-5.5} = \frac{-5.5}{-5.5}$$
$$\boxed{f^{-1}(f(1)) = 1}$$
QUESTION 45
12 marks
Hard
Let \(f(x) = \sqrt{x+1}\) for \(x\geq-1\), and let \(g\) be a function such that \((f\circ g)(x) = \sqrt{3x^2-2}\).
(a) By comparing \((f\circ g)(x) = \sqrt{g(x)+1}\) with the given expression, find \(g(x)\). [3]
(b) Given that \(f\) requires its input to satisfy \(g(x)\geq-1\), find the largest possible domain of \((f\circ g)(x)\). [4]
(c) For the domain \(x\geq0\), find \(g^{-1}(x)\), stating its domain and range. [3]
(d) Verify that \(g^{-1}(g(2)) = 2\). [2]
Show complete worked solution
**(a)**
Since $(f\circ g)(x) = \sqrt{g(x)+1}$, and this equals $\sqrt{3x^2-2}$:
$$\sqrt{g(x)+1} = \sqrt{3x^2-2}$$
Squaring both sides:
$$g(x)+1 = 3x^2-2$$
$$\boxed{g(x) = 3x^2-3}$$
**(b)**
Since $f$ requires $g(x)\geq-1$:
$$3x^2-3\geq-1$$
$$3x^2\geq2$$
$$x^2\geq\frac{2}{3}$$
Taking square roots:
$$x\leq-\sqrt{\frac{2}{3}} \quad \text{or} \quad x\geq\sqrt{\frac{2}{3}}$$
Simplifying $\sqrt{\dfrac23} = \dfrac{\sqrt2}{\sqrt3} = \dfrac{\sqrt6}{3}$:
$$\boxed{x\leq-\frac{\sqrt6}{3} \ \text{ or } \ x\geq\frac{\sqrt6}{3}}$$
**(c)**
Restricting to $x\geq0$, let $y=3x^2-3$ and solve for $x$:
$$y+3 = 3x^2$$
$$x^2 = \frac{y+3}{3}$$
Since $x\geq0$, take the positive root:
$$x = \sqrt{\frac{y+3}{3}}$$
Renaming $y$ as $x$:
$$\boxed{g^{-1}(x) = \sqrt{\frac{x+3}{3}}}$$
For $x\geq0$, $g(x)=3x^2-3$ has minimum value $g(0)=-3$ and is increasing thereafter, so its range is $g(x)\geq-3$. This becomes the domain of $g^{-1}$, and the range of $g^{-1}$ equals the restricted domain of $g$:
$$\boxed{\text{Domain: } x\geq-3, \quad \text{Range: } g^{-1}(x)\geq0}$$
**(d)**
$$g(2) = 3(2)^2-3 = 12-3 = 9$$
Substituting into $g^{-1}$:
$$g^{-1}(9) = \sqrt{\frac{9+3}{3}} = \sqrt{\frac{12}{3}} = \sqrt4$$
$$\boxed{g^{-1}(g(2)) = 2}$$
QUESTION 46
10 marks
Hard
Let \(f(x) = 3^{2x-1}+4\), \(x\in\mathbb{R}\).
(a) Find \(f^{-1}(x)\), stating its domain and range. [4]
(b) Find the exact value of \(f^{-1}(31)\). [3]
(c) Let \(g(x) = 2x+3\). Find \((f\circ g)(x)\) in the form \(3^{ax+b}+c\), stating the values of \(a\), \(b\) and \(c\). [3]
Show complete worked solution
**(a)**
Let $y=3^{2x-1}+4$ and solve for $x$:
$$y-4 = 3^{2x-1}$$
Taking $\log_3$ of both sides:
$$\log_3(y-4) = 2x-1$$
$$x = \frac{\log_3(y-4)+1}{2}$$
Renaming $y$ as $x$:
$$\boxed{f^{-1}(x) = \frac{\log_3(x-4)+1}{2}}$$
Since $3^{2x-1}>0$ for all real $x$, $f(x)>4$ for all $x$, so the range of $f$ is $f(x)>4$; this becomes the domain of $f^{-1}$, while the range of $f^{-1}$ is all real numbers:
$$\boxed{\text{Domain: } x>4, \quad \text{Range: } f^{-1}(x)\in\mathbb{R}}$$
**(b)**
Using the formula from part (a):
$$f^{-1}(31) = \frac{\log_3(31-4)+1}{2} = \frac{\log_3(27)+1}{2}$$
Since $3^3=27$, $\log_3(27)=3$:
$$f^{-1}(31) = \frac{3+1}{2} = \frac{4}{2}$$
$$\boxed{f^{-1}(31) = 2}$$
Check: $f(2) = 3^{2(2)-1}+4 = 3^3+4 = 27+4=31$, which agrees.
**(c)**
Since $(f\circ g)(x) = f(g(x))$, substitute $g(x)=2x+3$ into $f$:
$$(f\circ g)(x) = 3^{2(2x+3)-1}+4$$
Expanding the exponent:
$$2(2x+3)-1 = 4x+6-1 = 4x+5$$
$$\boxed{(f\circ g)(x) = 3^{4x+5}+4, \quad a=4,\ b=5,\ c=4}$$
QUESTION 47
9 marks
Hard
Let \(f(x) = \dfrac{1}{x-2}\), \(x\neq2\), and \(g(x) = \sqrt{x+1}\), \(x\geq-1\).
(a) Find \((f\circ g)(x)\), and state its largest possible domain, considering the restrictions imposed by both \(f\) and \(g\). [4]
(b) Find \((g\circ f)(x)\), and state its largest possible domain. [3]
(c) Evaluate \((g\circ f)(6)\), giving your answer in exact form. [2]
Show complete worked solution
**(a)**
Since $(f\circ g)(x) = f(g(x))$, substitute $g(x)=\sqrt{x+1}$ into $f$:
$$(f\circ g)(x) = \frac{1}{\sqrt{x+1}-2}$$
Two restrictions apply: first, $g$ requires $x\geq-1$; second, $f$ requires its input $\sqrt{x+1}$ to not equal $2$ (otherwise the denominator is zero):
$$\sqrt{x+1}\neq2 \implies x+1\neq4 \implies x\neq3$$
Combining both restrictions:
$$\boxed{(f\circ g)(x) = \frac{1}{\sqrt{x+1}-2}, \quad \text{domain: } x\geq-1, \ x\neq3}$$
**(b)**
Since $(g\circ f)(x) = g(f(x))$, substitute $f(x)=\dfrac{1}{x-2}$ into $g$:
$$(g\circ f)(x) = \sqrt{\frac{1}{x-2}+1} = \sqrt{\frac{1+(x-2)}{x-2}} = \sqrt{\frac{x-1}{x-2}}$$
We need $\dfrac{x-1}{x-2}\geq0$ and $x\neq2$. Using a sign table with critical values $x=1$ and $x=2$:
| Interval | $x<1$ | $x=1$ | $12$ |
|---|---|---|---|---|---|
| $x-1$ | $-$ | $0$ | $+$ | $+$ | $+$ |
| $x-2$ | $-$ | $-$ | $-$ | $0$ | $+$ |
| $\frac{x-1}{x-2}$ | $+$ | $0$ | $-$ | undef. | $+$ |
The expression is non-negative (and defined) for $x\leq1$ or $x>2$:
$$\boxed{(g\circ f)(x) = \sqrt{\frac{x-1}{x-2}}, \quad \text{domain: } x\leq1 \text{ or } x>2}$$
**(c)**
$$(g\circ f)(6) = \sqrt{\frac{6-1}{6-2}} = \sqrt{\frac{5}{4}}$$
$$\boxed{(g\circ f)(6) = \frac{\sqrt5}{2}}$$
QUESTION 48
11 marks
Hard
Let \(f(x) = \dfrac{ax+3}{x-a}\), \(x\neq a\), where \(a\) is a positive constant.
(a) By finding \(f^{-1}(x)\), show that \(f\) is self-inverse for all values of \(a\), i.e. that \(f^{-1}(x) = f(x)\). [4]
(b) Given that \(a=3\), find \(f^{-1}(6)\) using the result from part (a), and verify your answer by showing that \(f(f^{-1}(6)) = 6\). [4]
(c) State the value that must be excluded from the domain of \(f\) when \(a=3\), and explain why the graph of \(f^{-1}\) must have this in common with the graph of \(f\). [3]
Show complete worked solution
**(a)**
Let $y=\dfrac{ax+3}{x-a}$ and interchange the roles of $x$ and $y$, then solve for $x$:
$$y(x-a) = ax+3$$
$$yx-ay = ax+3$$
Collecting the terms containing $x$ on one side:
$$yx-ax = 3+ay$$
$$x(y-a) = 3+ay$$
Solving for $x$:
$$x = \frac{ay+3}{y-a}$$
Renaming $y$ as $x$:
$$f^{-1}(x) = \frac{ax+3}{x-a}$$
Comparing with the original function:
$$\boxed{f^{-1}(x) = \frac{ax+3}{x-a} = f(x) \text{ for every value of } a}$$
**(b)**
Since $f$ is self-inverse, $f^{-1}(x)=f(x)$ for every $a$, so when $a=3$:
$$f^{-1}(6) = f(6) = \frac{3(6)+3}{6-3} = \frac{21}{3}$$
$$\boxed{f^{-1}(6) = 7}$$
Verifying by evaluating $f$ at this result:
$$f(7) = \frac{3(7)+3}{7-3} = \frac{24}{4}$$
$$\boxed{f(f^{-1}(6)) = f(7) = 6}$$
**(c)**
When $a=3$, the denominator $x-3$ is zero at $x=3$, so:
$$\boxed{x\neq3}$$
Since $f^{-1}(x)=f(x)$ (from part (a)), the graphs of $f$ and $f^{-1}$ are identical, so they necessarily share every feature, including this excluded value (a vertical asymptote at $x=3$).
QUESTION 49
9 marks
Hard
Let \(f(x) = 2x-5\) and \(g(x) = \sqrt{x+7}\), \(x\geq-7\).
(a) Find \((g\circ f)(x)\) and state its domain. [3]
(b) Solve \((g\circ f)(x) = 4\). [3]
(c) By finding \(f^{-1}(x)\) and \(g^{-1}(x)\) and using the identity \((g\circ f)^{-1}(x) = (f^{-1}\circ g^{-1})(x)\), find \((g\circ f)^{-1}(x)\), and verify it agrees with part (b) when \(x=4\). [3]
Show complete worked solution
**(a)**
Since $(g\circ f)(x) = g(f(x))$, substitute $f(x)=2x-5$ into $g$:
$$(g\circ f)(x) = \sqrt{(2x-5)+7} = \sqrt{2x+2}$$
For this to be defined, we need $2x+2\geq0$:
$$\boxed{(g\circ f)(x) = \sqrt{2x+2}, \quad \text{domain: } x\geq-1}$$
**(b)**
$$\sqrt{2x+2} = 4$$
Squaring both sides:
$$2x+2 = 16$$
$$2x=14$$
$$\boxed{x=7}$$
(This satisfies the domain $x\geq-1$ found in part (a).)
**(c)**
Finding $f^{-1}(x)$: from $y=2x-5$, $x=\dfrac{y+5}{2}$, so $f^{-1}(x) = \dfrac{x+5}{2}$.
Finding $g^{-1}(x)$: from $y=\sqrt{x+7}$ ($y\geq0$), $y^2=x+7$, so $x=y^2-7$; hence $g^{-1}(x) = x^2-7$, $x\geq0$.
Using the identity for the inverse of a composite function:
$$(g\circ f)^{-1}(x) = (f^{-1}\circ g^{-1})(x) = f^{-1}(g^{-1}(x)) = f^{-1}(x^2-7)$$
$$= \frac{(x^2-7)+5}{2}$$
$$\boxed{(g\circ f)^{-1}(x) = \frac{x^2-2}{2}}$$
Substituting $x=4$:
$$(g\circ f)^{-1}(4) = \frac{4^2-2}{2} = \frac{16-2}{2} = \frac{14}{2}$$
$$\boxed{(g\circ f)^{-1}(4) = 7}$$
This matches the answer $x=7$ found in part (b).
QUESTION 50
10 marks
Hard
Let \(f(x) = x^2+1\), \(x\in\mathbb{R}\), and \(g(x) = \dfrac{1}{x}\), \(x\neq0\).
(a) Find \((g\circ f)(x)\), and explain why it is defined for every real value of \(x\). [3]
(b) By considering the range of \(f\), find the range of \((g\circ f)(x)\). [4]
(c) Hence, or otherwise, determine whether \((g\circ f)\) has an inverse function on the domain \(x\in\mathbb{R}\), justifying your answer. [3]
Show complete worked solution
**(a)**
Since $(g\circ f)(x) = g(f(x))$, substitute $f(x)=x^2+1$ into $g$:
$$\boxed{(g\circ f)(x) = \frac{1}{x^2+1}}$$
This is defined for every real $x$ because $x^2\geq0$ for all real $x$, so $x^2+1\geq1$, which means the denominator $x^2+1$ is never zero (it is always at least $1$):
$$\boxed{\text{Defined for all } x\in\mathbb{R} \text{, since } x^2+1 \geq 1 > 0}$$
**(b)**
Since $f(x)=x^2+1$ has minimum value $1$ (at $x=0$) and increases without bound, the range of $f$ is $f(x)\geq1$, i.e. the input to $g$ ranges over $[1,\infty)$.
The function $g(u)=\dfrac{1}{u}$ is decreasing for $u>0$: as $u$ increases from $1$ to $\infty$, $\dfrac1u$ decreases from $1$ towards (but never reaching) $0$. So as $f(x)$ ranges over $[1,\infty)$, $(g\circ f)(x) = \dfrac{1}{f(x)}$ ranges over $(0,1]$:
$$\boxed{\text{Range: } 0 < (g\circ f)(x) \leq 1}$$
**(c)**
Testing whether $(g\circ f)$ is one-to-one: since $f(x)=x^2+1$ is an even function, $f(-x)=f(x)$ for all $x$, and therefore:
$$(g\circ f)(-x) = \frac{1}{f(-x)} = \frac{1}{f(x)} = (g\circ f)(x)$$
For example, $(g\circ f)(1) = \dfrac{1}{2}$ and $(g\circ f)(-1) = \dfrac{1}{2}$, with $1\neq-1$.
$$\boxed{(g\circ f) \text{ is not one-to-one on } x\in\mathbb{R} \text{, so it has no inverse function on this domain}}$$
Linear and Quadratic Functions 50 questions
QUESTION 1
3 marks
Easy
Line \(L\) passes through the points \(A(-2, 5)\) and \(B(4, -7)\).
Find the equation of \(L\), giving your answer in the form \(y = mx + c\).
Show complete worked solution
Using the gradient formula between two points:
$$m = \frac{y_B - y_A}{x_B - x_A}$$
Substituting $A(-2,5)$ and $B(4,-7)$:
$$m = \frac{-7-5}{4-(-2)} = \frac{-12}{6} = -2$$
Using the point-gradient form of a line through $A(-2,5)$:
$$y - 5 = -2(x-(-2)) = -2(x+2)$$
Expanding and simplifying:
$$y - 5 = -2x - 4$$
$$y = -2x + 1$$
Checking with $B(4,-7)$: $y = -2(4)+1 = -7$, which agrees.
$$\boxed{y = -2x+1}$$
QUESTION 2
4 marks
Easy
A line has gradient \(\dfrac{3}{4}\) and passes through the point \((8, -1)\).
Find the equation of the line, giving your answer in the form \(ax + by + d = 0\), where \(a, b, d \in \mathbb{Z}\).
Show complete worked solution
Using the point-gradient form of a line, $y - y_1 = m(x-x_1)$, with $m=\dfrac{3}{4}$ and $(x_1,y_1)=(8,-1)$:
$$y - (-1) = \frac{3}{4}(x - 8)$$
Expanding:
$$y + 1 = \frac{3}{4}x - 6$$
$$y = \frac{3}{4}x - 7$$
Multiplying through by 4 to clear the fraction:
$$4y = 3x - 28$$
Rearranging into the required form $ax+by+d=0$:
$$\boxed{3x - 4y - 28 = 0}$$
Checking with $(8,-1)$: $3(8) - 4(-1) - 28 = 24 + 4 - 28 = 0$, which agrees.
QUESTION 3
5 marks
Medium
Line \(L_1\) has equation \(2x - 5y + 10 = 0\).
Line \(L_2\) is perpendicular to \(L_1\) and passes through the point \((3, -4)\).
Find the equation of \(L_2\) in the form \(ax + by + c = 0\), where \(a, b, c \in \mathbb{Z}\).
Show complete worked solution
Using the method of rearranging $L_1$ into the form $y=mx+c$ to find its gradient:
$$2x - 5y + 10 = 0 \implies 5y = 2x + 10 \implies y = \frac{2}{5}x + 2$$
So the gradient of $L_1$ is $m_1 = \dfrac{2}{5}$.
Using the perpendicular gradient condition $m_1 m_2 = -1$:
$$m_2 = -\frac{1}{m_1} = -\frac{5}{2}$$
Using the point-gradient form through $(3,-4)$:
$$y - (-4) = -\frac{5}{2}(x - 3)$$
Expanding and simplifying:
$$y + 4 = -\frac{5}{2}x + \frac{15}{2}$$
$$y = -\frac{5}{2}x + \frac{7}{2}$$
Multiplying through by 2 to clear the fractions:
$$2y = -5x + 7$$
Rearranging into the required form:
$$\boxed{5x + 2y - 7 = 0}$$
Checking with $(3,-4)$: $5(3)+2(-4)-7 = 15-8-7=0$, which agrees.
QUESTION 4
4 marks
Medium
A tour company offers two pricing plans for hiring a bicycle for \(n\) days.
Plan A: cost in dollars is \(C = 45 + 2.50n\)
Plan B: cost in dollars is \(C = 30 + 3.25n\)
Find the number of days \(n\) for which the two plans cost the same amount, and find this cost.
Show complete worked solution
Since both plans give the cost $C$ as a linear function of $n$, the two costs are equal when the two expressions for $C$ are set equal:
$$45 + 2.50n = 30 + 3.25n$$
Collecting like terms:
$$45 - 30 = 3.25n - 2.50n$$
$$15 = 0.75n$$
Solving for $n$:
$$n = \frac{15}{0.75} = 20$$
Substituting $n=20$ back into Plan A to find the common cost:
$$C = 45 + 2.50(20) = 45 + 50 = 95$$
Checking with Plan B: $C = 30 + 3.25(20) = 30+65=95$, which agrees.
$$\boxed{n = 20 \text{ days, cost } \$95}$$
QUESTION 5
7 marks
Hard
The points \(A(1, 2)\), \(B(5, k)\) and \(C(11, 14)\) are collinear.
(a) Find the value of \(k\).
(b) Find the equation of the perpendicular bisector of \([AC]\), giving your answer in the form \(ax+by+c=0\) with \(a,b,c \in \mathbb{Z}\).
Show complete worked solution
**(a)**
Since $A$, $B$, $C$ are collinear, the gradient of $[AB]$ equals the gradient of $[AC]$. Using the gradient formula:
$$\text{gradient of } AC = \frac{14-2}{11-1} = \frac{12}{10} = \frac{6}{5}$$
$$\text{gradient of } AB = \frac{k-2}{5-1} = \frac{k-2}{4}$$
Setting the two gradients equal:
$$\frac{k-2}{4} = \frac{6}{5}$$
Solving for $k$:
$$k - 2 = \frac{24}{5}$$
$$k = 2 + \frac{24}{5} = \frac{34}{5}$$
$$\boxed{k = \frac{34}{5} = 6.8}$$
**(b)**
Using the midpoint formula for $[AC]$:
$$M = \left(\frac{1+11}{2}, \frac{2+14}{2}\right) = (6, 8)$$
The perpendicular bisector has gradient equal to the negative reciprocal of the gradient of $AC$, which is $\dfrac{6}{5}$:
$$m_{\perp} = -\frac{5}{6}$$
Using the point-gradient form through $M(6,8)$:
$$y - 8 = -\frac{5}{6}(x-6)$$
Expanding and simplifying:
$$y = -\frac{5}{6}x + 5 + 8 = -\frac{5}{6}x + 13$$
Multiplying through by 6 and rearranging into the required form:
$$6y = -5x + 78$$
$$\boxed{5x + 6y - 78 = 0}$$
QUESTION 6
3 marks
Easy
Let \(f(x) = x^2 - 5x + 6\).
(a) Write \(f(x)\) in factorised form.
(b) Write down the \(x\)-intercepts of the graph of \(f\).
Show complete worked solution
**(a)**
Finding two numbers that multiply to give 6 and add to give $-5$: these are $-2$ and $-3$.
$$\boxed{f(x) = (x-2)(x-3)}$$
**(b)**
The $x$-intercepts occur where each factor is zero:
$$\boxed{x=2 \text{ and } x=3}$$
QUESTION 7
3 marks
Easy
A quadratic function is given in vertex form as \(f(x) = 2(x-1)^2 - 8\).
Write \(f(x)\) in the general form \(ax^2+bx+c\).
Show complete worked solution
Expanding $(x-1)^2$:
$$(x-1)^2 = x^2-2x+1$$
Multiplying by 2 and subtracting 8:
$$f(x) = 2(x^2-2x+1) - 8 = 2x^2-4x+2-8$$
$$\boxed{f(x) = 2x^2-4x-6}$$
QUESTION 8
5 marks
Medium
A parabola has \(x\)-intercepts at \(x=-3\) and \(x=5\), and passes through the point \((0,-30)\).
(a) Find the equation of the parabola in the form \(y = a(x-p)(x-q)\).
(b) Write the equation in the general form \(y=ax^2+bx+c\).
Show complete worked solution
**(a)**
Since the $x$-intercepts are $-3$ and $5$, write the equation in factorised form:
$$y = a(x-(-3))(x-5) = a(x+3)(x-5)$$
Using the point $(0,-30)$ to find $a$:
$$-30 = a(0+3)(0-5) = a(3)(-5) = -15a$$
Solving for $a$:
$$a = 2$$
$$\boxed{y = 2(x+3)(x-5)}$$
**(b)**
Expanding the factorised form:
$$(x+3)(x-5) = x^2-2x-15$$
$$y = 2(x^2-2x-15)$$
$$\boxed{y = 2x^2-4x-30}$$
QUESTION 9
5 marks
Medium
The graph of \(f(x) = 3x^2 - 12x + k\) passes through the point \((1,-2)\).
(a) Find the value of \(k\).
(b) Find the coordinates of the vertex of the graph of \(f\), and state the minimum value of \(f\).
Show complete worked solution
**(a)**
Substituting $x=1$, using $f(1)=-2$:
$$3(1)^2 - 12(1) + k = -2$$
Simplifying:
$$3-12+k=-2$$
$$-9+k=-2$$
Solving for $k$:
$$k=7$$
$$\boxed{k=7}$$
So $f(x) = 3x^2-12x+7$.
**(b)**
Using $x = -\dfrac{b}{2a}$ to find the $x$-coordinate of the vertex:
$$x = -\frac{-12}{2(3)} = \frac{12}{6} = 2$$
Substituting back to find $f(2)$:
$$f(2) = 3(4) - 12(2) + 7 = 12-24+7=-5$$
$$\boxed{\text{Vertex: } (2,-5); \text{ minimum value } = -5}$$
QUESTION 10
6 marks
Hard
A quadratic function \(f\) has vertex \((2,-9)\) and its graph passes through the point \((5,0)\).
(a) Write \(f(x)\) in vertex form.
(b) Show that \(f(x)\) can be written in factorised form as \(f(x) = (x+1)(x-5)\).
(c) Write down the other \(x\)-intercept and the \(y\)-intercept of the graph of \(f\).
Show complete worked solution
**(a)**
Writing the vertex form using vertex $(2,-9)$:
$$f(x) = a(x-2)^2 - 9$$
Using the point $(5,0)$ to find $a$:
$$0 = a(5-2)^2 - 9 = 9a - 9$$
Solving for $a$:
$$a = 1$$
$$\boxed{f(x) = (x-2)^2 - 9}$$
**(b)**
Expanding the vertex form:
$$f(x) = x^2-4x+4-9 = x^2-4x-5$$
Factorising: two numbers multiplying to $-5$ and adding to $-4$ are $-5$ and $1$:
$$f(x) = (x-5)(x+1) = (x+1)(x-5)$$
This confirms the required factorised form. $\blacksquare$
**(c)**
From the factorised form, the $x$-intercepts are $x=5$ and $x=-1$; the intercept other than the given point $(5,0)$ is:
$$\boxed{x=-1}$$
Using the fact that the $y$-intercept is $f(0)$:
$$f(0) = (0+1)(0-5) = -5$$
$$\boxed{y\text{-intercept: } (0,-5)}$$
QUESTION 11
3 marks
Easy
Solve the quadratic equation $2x^2 - 7x + 3 = 0$.
Show complete worked solution
Using the method of factorising the quadratic $2x^2-7x+3=0$, we seek factors of the form $(2x-a)(x-b)=0$ such that $ab=3$ and $a+2b=7$.
Trying $a=1,\ b=3$:
$$(2x-1)(x-3) = 2x^2-6x-x+3 = 2x^2-7x+3 \checkmark$$
So the equation factorises as:
$$(2x-1)(x-3)=0$$
Setting each factor to zero:
$$2x-1=0 \Rightarrow x=\frac{1}{2}$$
$$x-3=0 \Rightarrow x=3$$
$$\boxed{x=\frac{1}{2} \text{ or } x=3}$$
QUESTION 12
3 marks
Easy
Find the value of the discriminant of the equation $3x^2 + 2x + 5 = 0$, and hence state the number of real roots the equation has.
Show complete worked solution
Using the discriminant formula $\Delta = b^2-4ac$ with $a=3,\ b=2,\ c=5$:
$$\Delta = (2)^2 - 4(3)(5)$$
Simplifying:
$$\Delta = 4 - 60 = -56$$
Since $\Delta < 0$, the equation has no real roots.
$$\boxed{\Delta = -56,\ \text{no real roots}}$$
QUESTION 13
5 marks
Medium
Find the set of values of $k$, where $k \neq 0$, for which the equation $kx^2 - 4x + k = 0$ has two distinct real roots.
Show complete worked solution
For a quadratic $ax^2+bx+c=0$ to have two distinct real roots, the discriminant condition $\Delta = b^2-4ac>0$ must be satisfied (with $a \neq 0$).
Here $a=k,\ b=-4,\ c=k$:
$$\Delta = (-4)^2 - 4(k)(k)$$
Simplifying:
$$\Delta = 16-4k^2$$
Requiring $\Delta > 0$:
$$16-4k^2>0$$
$$4k^2<16$$
$$k^2<4$$
Solving this inequality:
$$-2
QUESTION 14
6 marks
Medium
A ball is thrown vertically upward. Its height above the ground, in metres, $t$ seconds after being thrown, is modelled by $h(t) = -5t^2+20t+2$ for $0 \le t \le 4.1$. Find, using algebra, the interval of time during which the ball is more than $17$ metres above the ground.
Show complete worked solution
Solving the inequality $h(t) > 17$:
$$-5t^2+20t+2>17$$
Rearranging so one side is zero:
$$-5t^2+20t-15>0$$
Dividing both sides by $-5$ (reversing the inequality sign since we divide by a negative number):
$$t^2-4t+3<0$$
Factorising:
$$(t-1)(t-3)<0$$
The critical values are $t=1$ and $t=3$. Since the coefficient of $t^2$ in $t^2-4t+3$ is positive, the expression is negative between the roots:
$$\boxed{1 < t < 3 \text{ (seconds)}}$$
QUESTION 15
7 marks
Hard
The quadratic equation $(p-2)x^2 - 2px + (p-5) = 0$, where $p \neq 2$, has two distinct real roots. Find the range of possible values of $p$.
Show complete worked solution
Applying the discriminant condition $\Delta > 0$ for two distinct real roots, with $a=p-2,\ b=-2p,\ c=p-5$:
$$\Delta = (-2p)^2 - 4(p-2)(p-5)$$
Expanding:
$$\Delta = 4p^2 - 4(p^2-7p+10)$$
Simplifying:
$$\Delta = 4p^2 - 4p^2+28p-40 = 28p-40$$
Requiring $\Delta>0$:
$$28p-40>0$$
Solving for $p$:
$$p>\frac{40}{28}=\frac{10}{7}$$
Combining with the given condition $p \neq 2$ (needed for $a \neq 0$):
$$\boxed{p>\frac{10}{7},\ p \neq 2}$$
QUESTION 16
3 marks
Easy
Find the gradient of the line joining the points $P(3, -1)$ and $Q(-2, 9)$.
Show complete worked solution
Using the gradient formula between two points:
$$m = \frac{y_Q - y_P}{x_Q - x_P}$$
Substituting $P(3,-1)$ and $Q(-2,9)$:
$$m = \frac{9-(-1)}{-2-3} = \frac{10}{-5} = -2$$
$$\boxed{m = -2}$$
QUESTION 17
5 marks
Easy
A line has gradient $-\dfrac{3}{5}$ and passes through the point $(10, 4)$.
Find the equation of the line, giving your answer in the form $ax+by+d=0$, where $a,b,d \in \mathbb{Z}$.
Show complete worked solution
Using the point-gradient form of a line, $y-y_1 = m(x-x_1)$, with $m=-\dfrac{3}{5}$ and $(x_1,y_1)=(10,4)$:
$$y - 4 = -\frac{3}{5}(x-10)$$
Expanding:
$$y - 4 = -\frac{3}{5}x + 6$$
$$y = -\frac{3}{5}x + 10$$
Multiplying through by 5 to clear the fraction:
$$5y = -3x + 50$$
Rearranging into the required form:
$$\boxed{3x + 5y - 50 = 0}$$
Checking with $(10,4)$: $3(10)+5(4)-50 = 30+20-50=0$, which agrees.
QUESTION 18
5 marks
Easy
Find the equation of the line that is parallel to the line $4x - 2y + 7 = 0$ and passes through the point $(3, -1)$. Give your answer in the form $y = mx+c$.
Show complete worked solution
Rearranging the given line into the form $y=mx+c$ to find its gradient:
$$4x - 2y + 7 = 0 \implies 2y = 4x+7 \implies y = 2x + 3.5$$
So the gradient of the given line is $2$. Since parallel lines have equal gradients, the required line also has gradient $2$.
Using the point-gradient form through $(3,-1)$:
$$y - (-1) = 2(x-3)$$
Expanding and simplifying:
$$y + 1 = 2x - 6$$
$$\boxed{y = 2x - 7}$$
Checking with $(3,-1)$: $y=2(3)-7=-1$, which agrees.
QUESTION 19
2 marks
Easy
A line has gradient $\dfrac{2}{7}$. Find the gradient of any line that is perpendicular to it.
Show complete worked solution
Using the perpendicular gradient condition $m_1 m_2 = -1$, with $m_1 = \dfrac{2}{7}$:
$$m_2 = -\frac{1}{m_1} = -\frac{7}{2}$$
$$\boxed{m_2 = -\frac{7}{2}}$$
QUESTION 20
4 marks
Easy
Solve the quadratic equation $x^2 + 2x - 15 = 0$.
Show complete worked solution
Using the method of factorising: we seek two numbers that multiply to give $-15$ and add to give $2$. These are $5$ and $-3$.
$$x^2+2x-15 = (x+5)(x-3)$$
So the equation factorises as:
$$(x+5)(x-3) = 0$$
Setting each factor to zero:
$$x+5=0 \Rightarrow x=-5$$
$$x-3=0 \Rightarrow x=3$$
$$\boxed{x=-5 \text{ or } x=3}$$
QUESTION 21
5 marks
Easy
Let $f(x) = x^2 - x - 12$.
(a) Find the $x$-intercepts of the graph of $f$. [3]
(b) Find the $y$-intercept of the graph of $f$. [2]
Show complete worked solution
**(a)**
Finding two numbers that multiply to give $-12$ and add to give $-1$: these are $-4$ and $3$.
$$f(x) = x^2-x-12 = (x-4)(x+3)$$
The $x$-intercepts occur where $f(x)=0$, i.e. where each factor is zero:
$$\boxed{x=4 \text{ and } x=-3}$$
**(b)**
Using the fact that the $y$-intercept is $f(0)$:
$$f(0) = (0)^2-(0)-12 = -12$$
$$\boxed{y\text{-intercept: } (0,-12)}$$
QUESTION 22
4 marks
Easy
A quadratic function is given in vertex form as $f(x) = -2(x+3)^2 + 5$.
Write $f(x)$ in the general form $ax^2+bx+c$.
Show complete worked solution
Expanding $(x+3)^2$:
$$(x+3)^2 = x^2+6x+9$$
Multiplying by $-2$ and adding $5$:
$$f(x) = -2(x^2+6x+9) + 5 = -2x^2-12x-18+5$$
$$\boxed{f(x) = -2x^2-12x-13}$$
QUESTION 23
5 marks
Easy
Let $f(x) = x^2 - 6x + 10$.
By completing the square, find the coordinates of the vertex of the graph of $f$.
Show complete worked solution
Using the method of completing the square on $x^2-6x+10$: halving the coefficient of $x$ gives $-3$, and $(-3)^2=9$.
$$f(x) = x^2-6x+10 = (x^2-6x+9) - 9 + 10$$
$$f(x) = (x-3)^2 + 1$$
Since the vertex form is $f(x)=(x-h)^2+k$ with vertex $(h,k)$:
$$\boxed{\text{Vertex} = (3, 1)}$$
QUESTION 24
3 marks
Easy
Find the equation of the axis of symmetry of the graph of $f(x) = 2x^2 - 8x + 1$.
Show complete worked solution
Using the formula for the axis of symmetry of a quadratic $ax^2+bx+c$, $x = -\dfrac{b}{2a}$, with $a=2,\ b=-8$:
$$x = -\frac{-8}{2(2)} = \frac{8}{4} = 2$$
$$\boxed{x = 2}$$
QUESTION 25
4 marks
Easy
Without solving the equation, find the sum and the product of the roots of $4x^2 - 12x + 5 = 0$.
Show complete worked solution
Using the formulas for the sum and product of the roots of a quadratic $ax^2+bx+c=0$, namely $\alpha+\beta = -\dfrac{b}{a}$ and $\alpha\beta = \dfrac{c}{a}$, with $a=4,\ b=-12,\ c=5$:
$$\alpha + \beta = -\frac{-12}{4} = 3$$
$$\alpha \beta = \frac{5}{4}$$
$$\boxed{\text{Sum} = 3, \ \text{Product} = \frac{5}{4}}$$
QUESTION 26
5 marks
Easy
Solve the quadratic equation $x^2 + 4x + 1 = 0$, giving your answers in exact form.
Show complete worked solution
Using the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ with $a=1,\ b=4,\ c=1$:
$$x = \frac{-4 \pm \sqrt{4^2-4(1)(1)}}{2(1)}$$
Simplifying inside the square root:
$$x = \frac{-4 \pm \sqrt{16-4}}{2} = \frac{-4 \pm \sqrt{12}}{2}$$
Since $\sqrt{12} = 2\sqrt{3}$:
$$x = \frac{-4 \pm 2\sqrt{3}}{2}$$
$$\boxed{x = -2 \pm \sqrt{3}}$$
QUESTION 27
3 marks
Easy
Determine whether the point $(4, -1)$ lies on the line with equation $3x + 5y - 7 = 0$.
Show complete worked solution
Substituting $x=4,\ y=-1$ into the left-hand side of the equation:
$$3(4) + 5(-1) - 7 = 12 - 5 - 7$$
$$= 0$$
Since this equals $0$, the point satisfies the equation of the line.
$$\boxed{\text{Yes, } (4,-1) \text{ lies on the line.}}$$
QUESTION 28
3 marks
Easy
A quadratic function is given by $f(x) = 3(x-2)^2 - 7$.
Find the $y$-intercept of the graph of $f$.
Show complete worked solution
Using the fact that the $y$-intercept is $f(0)$:
$$f(0) = 3(0-2)^2 - 7 = 3(4) - 7$$
$$= 12 - 7 = 5$$
$$\boxed{y\text{-intercept: } (0, 5)}$$
QUESTION 29
6 marks
Easy
A linear function is given by $f(x) = -4x + 9$.
(a) Find the value of $f(-2)$. [2]
(b) Find the value of $x$ for which $f(x) = 0$. [4]
Show complete worked solution
**(a)**
Substituting $x=-2$:
$$f(-2) = -4(-2) + 9 = 8+9$$
$$\boxed{f(-2) = 17}$$
**(b)**
Setting $f(x)=0$:
$$-4x+9 = 0$$
Solving for $x$:
$$-4x = -9$$
$$x = \frac{9}{4}$$
$$\boxed{x = \frac{9}{4} = 2.25}$$
QUESTION 30
5 marks
Medium
Solve the simultaneous equations $y = 2x - 3$ and $3x + y = 12$ to find the point of intersection of the two lines.
Show complete worked solution
Since $y=2x-3$, substituting this expression for $y$ into the second equation $3x+y=12$:
$$3x + (2x-3) = 12$$
Simplifying:
$$5x - 3 = 12$$
$$5x = 15$$
Solving for $x$:
$$x = 3$$
Substituting $x=3$ back into $y=2x-3$:
$$y = 2(3) - 3 = 3$$
Checking with $3x+y=12$: $3(3)+3=12$, which agrees.
$$\boxed{(3, 3)}$$
QUESTION 31
7 marks
Medium
A company finds that if it sells $x$ units of a product, the price it can charge, in dollars per unit, is $p = 300 - 5x$. The revenue, in dollars, is given by $R(x) = xp$.
(a) Show that $R(x) = 300x - 5x^2$, and find the number of units $x$ that maximises the revenue. [4]
(b) Find the maximum revenue. [3]
Show complete worked solution
**(a)**
Since $R(x) = xp$ and $p = 300-5x$:
$$R(x) = x(300-5x) = 300x - 5x^2$$
This confirms the required expression. $\blacksquare$
Writing $R(x) = -5x^2+300x$, this is a quadratic with $a=-5,\ b=300$, which is maximised at its vertex. Using $x = -\dfrac{b}{2a}$:
$$x = -\frac{300}{2(-5)} = \frac{300}{10} = 30$$
$$\boxed{x = 30 \text{ units}}$$
**(b)**
Substituting $x=30$ into $R(x)$:
$$R(30) = 300(30) - 5(30)^2 = 9000 - 4500$$
$$\boxed{\text{Maximum revenue} = \$4500}$$
QUESTION 32
7 marks
Medium
A ball is thrown vertically upward from a platform. Its height above the ground, in metres, $t$ seconds after being thrown, is modelled by $h(t) = -5t^2 + 30t + 5$.
(a) Find the time at which the ball reaches its maximum height. [3]
(b) Find the maximum height of the ball above the ground. [4]
Show complete worked solution
**(a)**
Since $h(t)$ is a quadratic in $t$ with $a=-5,\ b=30$, opening downward, its maximum occurs at the vertex. Using $t=-\dfrac{b}{2a}$:
$$t = -\frac{30}{2(-5)} = \frac{30}{10} = 3$$
$$\boxed{t = 3 \text{ seconds}}$$
**(b)**
Substituting $t=3$ into $h(t)$:
$$h(3) = -5(3)^2 + 30(3) + 5$$
$$= -45 + 90 + 5$$
$$\boxed{\text{Maximum height} = 50 \text{ metres}}$$
QUESTION 33
5 marks
Medium
Find the value of $k$ for which the line $kx - 3y + 6 = 0$ is perpendicular to the line $2x + y - 5 = 0$.
Show complete worked solution
Rearranging the second line into the form $y=mx+c$:
$$2x+y-5=0 \implies y = -2x+5$$
So its gradient is $m_1=-2$.
Rearranging the first line:
$$kx - 3y+6=0 \implies 3y = kx+6 \implies y = \frac{k}{3}x+2$$
So its gradient is $m_2 = \dfrac{k}{3}$.
Using the perpendicular gradient condition $m_1 m_2 = -1$:
$$(-2)\left(\frac{k}{3}\right) = -1$$
Solving for $k$:
$$-\frac{2k}{3} = -1$$
$$2k = 3$$
$$\boxed{k = \frac{3}{2}}$$
QUESTION 34
5 marks
Medium
Find the values of $k$ for which the equation $x^2 + kx + 9 = 0$ has two equal real roots.
Show complete worked solution
For a quadratic $ax^2+bx+c=0$ to have two equal real roots, the discriminant condition $\Delta = b^2-4ac = 0$ must be satisfied.
Here $a=1,\ b=k,\ c=9$:
$$\Delta = k^2 - 4(1)(9) = k^2 - 36$$
Setting $\Delta=0$:
$$k^2 - 36 = 0$$
$$k^2 = 36$$
$$\boxed{k = 6 \text{ or } k = -6}$$
QUESTION 35
6 marks
Medium
The equation $x^2 - kx + (k+3) = 0$ has $x=5$ as one of its roots.
(a) Find the value of $k$. [3]
(b) Find the other root of the equation. [3]
Show complete worked solution
**(a)**
Since $x=5$ satisfies the equation, substituting $x=5$:
$$(5)^2 - k(5) + (k+3) = 0$$
Simplifying:
$$25 - 5k + k + 3 = 0$$
$$28 - 4k = 0$$
Solving for $k$:
$$\boxed{k = 7}$$
**(b)**
Substituting $k=7$, the equation becomes:
$$x^2 - 7x + 10 = 0$$
Factorising: two numbers that multiply to give $10$ and add to give $-7$ are $-5$ and $-2$:
$$(x-5)(x-2) = 0$$
So the roots are $x=5$ (given) and $x=2$.
$$\boxed{\text{Other root: } x = 2}$$
QUESTION 36
7 marks
Medium
A rectangular garden plot has a perimeter of $48$ metres. If the length of the plot is $x$ metres, the width is $(24-x)$ metres.
Find, using algebra, the range of values of $x$ for which the area of the plot is at least $128$ m$^2$.
Show complete worked solution
The area of the plot is $A(x) = x(24-x)$. Solving the inequality $A(x) \ge 128$:
$$x(24-x) \ge 128$$
Expanding:
$$24x - x^2 \ge 128$$
Rearranging so one side is zero:
$$0 \ge x^2 - 24x + 128$$
i.e.
$$x^2 - 24x + 128 \le 0$$
Using the quadratic formula to find the critical values, with $a=1,\ b=-24,\ c=128$:
$$x = \frac{24 \pm \sqrt{(-24)^2-4(1)(128)}}{2} = \frac{24 \pm \sqrt{576-512}}{2} = \frac{24\pm\sqrt{64}}{2} = \frac{24\pm 8}{2}$$
So the critical values are $x=8$ and $x=16$. Since the coefficient of $x^2$ is positive, the expression $x^2-24x+128$ is negative or zero between the roots:
$$\boxed{8 \le x \le 16 \text{ (metres)}}$$
QUESTION 37
6 marks
Medium
By completing the square, solve the equation $2x^2 - 8x - 3 = 0$, giving your answers in exact form.
Show complete worked solution
Using the method of completing the square, first factor out the coefficient of $x^2$ from the first two terms:
$$2x^2-8x-3 = 2(x^2-4x) - 3$$
Completing the square inside the brackets (halving the coefficient of $x$ gives $-2$, and $(-2)^2=4$):
$$2(x^2-4x) = 2\big[(x-2)^2-4\big] = 2(x-2)^2 - 8$$
So the equation becomes:
$$2(x-2)^2 - 8 - 3 = 0$$
$$2(x-2)^2 = 11$$
Dividing by 2:
$$(x-2)^2 = \frac{11}{2}$$
Taking square roots of both sides:
$$x - 2 = \pm\sqrt{\frac{11}{2}}$$
Since $\sqrt{\dfrac{11}{2}} = \dfrac{\sqrt{22}}{2}$:
$$\boxed{x = 2 \pm \frac{\sqrt{22}}{2}}$$
QUESTION 38
7 marks
Medium
The graph of a quadratic function $f$ has axis of symmetry $x=3$, one $x$-intercept at $x=-1$, and $y$-intercept $(0,-14)$.
(a) Find the other $x$-intercept of the graph of $f$. [2]
(b) Find $f(x)$ in the general form $ax^2+bx+c$. [5]
Show complete worked solution
**(a)**
Since the axis of symmetry lies exactly halfway between the two $x$-intercepts, and one intercept is $x=-1$ with axis of symmetry $x=3$:
$$\text{other intercept} = 2(3) - (-1) = 6+1$$
$$\boxed{x = 7}$$
**(b)**
Using the factorised form with $x$-intercepts $-1$ and $7$:
$$f(x) = a(x+1)(x-7)$$
Using the $y$-intercept $f(0)=-14$:
$$-14 = a(0+1)(0-7) = -7a$$
Solving for $a$:
$$a = 2$$
So $f(x) = 2(x+1)(x-7)$. Expanding:
$$(x+1)(x-7) = x^2-6x-7$$
$$f(x) = 2(x^2-6x-7)$$
$$\boxed{f(x) = 2x^2-12x-14}$$
Checking the axis of symmetry: $x=-\dfrac{b}{2a} = -\dfrac{-12}{4}=3$, which agrees.
QUESTION 39
8 marks
Medium
A farmer has $120$ m of fencing to build a rectangular pen along the side of an existing straight wall, so that the wall forms one side of the pen and no fencing is needed there. The two sides perpendicular to the wall each have length $y$ metres, and the side parallel to the wall (opposite the wall) has length $x$ metres, as shown in the diagram.
(a) Show that $x = 120 - 2y$, and express the area $A$ of the pen as a function of $y$ only. [2]
(b) Find the value of $y$ that maximises the area $A$. [4]
(c) Find the dimensions of the pen and the maximum possible area. [2]
Show complete worked solution
**(a)**
Since the total fencing used is $x + 2y = 120$ (one length and two widths):
$$\boxed{x = 120-2y}$$
Using $A=xy$:
$$A(y) = (120-2y)y = 120y - 2y^2$$
**(b)**
Writing $A(y) = -2y^2+120y$, this is a quadratic in $y$ with $a=-2,\ b=120$, which is maximised at its vertex. Using $y=-\dfrac{b}{2a}$:
$$y = -\frac{120}{2(-2)} = \frac{120}{4} = 30$$
$$\boxed{y = 30 \text{ m}}$$
**(c)**
Substituting $y=30$ into $x=120-2y$:
$$x = 120-2(30) = 60$$
Substituting into $A=xy$:
$$A = 60 \times 30 = 1800$$
$$\boxed{\text{Dimensions: } 60 \text{ m} \times 30 \text{ m}, \ \text{Maximum area} = 1800 \text{ m}^2}$$
QUESTION 40
6 marks
Medium
Let $f(x) = -(x-3)^2 + 7$ for $-1 \le x \le 6$.
(a) Write down the maximum value of $f$. [2]
(b) Find the range of $f$ over the given domain. [4]
Show complete worked solution
**(a)**
Since $f(x) = -(x-3)^2+7$ is in vertex form with vertex $(3,7)$, and the coefficient of the squared term is negative (so the graph opens downward), the maximum value of $f$ occurs at the vertex, and the vertex $x=3$ lies within the domain $-1 \le x \le 6$:
$$\boxed{\text{Maximum value} = 7}$$
**(b)**
To find the range, evaluate $f$ at the endpoints of the domain, since the minimum of $f$ over a restricted domain occurs at the endpoint furthest from the vertex.
$$f(-1) = -(-1-3)^2+7 = -(16)+7 = -9$$
$$f(6) = -(6-3)^2+7 = -(9)+7 = -2$$
Since $-9 < -2$, the minimum value on the domain is $-9$, occurring at $x=-1$. Combined with the maximum value of $7$ from part (a):
$$\boxed{-9 \le f(x) \le 7}$$
QUESTION 41
6 marks
Medium
Find the coordinates of the points of intersection of the line $y = 2x - 3$ and the curve $y = x^2 - 3x + 1$.
Show complete worked solution
At points of intersection, the $y$-values are equal, so:
$$x^2-3x+1 = 2x-3$$
Rearranging so one side is zero:
$$x^2 - 5x + 4 = 0$$
Factorising: two numbers that multiply to give $4$ and add to give $-5$ are $-1$ and $-4$:
$$(x-1)(x-4) = 0$$
So $x=1$ or $x=4$.
Substituting into $y=2x-3$ to find the corresponding $y$-values:
$$x=1: \ y = 2(1)-3 = -1$$
$$x=4: \ y = 2(4)-3 = 5$$
$$\boxed{(1,-1) \text{ and } (4,5)}$$
QUESTION 42
5 marks
Medium
The roots of the equation $2x^2 - 7x + 4 = 0$ are $\alpha$ and $\beta$.
Without solving the equation, find the value of $\alpha^2+\beta^2$.
Show complete worked solution
Using the formulas for the sum and product of roots, $\alpha+\beta = -\dfrac{b}{a}$ and $\alpha\beta = \dfrac{c}{a}$, with $a=2,\ b=-7,\ c=4$:
$$\alpha+\beta = -\frac{-7}{2} = \frac{7}{2}, \qquad \alpha\beta = \frac{4}{2} = 2$$
Using the identity $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$:
$$\alpha^2+\beta^2 = \left(\frac{7}{2}\right)^2 - 2(2)$$
$$= \frac{49}{4} - 4 = \frac{49}{4} - \frac{16}{4}$$
$$\boxed{\alpha^2+\beta^2 = \frac{33}{4}}$$
QUESTION 43
7 marks
Medium
A small manufacturer has fixed costs of \$850 and a cost of \$15 per unit produced. Each unit is sold for \$40. Let $x$ be the number of units produced and sold. The cost function is $C(x) = 850 + 15x$ and the revenue function is $R(x) = 40x$.
(a) Find the break-even number of units, i.e. the value of $x$ for which $C(x) = R(x)$. [3]
(b) Find the profit made if $60$ units are produced and sold. [4]
Show complete worked solution
**(a)**
Setting $C(x)=R(x)$:
$$850+15x = 40x$$
Collecting like terms:
$$850 = 40x-15x$$
$$850 = 25x$$
Solving for $x$:
$$\boxed{x = 34 \text{ units}}$$
**(b)**
Using $\text{Profit}(x) = R(x) - C(x)$ with $x=60$:
$$R(60) = 40(60) = 2400$$
$$C(60) = 850+15(60) = 850+900 = 1750$$
$$\text{Profit} = 2400-1750$$
$$\boxed{\text{Profit} = \$650}$$
QUESTION 44
9 marks
Hard
The line $y = kx + 3$ is tangent to the curve $y = x^2 - 4x + 7$.
(a) Find the two possible values of $k$. [5]
(b) For the value of $k$ with negative gradient, find the coordinates of the point of tangency. [4]
Show complete worked solution
**(a)**
At points of intersection of the line and curve:
$$x^2-4x+7 = kx+3$$
Rearranging so one side is zero:
$$x^2 - (4+k)x + 4 = 0$$
For the line to be tangent to the curve, this equation must have exactly one (repeated) solution, so the discriminant must equal zero. Using $\Delta = b^2-4ac$ with $a=1,\ b=-(4+k),\ c=4$:
$$\Delta = (4+k)^2 - 4(1)(4) = 0$$
$$(4+k)^2 = 16$$
Taking square roots:
$$4+k = \pm 4$$
$$\boxed{k = 0 \text{ or } k = -8}$$
**(b)**
The value of $k$ with negative gradient is $k=-8$. Substituting into $x^2-(4+k)x+4=0$:
$$x^2 - (4-8)x + 4 = 0$$
$$x^2 + 4x + 4 = 0$$
Factorising:
$$(x+2)^2 = 0 \implies x = -2$$
Substituting $x=-2$ into the line $y=-8x+3$:
$$y = -8(-2)+3 = 16+3 = 19$$
$$\boxed{\text{Point of tangency: } (-2, 19)}$$
QUESTION 45
10 marks
Hard
Find the set of values of $m$, where $m \neq 0$, for which the equation $mx^2 - (m+2)x + 3 = 0$ has two distinct positive real roots.
Show complete worked solution
For the equation to have two distinct positive real roots $\alpha, \beta$, three conditions must all be satisfied: the discriminant is positive ($\Delta>0$), the sum of roots is positive ($\alpha+\beta>0$), and the product of roots is positive ($\alpha\beta>0$).
**Discriminant condition:** With $a=m,\ b=-(m+2),\ c=3$:
$$\Delta = (m+2)^2 - 4(m)(3) = m^2+4m+4-12m = m^2-8m+4$$
Requiring $\Delta>0$, using the quadratic formula to find where $m^2-8m+4=0$:
$$m = \frac{8\pm\sqrt{64-16}}{2} = \frac{8\pm\sqrt{48}}{2} = 4\pm2\sqrt{3}$$
Since the coefficient of $m^2$ is positive, $m^2-8m+4>0$ outside these roots:
$$m < 4-2\sqrt{3} \quad \text{or} \quad m > 4+2\sqrt{3}$$
**Product condition:** $\alpha\beta = \dfrac{3}{m} > 0 \implies m>0$.
**Sum condition:** $\alpha+\beta = \dfrac{m+2}{m} > 0$. Since $m>0$ from the product condition, $m+2>0$ automatically, so this condition is already satisfied whenever $m>0$.
**Combining all conditions:** intersecting $m>0$ with the discriminant condition:
$$\boxed{0 < m < 4-2\sqrt{3} \quad \text{or} \quad m > 4+2\sqrt{3}}$$
QUESTION 46
8 marks
Hard
The roots of the equation $2x^2 - 5x + 3 = 0$ are $\alpha$ and $\beta$.
(a) Find the values of $\alpha+\beta$ and $\alpha\beta$. [2]
(b) Find the values of $(\alpha+1)+(\beta+1)$ and $(\alpha+1)(\beta+1)$. [3]
(c) Hence find a quadratic equation, with integer coefficients, that has roots $\alpha+1$ and $\beta+1$. [3]
Show complete worked solution
**(a)**
Using the formulas for the sum and product of roots with $a=2,\ b=-5,\ c=3$:
$$\alpha+\beta = -\frac{-5}{2} = \frac{5}{2}, \qquad \alpha\beta = \frac{3}{2}$$
$$\boxed{\alpha+\beta = \frac{5}{2}, \ \alpha\beta = \frac{3}{2}}$$
**(b)**
For the sum:
$$(\alpha+1)+(\beta+1) = (\alpha+\beta) + 2 = \frac{5}{2}+2 = \frac{9}{2}$$
For the product, expanding:
$$(\alpha+1)(\beta+1) = \alpha\beta + \alpha+\beta + 1 = \frac{3}{2} + \frac{5}{2} + 1$$
$$= \frac{8}{2}+1 = 4+1 = 5$$
$$\boxed{\text{Sum} = \frac{9}{2}, \ \text{Product} = 5}$$
**(c)**
A quadratic with roots summing to $S$ and multiplying to $P$ can be written $x^2-Sx+P=0$:
$$x^2 - \frac{9}{2}x + 5 = 0$$
Multiplying through by $2$ to obtain integer coefficients:
$$\boxed{2x^2 - 9x + 10 = 0}$$
QUESTION 47
11 marks
Hard
A farmer builds a rectangular enclosure using $180$ m of fencing along one side of a straight wall, so that the wall forms one full side of the enclosure. An internal fence, parallel to the wall, is also built to divide the enclosure into two equal rectangular sections, as shown in the diagram. The side parallel to the wall has length $x$ metres, and each of the three fence sections perpendicular to the wall (the two outer sides and the internal divider) has length $y$ metres.
(a) Show that $x + 3y = 180$, and express the total enclosed area $A$ as a function of $y$ only. [2]
(b) Find the value of $y$ that maximises $A$. [5]
(c) Find the corresponding value of $x$ and the maximum area. [4]
Show complete worked solution
**(a)**
The total fencing used consists of the side parallel to the wall (length $x$) plus three perpendicular sections each of length $y$ (two outer sides and one internal divider):
$$\boxed{x+3y = 180}$$
So $x = 180-3y$. Using $A = xy$:
$$A(y) = (180-3y)y = 180y - 3y^2$$
**(b)**
Writing $A(y) = -3y^2+180y$, this is a quadratic in $y$ with $a=-3,\ b=180$, maximised at its vertex. Using $y = -\dfrac{b}{2a}$:
$$y = -\frac{180}{2(-3)} = \frac{180}{6} = 30$$
$$\boxed{y = 30 \text{ m}}$$
**(c)**
Substituting $y=30$ into $x=180-3y$:
$$x = 180-3(30) = 180-90 = 90$$
Substituting into $A=xy$:
$$A = 90 \times 30 = 2700$$
$$\boxed{x=90 \text{ m}, \ \text{Maximum area} = 2700 \text{ m}^2}$$
QUESTION 48
12 marks
Hard
An object is launched vertically upward. Its height above the ground, in metres, $t$ seconds after launch, is modelled by $h(t) = -5t^2 + 25t + 4$ for $t \ge 0$.
(a) Find the maximum height reached by the object. [3]
(b) Find the time at which the object hits the ground, giving your answer correct to two decimal places. [4]
(c) Find, using algebra, the interval of time during which the object is more than $20$ metres above the ground, giving your answers correct to two decimal places. [5]
Show complete worked solution
**(a)**
Since $h(t)$ has $a=-5,\ b=25$, its maximum occurs at the vertex. Using $t=-\dfrac{b}{2a}$:
$$t = -\frac{25}{2(-5)} = \frac{25}{10} = 2.5$$
$$h(2.5) = -5(2.5)^2+25(2.5)+4 = -31.25+62.5+4$$
$$\boxed{\text{Maximum height} = 35.25 \text{ metres, at } t=2.5 \text{ s}}$$
**(b)**
The object hits the ground when $h(t)=0$:
$$-5t^2+25t+4=0$$
Multiplying by $-1$:
$$5t^2-25t-4=0$$
Using the quadratic formula with $a=5,\ b=-25,\ c=-4$:
$$t = \frac{25\pm\sqrt{(-25)^2-4(5)(-4)}}{2(5)} = \frac{25\pm\sqrt{625+80}}{10} = \frac{25\pm\sqrt{705}}{10}$$
Since $t \ge 0$, only the positive root is valid ($\sqrt{705}\approx 26.5498$):
$$t = \frac{25+26.5498}{10} \approx 5.15498$$
$$\boxed{t \approx 5.15 \text{ s}}$$
**(c)**
Solving the inequality $h(t) > 20$:
$$-5t^2+25t+4 > 20$$
Rearranging so one side is zero:
$$-5t^2+25t-16>0$$
Multiplying by $-1$ (reversing the inequality sign):
$$5t^2-25t+16<0$$
Using the quadratic formula with $a=5,\ b=-25,\ c=16$:
$$t = \frac{25\pm\sqrt{625-320}}{10} = \frac{25\pm\sqrt{305}}{10}$$
Since $\sqrt{305}\approx 17.4642$:
$$t = \frac{25-17.4642}{10} \approx 0.7536, \qquad t = \frac{25+17.4642}{10} \approx 4.2464$$
Since the coefficient of $t^2$ in $5t^2-25t+16$ is positive, the expression is negative between the roots:
$$\boxed{0.75 \text{ s} < t < 4.25 \text{ s (to 2 d.p.)}}$$
QUESTION 49
11 marks
Hard
(a) Find the point of intersection of the lines $y = x+2$ and $y=-2x+11$. [4]
(b) A quadratic function $f(x) = ax^2+bx+c$ has axis of symmetry equal to the $x$-coordinate of the point found in part (a), and its graph passes through the points $(0,4)$ and $(5,-6)$. Find the values of $a$, $b$ and $c$. [7]
Show complete worked solution
**(a)**
Setting the two expressions for $y$ equal:
$$x+2 = -2x+11$$
Solving for $x$:
$$3x = 9$$
$$x=3$$
Substituting into $y=x+2$:
$$y = 3+2 = 5$$
$$\boxed{(3,5)}$$
**(b)**
The axis of symmetry is $x=3$ (the $x$-coordinate found in part (a)). Using $x=-\dfrac{b}{2a}=3$:
$$b = -6a \quad \text{...(1)}$$
Using the point $(0,4)$: since $f(0)=c$:
$$c = 4 \quad \text{...(2)}$$
Using the point $(5,-6)$:
$$25a+5b+c = -6$$
Substituting (1) and (2):
$$25a + 5(-6a) + 4 = -6$$
Simplifying:
$$25a-30a+4=-6$$
$$-5a = -10$$
Solving for $a$:
$$a=2$$
Substituting back into (1):
$$b = -6(2) = -12$$
$$\boxed{a=2, \ b=-12, \ c=4}$$
Checking: $f(x)=2x^2-12x+4$; axis of symmetry $-\dfrac{-12}{4}=3$, and $f(5)=50-60+4=-6$, both of which agree.
QUESTION 50
8 marks
Hard
The equation $x^2 - (3k-1)x + (2k^2-k-1) = 0$ has two roots that differ by $3$. Find the possible values of $k$.
Show complete worked solution
Let the roots be $\alpha$ and $\beta$ with $\alpha-\beta = 3$ (taking the difference as positive without loss of generality), so $(\alpha-\beta)^2=9$.
Using the sum and product of roots, with $a=1,\ b=-(3k-1),\ c=2k^2-k-1$:
$$\alpha+\beta = 3k-1, \qquad \alpha\beta = 2k^2-k-1$$
Using the identity $(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta$:
$$9 = (3k-1)^2 - 4(2k^2-k-1)$$
Expanding $(3k-1)^2 = 9k^2-6k+1$:
$$9 = 9k^2-6k+1 - 8k^2+4k+4$$
Simplifying the right-hand side:
$$9 = k^2-2k+5$$
Rearranging so one side is zero:
$$k^2-2k-4=0$$
Using the quadratic formula with $a=1,\ b=-2,\ c=-4$:
$$k = \frac{2\pm\sqrt{4+16}}{2} = \frac{2\pm\sqrt{20}}{2} = \frac{2\pm2\sqrt{5}}{2}$$
$$\boxed{k = 1+\sqrt{5} \ \text{ or } \ k = 1-\sqrt{5}}$$
Rational Functions 50 questions
QUESTION 1
3 marks
Easy
Consider the function $f(x) = \dfrac{3}{x-2}+1$. Write down the equations of the vertical and horizontal asymptotes of the graph of $f$.
Show complete worked solution
Using the fact that for $f(x)=\dfrac{a}{x-h}+k$, the vertical asymptote is $x=h$ and the horizontal asymptote is $y=k$:
Vertical asymptote: setting the denominator equal to zero,
$$x-2=0 \Rightarrow x=2$$
Horizontal asymptote: as $x \to \pm\infty$, $\dfrac{3}{x-2} \to 0$, so $f(x) \to 1$.
$$\boxed{x=2 \text{ and } y=1}$$
QUESTION 2
4 marks
Easy
Consider the function $f(x) = \dfrac{2x+5}{x-3}$. Find the equations of the vertical and horizontal asymptotes of the graph of $f$.
Show complete worked solution
The vertical asymptote occurs where the denominator is zero:
$$x-3=0 \Rightarrow x=3$$
For the horizontal asymptote, rewrite $f$ by division. Writing the numerator in terms of $(x-3)$:
$$2x+5 = 2(x-3)+11$$
so
$$f(x) = 2+\frac{11}{x-3}$$
As $x \to \pm\infty$, $\dfrac{11}{x-3} \to 0$, so $f(x) \to 2$.
$$\boxed{x=3 \text{ and } y=2}$$
QUESTION 3
5 marks
Medium
Consider the function $f(x) = \dfrac{4}{2x+6}-3$.
(a) Find the equations of the vertical and horizontal asymptotes of the graph of $f$.
(b) Find the coordinates of the points where the graph of $f$ crosses the $x$-axis and the $y$-axis.
Show complete worked solution
**(a)**
Simplifying $f$ first:
$$f(x) = \frac{4}{2x+6}-3 = \frac{4}{2(x+3)}-3 = \frac{2}{x+3}-3$$
Vertical asymptote: setting the denominator equal to zero,
$$x+3=0 \Rightarrow x=-3$$
Horizontal asymptote: as $x \to \pm\infty$, $\dfrac{2}{x+3} \to 0$, so $f(x) \to -3$.
So the asymptotes are $x=-3$ and $y=-3$.
**(b)**
Finding the $y$-intercept by substituting $x=0$:
$$f(0) = \frac{2}{3}-3 = -\frac{7}{3}$$
giving $\left(0,-\dfrac{7}{3}\right)$.
Finding the $x$-intercept by setting $f(x)=0$:
$$\frac{2}{x+3}=3 \Rightarrow x+3=\frac{2}{3} \Rightarrow x=\frac{2}{3}-3=-\frac{7}{3}$$
giving $\left(-\dfrac{7}{3},0\right)$.
$$\boxed{\text{Asymptotes: } x=-3,\ y=-3;\quad x\text{-intercept } \left(-\tfrac{7}{3},0\right);\quad y\text{-intercept } \left(0,-\tfrac{7}{3}\right)}$$
QUESTION 4
6 marks
Medium
The function $f(x) = \dfrac{k}{x+a}+b$ has a vertical asymptote at $x=4$, a horizontal asymptote at $y=-2$, and its graph passes through the point $(6,1)$. Find the values of $k$, $a$ and $b$.
Show complete worked solution
Using the asymptotes to find $a$ and $b$, then substituting the given point to find $k$.
A vertical asymptote at $x=4$ means $x+a=0$ when $x=4$, so:
$$a=-4$$
The horizontal asymptote is $y=b$, so:
$$b=-2$$
So $f(x) = \dfrac{k}{x-4}-2$. Substituting the point $(6,1)$:
$$\frac{k}{6-4}-2=1$$
Solving for $k$:
$$\frac{k}{2}=3$$
$$k=6$$
$$\boxed{k=6,\ a=-4,\ b=-2}$$
QUESTION 5
7 marks
Hard
A function is defined by $f(x) = \dfrac{2x+p}{x-q}$, where $p$ and $q$ are constants. The graph of $f$ has a vertical asymptote at $x=5$ and passes through the point $(0,-3)$.
(a) Find the values of $p$ and $q$.
(b) Write down the equation of the horizontal asymptote of the graph of $f$.
(c) Find the $x$-intercept of the graph of $f$.
Show complete worked solution
**(a)**
The vertical asymptote occurs where $x-q=0$, so:
$$q=5$$
Substituting the point $(0,-3)$ into $f(x) = \dfrac{2x+p}{x-5}$:
$$\frac{0+p}{0-5}=-3 \Rightarrow \frac{p}{-5}=-3$$
Solving for $p$:
$$p=15$$
$$\boxed{p=15,\ q=5}$$
So $f(x) = \dfrac{2x+15}{x-5}$.
**(b)**
As $x \to \pm\infty$, the ratio of the leading terms dominates:
$$f(x) \to \frac{2x}{x}=2$$
$$\boxed{y=2}$$
**(c)**
Setting $f(x)=0$, which requires the numerator to be zero:
$$2x+15=0$$
Solving for $x$:
$$x=-7.5$$
$$\boxed{x\text{-intercept at } x=-7.5}$$
QUESTION 6
3 marks
Easy
Consider the function $f(x) = \dfrac{5}{x+4}-2$. Write down the equations of the vertical and horizontal asymptotes of the graph of $f$.
Show complete worked solution
Using the fact that for $f(x)=\dfrac{a}{x-h}+k$, the vertical asymptote is $x=h$ and the horizontal asymptote is $y=k$:
Vertical asymptote: setting the denominator equal to zero,
$$x+4=0 \Rightarrow x=-4$$
Horizontal asymptote: as $x \to \pm\infty$, $\dfrac{5}{x+4} \to 0$, so $f(x) \to -2$.
$$\boxed{x=-4 \text{ and } y=-2}$$
QUESTION 7
4 marks
Easy
Consider the function $f(x) = \dfrac{3x-1}{x+2}$. Find the equations of the vertical and horizontal asymptotes of the graph of $f$.
Show complete worked solution
The vertical asymptote occurs where the denominator is zero:
$$x+2=0 \Rightarrow x=-2$$
For the horizontal asymptote, rewrite $f$ by division. Writing the numerator in terms of $(x+2)$:
$$3x-1 = 3(x+2)-7$$
so
$$f(x) = 3-\frac{7}{x+2}$$
As $x \to \pm\infty$, $\dfrac{7}{x+2} \to 0$, so $f(x) \to 3$.
$$\boxed{x=-2 \text{ and } y=3}$$
QUESTION 8
3 marks
Easy
Consider the function $f(x) = \dfrac{1}{x}$. State the equations of the vertical and horizontal asymptotes of the graph of $f$, and write down the domain of $f$.
Show complete worked solution
For $f(x) = \dfrac{1}{x}$, the denominator is zero when $x=0$, so:
$$\text{Vertical asymptote: } x=0$$
As $x \to \pm\infty$, $\dfrac{1}{x} \to 0$, so:
$$\text{Horizontal asymptote: } y=0$$
Since division by zero is undefined, $x=0$ must be excluded from the domain.
$$\boxed{x=0,\ y=0,\ \text{domain: } x \in \mathbb{R},\ x \neq 0}$$
QUESTION 9
4 marks
Easy
Consider the function $f(x) = \dfrac{-2}{x-1}+3$.
(a) Find the coordinates of the point where the graph of $f$ crosses the $y$-axis. [2]
(b) Find the coordinates of the point where the graph of $f$ crosses the $x$-axis. [2]
Show complete worked solution
**(a)**
The $y$-intercept is found by substituting $x=0$:
$$f(0) = \frac{-2}{0-1}+3 = \frac{-2}{-1}+3 = 2+3 = 5$$
$$\boxed{(0,5)}$$
**(b)**
The $x$-intercept is found by setting $f(x)=0$:
$$\frac{-2}{x-1}+3=0 \Rightarrow \frac{-2}{x-1}=-3$$
Multiplying both sides by $(x-1)$:
$$-2 = -3(x-1) = -3x+3$$
Solving for $x$:
$$3x = 5 \Rightarrow x=\frac{5}{3}$$
$$\boxed{\left(\frac{5}{3},0\right)}$$
QUESTION 10
4 marks
Easy
Write down the domain and range of the function $f(x) = \dfrac{4}{x+5}-1$.
Show complete worked solution
The domain excludes any value of $x$ that makes the denominator zero.
$$x+5=0 \Rightarrow x=-5$$
So the domain is all real $x$ except $x=-5$.
The range excludes the value of $f(x)$ given by the horizontal asymptote, since $\dfrac{4}{x+5}$ can never equal zero.
$$\text{As } x \to \pm\infty,\ f(x) \to -1$$
So the range is all real $y$ except $y=-1$.
$$\boxed{\text{Domain: } x \in \mathbb{R},\ x\neq -5; \quad \text{Range: } y \in \mathbb{R},\ y\neq -1}$$
QUESTION 11
4 marks
Easy
Consider the function $f(x) = \dfrac{x+6}{x-2}$.
(a) Find the value of $f(4)$. [2]
(b) Find the value of $f(0)$. [2]
Show complete worked solution
**(a)**
Substituting $x=4$ into $f(x)=\dfrac{x+6}{x-2}$:
$$f(4) = \frac{4+6}{4-2} = \frac{10}{2}$$
$$\boxed{f(4)=5}$$
**(b)**
Substituting $x=0$:
$$f(0) = \frac{0+6}{0-2} = \frac{6}{-2}$$
$$\boxed{f(0)=-3}$$
QUESTION 12
2 marks
Easy
State the equation of the vertical asymptote of the graph of $g(x) = \dfrac{7}{2x-6}$.
Show complete worked solution
The vertical asymptote occurs where the denominator is equal to zero.
$$2x-6=0$$
Solving for $x$:
$$2x=6 \Rightarrow x=3$$
$$\boxed{x=3}$$
QUESTION 13
3 marks
Easy
Consider the function $f(x) = \dfrac{6}{3x+9}+2$.
(a) Show that $f(x)$ can be written as $\dfrac{2}{x+3}+2$. [2]
(b) Write down the equation of the horizontal asymptote of the graph of $f$. [1]
Show complete worked solution
**(a)**
Factorising the denominator $3x+9$:
$$3x+9 = 3(x+3)$$
so
$$f(x) = \frac{6}{3(x+3)}+2 = \frac{6}{3}\cdot\frac{1}{x+3}+2$$
$$\boxed{f(x) = \frac{2}{x+3}+2}$$
**(b)**
As $x \to \pm\infty$, $\dfrac{2}{x+3} \to 0$, so $f(x) \to 2$.
$$\boxed{y=2}$$
QUESTION 14
3 marks
Easy
The graph of $y = \dfrac{k}{x}$, where $k>0$, passes through the point $(2,5)$. Find the value of $k$.
Show complete worked solution
Substituting the coordinates $x=2$, $y=5$ into $y=\dfrac{k}{x}$:
$$5 = \frac{k}{2}$$
Solving for $k$ by multiplying both sides by $2$:
$$k = 5 \times 2$$
$$\boxed{k=10}$$
QUESTION 15
2 marks
Easy
Consider the function $f(x) = \dfrac{x-3}{x+1}$. Find the coordinates of the point where the graph of $f$ crosses the $y$-axis.
Show complete worked solution
The $y$-intercept is found by substituting $x=0$:
$$f(0) = \frac{0-3}{0+1} = \frac{-3}{1}$$
$$\boxed{(0,-3)}$$
QUESTION 16
4 marks
Easy
Write down the domain and range of the function $f(x) = \dfrac{-3}{x+2}-4$.
Show complete worked solution
The domain excludes the value of $x$ that makes the denominator zero.
$$x+2=0 \Rightarrow x=-2$$
So the domain is all real $x$ except $x=-2$.
For the range, since $\dfrac{-3}{x+2}$ can never be equal to $0$, $f(x)$ can never equal the horizontal asymptote value:
$$\text{As } x \to \pm\infty,\ f(x) \to -4$$
$$\boxed{\text{Domain: } x \in \mathbb{R},\ x\neq -2; \quad \text{Range: } y \in \mathbb{R},\ y\neq -4}$$
QUESTION 17
4 marks
Easy
Sketch the graph of $y = \dfrac{1}{x}$ for $-3 \leq x \leq 3$, $x \neq 0$, clearly showing the vertical and horizontal asymptotes as dashed lines.
Show complete worked solution
The function $y=\dfrac{1}{x}$ has a vertical asymptote where the denominator is zero, at $x=0$, and a horizontal asymptote where $y \to 0$ as $x \to \pm\infty$, at $y=0$.
Testing a few values: $f(1)=1$, $f(-1)=-1$, $f(2)=0.5$, $f(-2)=-0.5$, $f(3) \approx 0.33$, $f(-3) \approx -0.33$.
The graph consists of two separate branches, one in the region $x>0$ (lying in the first quadrant, decreasing from $+\infty$ towards $0$ as $x$ increases) and one in the region $x<0$ (lying in the third quadrant, decreasing from $0$ towards $-\infty$ as $x \to 0^-$).
$$\boxed{\text{Two branches, asymptotes } x=0 \text{ and } y=0 \text{ shown dashed (see diagram)}}$$
QUESTION 18
3 marks
Easy
Consider the function $f(x) = \dfrac{2}{x-5}$. Find the value of $x$ for which $f(x) = -1$.
Show complete worked solution
Setting $f(x)=-1$:
$$\frac{2}{x-5} = -1$$
Multiplying both sides by $(x-5)$:
$$2 = -1(x-5) = -x+5$$
Solving for $x$:
$$x = 5-2$$
$$\boxed{x=3}$$
QUESTION 19
3 marks
Easy
State the equations of the vertical and horizontal asymptotes of the graph of $y = \dfrac{4x}{x-1}$.
Show complete worked solution
The vertical asymptote occurs where the denominator is zero:
$$x-1=0 \Rightarrow x=1$$
For the horizontal asymptote, as $x \to \pm\infty$ the ratio of the leading terms dominates:
$$y \to \frac{4x}{x} = 4$$
$$\boxed{x=1 \text{ and } y=4}$$
QUESTION 20
4 marks
Easy
Consider the function $f(x) = \dfrac{1}{x+3}+5$.
(a) Find the value of $f(-2)$. [2]
(b) Find the value of $f(-4)$. [2]
Show complete worked solution
**(a)**
Substituting $x=-2$:
$$f(-2) = \frac{1}{-2+3}+5 = \frac{1}{1}+5$$
$$\boxed{f(-2)=6}$$
**(b)**
Substituting $x=-4$:
$$f(-4) = \frac{1}{-4+3}+5 = \frac{1}{-1}+5 = -1+5$$
$$\boxed{f(-4)=4}$$
QUESTION 21
4 marks
Easy
Consider the function $f(x) = \dfrac{5-x}{x+1}$. Find the equations of the vertical and horizontal asymptotes of the graph of $f$.
Show complete worked solution
The vertical asymptote occurs where the denominator is zero:
$$x+1=0 \Rightarrow x=-1$$
For the horizontal asymptote, as $x \to \pm\infty$ the ratio of the leading terms dominates. Writing the numerator as $-x+5$:
$$f(x) \to \frac{-x}{x} = -1$$
$$\boxed{x=-1 \text{ and } y=-1}$$
QUESTION 22
4 marks
Easy
Sketch the graph of $y = -\dfrac{1}{x}$ for $-3 \leq x \leq 3$, $x \neq 0$, clearly showing the vertical and horizontal asymptotes as dashed lines and marking one point on each branch.
Show complete worked solution
The function $y=-\dfrac{1}{x}$ has a vertical asymptote at $x=0$ (where the denominator is zero) and a horizontal asymptote at $y=0$ (since $y \to 0$ as $x \to \pm\infty$).
Testing a few values: $f(1)=-1$, $f(-1)=1$, $f(2)=-0.5$, $f(-2)=0.5$.
Since $y=-\dfrac{1}{x}$ is the reflection of $y=\dfrac{1}{x}$ in the $x$-axis, the graph has one branch in the second quadrant ($x<0$, $y>0$) and one branch in the fourth quadrant ($x>0$, $y<0$).
$$\boxed{\text{Two branches, asymptotes } x=0 \text{ and } y=0 \text{ shown dashed (see diagram)}}$$
QUESTION 23
3 marks
Easy
Consider the function $f(x) = \dfrac{3}{x-1}-2$. Determine, showing your reasoning, whether the point $(2,1)$ lies on the graph of $f$.
Show complete worked solution
Substituting $x=2$ into $f(x) = \dfrac{3}{x-1}-2$:
$$f(2) = \frac{3}{2-1}-2 = \frac{3}{1}-2 = 3-2$$
$$f(2) = 1$$
Since $f(2)=1$ matches the given $y$-coordinate of the point $(2,1)$:
$$\boxed{\text{Yes, the point } (2,1) \text{ lies on the graph of } f}$$
QUESTION 24
6 marks
Medium
Consider the function $f(x) = \dfrac{3x+2}{x-1}$.
(a) Find the equations of the vertical and horizontal asymptotes of the graph of $f$. [3]
(b) Find the coordinates of the points where the graph of $f$ crosses the $x$-axis and the $y$-axis. [3]
Show complete worked solution
**(a)**
The vertical asymptote occurs where the denominator is zero:
$$x-1=0 \Rightarrow x=1$$
For the horizontal asymptote, as $x \to \pm\infty$ the ratio of the leading terms dominates:
$$f(x) \to \frac{3x}{x} = 3$$
$$\boxed{x=1 \text{ and } y=3}$$
**(b)**
The $y$-intercept is found by substituting $x=0$:
$$f(0) = \frac{3(0)+2}{0-1} = \frac{2}{-1} = -2$$
giving $(0,-2)$.
The $x$-intercept is found by setting the numerator equal to zero:
$$3x+2=0 \Rightarrow x=-\frac{2}{3}$$
giving $\left(-\dfrac{2}{3},0\right)$.
$$\boxed{x\text{-intercept } \left(-\tfrac{2}{3},0\right); \quad y\text{-intercept } (0,-2)}$$
QUESTION 25
6 marks
Medium
The function $f(x) = \dfrac{k}{x-3}+2$ has a graph that passes through the point $(5,5)$.
(a) Find the value of $k$. [3]
(b) Find the coordinates of the point where the graph of $f$ crosses the $x$-axis. [3]
Show complete worked solution
**(a)**
Substituting the point $(5,5)$ into $f(x) = \dfrac{k}{x-3}+2$:
$$5 = \frac{k}{5-3}+2 = \frac{k}{2}+2$$
Solving for $k$:
$$3 = \frac{k}{2}$$
$$\boxed{k=6}$$
**(b)**
With $k=6$, $f(x) = \dfrac{6}{x-3}+2$. Setting $f(x)=0$:
$$\frac{6}{x-3}+2=0 \Rightarrow \frac{6}{x-3}=-2$$
Multiplying both sides by $(x-3)$:
$$6=-2(x-3)=-2x+6$$
Solving for $x$:
$$2x=0 \Rightarrow x=0$$
$$\boxed{(0,0)}$$
QUESTION 26
6 marks
Medium
The function $f(x) = \dfrac{2x+a}{x+b}$ has a vertical asymptote at $x=-3$ and its graph passes through the point $(0,4)$.
(a) Find the value of $b$. [3]
(b) Find the value of $a$. [3]
Show complete worked solution
**(a)**
The vertical asymptote occurs where $x+b=0$. Since this happens at $x=-3$:
$$-3+b=0$$
$$\boxed{b=3}$$
**(b)**
So $f(x) = \dfrac{2x+a}{x+3}$. Substituting the point $(0,4)$:
$$4 = \frac{2(0)+a}{0+3} = \frac{a}{3}$$
Solving for $a$:
$$\boxed{a=12}$$
QUESTION 27
6 marks
Medium
Find the domain and range of the function $f(x) = \dfrac{4-2x}{x+5}$.
Show complete worked solution
**Domain**
The domain excludes the value of $x$ that makes the denominator zero.
$$x+5=0 \Rightarrow x=-5$$
$$\boxed{\text{Domain: } x \in \mathbb{R},\ x \neq -5}$$
**Range**
The range excludes the value of the horizontal asymptote. As $x \to \pm\infty$, using the ratio of the leading terms (writing the numerator as $-2x+4$):
$$f(x) \to \frac{-2x}{x} = -2$$
$$\boxed{\text{Range: } y \in \mathbb{R},\ y \neq -2}$$
QUESTION 28
5 marks
Medium
Solve the equation $\dfrac{x+4}{x-2} = 3$ algebraically, and verify your answer by substitution.
Show complete worked solution
**Solving**
Multiplying both sides by $(x-2)$:
$$x+4 = 3(x-2)$$
Expanding the right-hand side:
$$x+4 = 3x-6$$
Collecting terms:
$$4+6 = 3x-x$$
$$10=2x$$
$$x=5$$
Note that $x=5$ does not violate the domain restriction $x \neq 2$, so it is a valid solution.
**Verifying**
Substituting $x=5$ back into the original expression:
$$\frac{5+4}{5-2} = \frac{9}{3} = 3 \checkmark$$
$$\boxed{x=5}$$
QUESTION 29
7 marks
Medium
Consider the function $f(x) = \dfrac{2}{x+1}-3$.
(a) Find the equations of the vertical and horizontal asymptotes of the graph of $f$. [2]
(b) Find the coordinates of the points where the graph of $f$ crosses the $x$-axis and the $y$-axis. [2]
(c) Sketch the graph of $f$, showing the asymptotes as dashed lines and labelling the intercepts found in part (b). [3]
Show complete worked solution
**(a)**
Vertical asymptote: setting the denominator equal to zero,
$$x+1=0 \Rightarrow x=-1$$
Horizontal asymptote: as $x \to \pm\infty$, $\dfrac{2}{x+1} \to 0$, so $f(x) \to -3$.
$$\boxed{x=-1 \text{ and } y=-3}$$
**(b)**
The $y$-intercept is found by substituting $x=0$:
$$f(0) = \frac{2}{0+1}-3 = 2-3=-1$$
giving $(0,-1)$.
The $x$-intercept is found by setting $f(x)=0$:
$$\frac{2}{x+1}-3=0 \Rightarrow \frac{2}{x+1}=3$$
$$2=3(x+1) = 3x+3$$
$$3x=-1 \Rightarrow x=-\frac{1}{3}$$
giving $\left(-\dfrac{1}{3},0\right)$.
$$\boxed{x\text{-intercept } \left(-\tfrac{1}{3},0\right); \quad y\text{-intercept } (0,-1)}$$
**(c)**
Using the asymptotes $x=-1$ (dashed vertical line) and $y=-3$ (dashed horizontal line) together with the intercepts found above, the graph is drawn with one branch in the region $x<-1$ (lying below $y=-3$, approaching both asymptotes) and one branch in the region $x>-1$ (approaching $y=-3$ from above as $x \to \infty$, and rising steeply as $x \to -1^+$). See diagram.
$$\boxed{\text{Sketch as shown, with asymptotes } x=-1,\ y=-3}$$
QUESTION 30
7 marks
Medium
A company's total production cost, in appropriate cost units, for making $x$ items ($x>0$) is modeled by $T(x) = 10x+250$. The average cost per item is $A(x) = \dfrac{T(x)}{x}$.
(a) Show that $A(x) = 10+\dfrac{250}{x}$. [2]
(b) Find the value of $A(25)$. [2]
(c) State the equation of the horizontal asymptote of the graph of $A$, and interpret its meaning in this context. [3]
Show complete worked solution
**(a)**
Writing $A(x)$ as a single fraction split into two terms:
$$A(x) = \frac{T(x)}{x} = \frac{10x+250}{x} = \frac{10x}{x}+\frac{250}{x}$$
$$\boxed{A(x) = 10+\frac{250}{x}}$$
**(b)**
Substituting $x=25$:
$$A(25) = 10+\frac{250}{25} = 10+10$$
$$\boxed{A(25)=20 \text{ cost units per item}}$$
**(c)**
As $x \to \infty$, $\dfrac{250}{x} \to 0$, so $A(x) \to 10$.
$$\boxed{y=10}$$
This means that as the number of items produced becomes very large, the average cost per item approaches $10$ cost units -- the fixed cost of $250$ becomes negligible per item, and the average cost tends towards the marginal (variable) cost of $10$ per item.
QUESTION 31
5 marks
Medium
Solve the equation $\dfrac{3x+1}{x-2} = 4$ algebraically, and state the value of $x$ that must be excluded from the domain of $f(x)=\dfrac{3x+1}{x-2}$.
Show complete worked solution
**Solving**
Multiplying both sides by $(x-2)$:
$$3x+1 = 4(x-2)$$
Expanding the right-hand side:
$$3x+1 = 4x-8$$
Collecting terms:
$$1+8 = 4x-3x$$
$$\boxed{x=9}$$
**Domain restriction**
The denominator $x-2$ cannot equal zero, so:
$$\boxed{x=2 \text{ must be excluded from the domain}}$$
(Note $x=9 \neq 2$, so the solution found above is valid.)
QUESTION 32
6 marks
Medium
The function $f(x) = \dfrac{x+p}{x-3}$ has a $y$-intercept at $(0,-2)$.
(a) Find the value of $p$. [3]
(b) Find the coordinates of the point where the graph of $f$ crosses the $x$-axis. [3]
Show complete worked solution
**(a)**
The $y$-intercept is $f(0)$:
$$f(0) = \frac{0+p}{0-3} = \frac{p}{-3}$$
Setting this equal to $-2$:
$$\frac{p}{-3} = -2$$
$$\boxed{p=6}$$
**(b)**
With $p=6$, $f(x) = \dfrac{x+6}{x-3}$. Setting the numerator equal to zero to find the $x$-intercept:
$$x+6=0$$
$$\boxed{(-6,0)}$$
QUESTION 33
6 marks
Medium
The function $f(x) = \dfrac{a}{x-2}-1$ has a graph that passes through the point $(4,1)$.
(a) Find the value of $a$. [3]
(b) Write down the full expression for $f(x)$ and hence find the value of $f(0)$. [3]
Show complete worked solution
**(a)**
Substituting the point $(4,1)$ into $f(x) = \dfrac{a}{x-2}-1$:
$$1 = \frac{a}{4-2}-1 = \frac{a}{2}-1$$
Solving for $a$:
$$2 = \frac{a}{2}$$
$$\boxed{a=4}$$
**(b)**
$$f(x) = \frac{4}{x-2}-1$$
Substituting $x=0$:
$$f(0) = \frac{4}{0-2}-1 = -2-1$$
$$\boxed{f(0)=-3}$$
QUESTION 34
5 marks
Medium
Consider the function $f(x) = \dfrac{5x-10}{2x+6}$.
(a) Find the equations of the vertical and horizontal asymptotes of the graph of $f$. [3]
(b) Find the coordinates of the point at which the two asymptotes intersect. [2]
Show complete worked solution
**(a)**
Vertical asymptote: setting the denominator equal to zero,
$$2x+6=0 \Rightarrow x=-3$$
Horizontal asymptote: as $x \to \pm\infty$, the ratio of the leading terms dominates,
$$f(x) \to \frac{5x}{2x} = \frac{5}{2}$$
$$\boxed{x=-3 \text{ and } y=\frac{5}{2}}$$
**(b)**
The two asymptotes are the vertical line $x=-3$ and the horizontal line $y=\dfrac{5}{2}$. Their point of intersection has $x$-coordinate $-3$ and $y$-coordinate $\dfrac{5}{2}$.
$$\boxed{\left(-3,\ \frac{5}{2}\right)}$$
QUESTION 35
7 marks
Medium
A car travels a fixed distance of $180$ km. Including a fixed $0.5$ hour stop, its average speed for the whole journey is modeled by $S(t) = \dfrac{180}{t+0.5}$ km/h, where $t$ is the driving time in hours, $t>0$.
(a) Find the value of $S(2)$. [2]
(b) Find the value of $t$ for which $S(t)=60$. [3]
(c) State the equation of the horizontal asymptote of the graph of $S$, and describe what this means in the context of the journey. [2]
Show complete worked solution
**(a)**
Substituting $t=2$:
$$S(2) = \frac{180}{2+0.5} = \frac{180}{2.5}$$
$$\boxed{S(2)=72 \text{ km/h}}$$
**(b)**
Setting $S(t)=60$:
$$\frac{180}{t+0.5} = 60$$
Multiplying both sides by $(t+0.5)$:
$$180 = 60(t+0.5) = 60t+30$$
Solving for $t$:
$$150 = 60t$$
$$\boxed{t=2.5 \text{ hours}}$$
**(c)**
As $t \to \infty$, $\dfrac{180}{t+0.5} \to 0$, so $S(t) \to 0$.
$$\boxed{y=0}$$
In context, this means that as the driving time becomes extremely long, the average speed for the whole journey (including the fixed stop) approaches $0$ km/h -- a very long journey time makes the average speed negligible.
QUESTION 36
6 marks
Medium
Find the domain and range of the function $f(x) = \dfrac{6-3x}{2x+4}$.
Show complete worked solution
**Domain**
The domain excludes the value of $x$ that makes the denominator zero.
$$2x+4=0 \Rightarrow x=-2$$
$$\boxed{\text{Domain: } x \in \mathbb{R},\ x \neq -2}$$
**Range**
The range excludes the value of the horizontal asymptote. As $x \to \pm\infty$, using the ratio of the leading terms (writing the numerator as $-3x+6$):
$$f(x) \to \frac{-3x}{2x} = -\frac{3}{2}$$
$$\boxed{\text{Range: } y \in \mathbb{R},\ y \neq -\frac{3}{2}}$$
QUESTION 37
5 marks
Medium
Consider the function $f(x) = \dfrac{2x-3}{x+1}$.
(a) Show that $f(x)$ can be written in the form $f(x) = 2-\dfrac{5}{x+1}$. [3]
(b) Hence state the equations of the vertical and horizontal asymptotes of the graph of $f$. [2]
Show complete worked solution
**(a)**
Combining the right-hand side into a single fraction:
$$2-\frac{5}{x+1} = \frac{2(x+1)}{x+1}-\frac{5}{x+1} = \frac{2(x+1)-5}{x+1}$$
Expanding the numerator:
$$2(x+1)-5 = 2x+2-5 = 2x-3$$
So:
$$2-\frac{5}{x+1} = \frac{2x-3}{x+1} = f(x) \checkmark$$
$$\boxed{f(x) = 2-\frac{5}{x+1}}$$
**(b)**
Using the form $f(x) = 2-\dfrac{5}{x+1}$: the vertical asymptote occurs where $x+1=0$, and the horizontal asymptote is $y=2$ (the constant term, since $\dfrac{5}{x+1}\to 0$ as $x\to\pm\infty$).
$$\boxed{x=-1 \text{ and } y=2}$$
QUESTION 38
6 marks
Medium
The function $f(x) = \dfrac{ax+3}{2x+b}$ has a vertical asymptote at $x=-4$ and a horizontal asymptote at $y=5$.
(a) Find the values of $a$ and $b$. [4]
(b) Find the coordinates of the point where the graph of $f$ crosses the $y$-axis. [2]
Show complete worked solution
**(a)**
The vertical asymptote occurs where $2x+b=0$. Since this happens at $x=-4$:
$$2(-4)+b=0 \Rightarrow -8+b=0$$
$$\boxed{b=8}$$
The horizontal asymptote is given by the ratio of the leading coefficients:
$$\frac{a}{2} = 5$$
$$\boxed{a=10}$$
**(b)**
So $f(x) = \dfrac{10x+3}{2x+8}$. Substituting $x=0$:
$$f(0) = \frac{10(0)+3}{2(0)+8} = \frac{3}{8}$$
$$\boxed{\left(0,\frac{3}{8}\right)}$$
QUESTION 39
7 marks
Medium
Consider the function $f(x) = \dfrac{x-4}{x+2}$.
(a) Find the equations of the vertical and horizontal asymptotes of the graph of $f$. [2]
(b) Find the coordinates of the $x$-intercept and the $y$-intercept of the graph of $f$. [2]
(c) Sketch the graph of $f$, showing the asymptotes as dashed lines and labelling the intercepts found in part (b). [3]
Show complete worked solution
**(a)**
Vertical asymptote: setting the denominator equal to zero,
$$x+2=0 \Rightarrow x=-2$$
Horizontal asymptote: as $x \to \pm\infty$, the ratio of the leading terms dominates,
$$f(x) \to \frac{x}{x} = 1$$
$$\boxed{x=-2 \text{ and } y=1}$$
**(b)**
The $x$-intercept occurs where the numerator is zero:
$$x-4=0 \Rightarrow x=4$$
giving $(4,0)$.
The $y$-intercept is found by substituting $x=0$:
$$f(0) = \frac{0-4}{0+2} = \frac{-4}{2}=-2$$
giving $(0,-2)$.
$$\boxed{x\text{-intercept } (4,0); \quad y\text{-intercept } (0,-2)}$$
**(c)**
Using the asymptotes $x=-2$ and $y=1$ (both dashed) together with the intercepts $(4,0)$ and $(0,-2)$, the graph is drawn with one branch in the region $x<-2$ and one branch in the region $x>-2$, each approaching both asymptotes without crossing the vertical one. See diagram.
$$\boxed{\text{Sketch as shown, with asymptotes } x=-2,\ y=1}$$
QUESTION 40
5 marks
Medium
Consider the function $f(x) = \dfrac{7-2x}{x-3}$. Find the coordinates of the point at which the two asymptotes of the graph of $f$ intersect.
Show complete worked solution
First find the vertical asymptote, where the denominator is zero:
$$x-3=0 \Rightarrow x=3$$
Next find the horizontal asymptote. Writing the numerator as $-2x+7$, as $x \to \pm\infty$ the ratio of the leading terms dominates:
$$f(x) \to \frac{-2x}{x} = -2$$
So the asymptotes are $x=3$ and $y=-2$. Their point of intersection has these coordinates:
$$\boxed{(3,-2)}$$
QUESTION 41
5 marks
Medium
The graph of $f(x) = \dfrac{x+k}{x-2}$ passes through the origin.
(a) Find the value of $k$. [2]
(b) State the equations of the vertical and horizontal asymptotes of the graph of $f$. [3]
Show complete worked solution
**(a)**
Since the graph passes through the origin, $f(0)=0$:
$$\frac{0+k}{0-2} = 0$$
For this fraction to equal zero, the numerator must be zero:
$$\boxed{k=0}$$
**(b)**
With $k=0$, $f(x) = \dfrac{x}{x-2}$.
Vertical asymptote: setting the denominator equal to zero,
$$x-2=0 \Rightarrow x=2$$
Horizontal asymptote: as $x \to \pm\infty$, the ratio of the leading terms dominates,
$$f(x) \to \frac{x}{x} = 1$$
$$\boxed{x=2 \text{ and } y=1}$$
QUESTION 42
10 marks
Hard
Consider the function $f(x) = \dfrac{2x-1}{x+3}$.
(a) Find the equations of the vertical and horizontal asymptotes of the graph of $f$. [2]
(b) Find the coordinates of the points where the graph of $f$ crosses the $x$-axis and the $y$-axis. [3]
(c) Sketch the graph of $f$, showing the asymptotes as dashed lines and the intercepts found in part (b). [3]
(d) Write down the range of $f$. [2]
Show complete worked solution
**(a)**
Vertical asymptote: setting the denominator equal to zero,
$$x+3=0 \Rightarrow x=-3$$
Horizontal asymptote: as $x \to \pm\infty$, the ratio of the leading terms dominates,
$$f(x) \to \frac{2x}{x} = 2$$
$$\boxed{x=-3 \text{ and } y=2}$$
**(b)**
The $x$-intercept occurs where the numerator is zero:
$$2x-1=0 \Rightarrow x=\frac{1}{2}$$
giving $\left(\dfrac{1}{2},0\right)$.
The $y$-intercept is found by substituting $x=0$:
$$f(0) = \frac{2(0)-1}{0+3} = \frac{-1}{3}$$
giving $\left(0,-\dfrac{1}{3}\right)$.
$$\boxed{x\text{-intercept } \left(\tfrac{1}{2},0\right); \quad y\text{-intercept } \left(0,-\tfrac{1}{3}\right)}$$
**(c)**
Using the asymptotes $x=-3$ and $y=2$ (drawn dashed) together with the intercepts from part (b), the graph is sketched with one branch in the region $x<-3$ (lying above $y=2$ and rising steeply as $x \to -3^-$) and one branch in the region $x>-3$ (passing through both intercepts, lying below $y=2$, and falling steeply as $x \to -3^+$). See diagram.
**(d)**
Since $f$ can take every real value except the horizontal asymptote value:
$$\boxed{\text{Range: } y \in \mathbb{R},\ y \neq 2}$$
QUESTION 43
9 marks
Hard
The function $f(x) = \dfrac{ax+b}{x-2}$ has a graph that crosses the $y$-axis at $(0,-3)$ and passes through the point $(4,1)$.
(a) Find the values of $a$ and $b$. [4]
(b) Find the equations of the vertical and horizontal asymptotes of the graph of $f$. [2]
(c) Find the coordinates of the point where the two asymptotes intersect. [3]
Show complete worked solution
**(a)**
Using the $y$-intercept $(0,-3)$: substituting $x=0$ into $f(x)=\dfrac{ax+b}{x-2}$,
$$-3 = \frac{a(0)+b}{0-2} = \frac{b}{-2}$$
Solving for $b$:
$$b = 6$$
Using the point $(4,1)$: substituting $x=4$,
$$1 = \frac{4a+6}{4-2} = \frac{4a+6}{2}$$
Solving for $a$:
$$2 = 4a+6$$
$$4a=-4$$
$$a=-1$$
$$\boxed{a=-1,\ b=6}$$
**(b)**
So $f(x) = \dfrac{-x+6}{x-2}$.
Vertical asymptote: setting the denominator equal to zero,
$$x-2=0 \Rightarrow x=2$$
Horizontal asymptote: as $x \to \pm\infty$, the ratio of the leading terms dominates,
$$f(x) \to \frac{-x}{x} = -1$$
$$\boxed{x=2 \text{ and } y=-1}$$
**(c)**
The two asymptotes $x=2$ and $y=-1$ intersect at:
$$\boxed{(2,-1)}$$
QUESTION 44
7 marks
Hard
Solve the equation $\dfrac{x+6}{x-2} = x+1$ algebraically.
(a) Show that this equation can be rearranged to give $x^2-2x-8=0$. [3]
(b) Solve this quadratic equation by factorisation. [2]
(c) State whether either solution must be rejected, giving a reason. [2]
Show complete worked solution
**(a)**
Multiplying both sides of $\dfrac{x+6}{x-2} = x+1$ by $(x-2)$:
$$x+6 = (x+1)(x-2)$$
Expanding the right-hand side:
$$(x+1)(x-2) = x^2-2x+x-2 = x^2-x-2$$
So:
$$x+6 = x^2-x-2$$
Rearranging so that all terms are on one side:
$$0 = x^2-x-2-x-6$$
$$\boxed{x^2-2x-8=0}$$
**(b)**
Factorising: we need two numbers that multiply to $-8$ and add to $-2$, which are $-4$ and $2$.
$$x^2-2x-8 = (x-4)(x+2)$$
Setting each factor equal to zero:
$$x-4=0 \Rightarrow x=4 \qquad \text{or} \qquad x+2=0 \Rightarrow x=-2$$
$$\boxed{x=4 \text{ or } x=-2}$$
**(c)**
The original equation is undefined only when $x-2=0$, i.e. at $x=2$. Since neither $x=4$ nor $x=-2$ equals $2$, both solutions are valid.
$$\boxed{\text{Neither solution is rejected; both } x=4 \text{ and } x=-2 \text{ are valid}}$$
QUESTION 45
9 marks
Hard
A company's total production cost, in appropriate cost units, for making $x$ units ($x>0$) is modeled by $T(x) = 400+15x$. The average cost per unit is $A(x) = \dfrac{T(x)}{x}$.
(a) Show that $A(x) = 15+\dfrac{400}{x}$. [2]
(b) State the equations of the vertical and horizontal asymptotes of the graph of $A$. [2]
(c) Find the average cost per unit when $x=50$. [2]
(d) Interpret, in the context of this model, the meaning of the horizontal asymptote found in part (b). [2]
(e) Explain briefly why, in this context, the domain of $A$ is restricted to $x>0$. [1]
Show complete worked solution
**(a)**
Writing $A(x)$ as a single fraction split into two terms:
$$A(x) = \frac{T(x)}{x} = \frac{400+15x}{x} = \frac{15x}{x}+\frac{400}{x}$$
$$\boxed{A(x) = 15+\frac{400}{x}}$$
**(b)**
Vertical asymptote: the denominator of $\dfrac{400}{x}$ is zero at $x=0$, so:
$$\boxed{x=0}$$
Horizontal asymptote: as $x \to \infty$, $\dfrac{400}{x} \to 0$, so $A(x) \to 15$:
$$\boxed{y=15}$$
**(c)**
Substituting $x=50$:
$$A(50) = 15+\frac{400}{50} = 15+8$$
$$\boxed{A(50)=23 \text{ cost units per unit}}$$
**(d)**
The horizontal asymptote $y=15$ means that as the number of units produced, $x$, becomes very large, the average cost per unit approaches $15$ cost units. This is the marginal (variable) cost per unit -- as production increases, the fixed cost of $400$ is spread across more units, so its contribution to the average cost per unit becomes negligible.
**(e)**
$$\boxed{\text{The model represents the number of units produced, which must be a positive quantity, so } x>0}$$
QUESTION 46
8 marks
Hard
The concentration of a chemical in a mixing tank, in mg per litre, is modeled by $C(t) = \dfrac{3t+20}{t+5}$, where $t \geq 0$ is the time in minutes since mixing began.
(a) Find $C(0)$ and interpret its meaning in context. [2]
(b) Find the equation of the horizontal asymptote of the graph of $C$, and interpret its meaning as $t \to \infty$. [3]
(c) Find the time $t$ at which $C(t) = 3.5$. [3]
Show complete worked solution
**(a)**
Substituting $t=0$:
$$C(0) = \frac{3(0)+20}{0+5} = \frac{20}{5}$$
$$\boxed{C(0)=4 \text{ mg/L}}$$
This represents the initial concentration of the chemical in the tank, before any mixing has taken place.
**(b)**
As $t \to \infty$, dividing numerator and denominator by the dominant behaviour (or using the ratio of the leading terms):
$$C(t) \to \frac{3t}{t} = 3$$
$$\boxed{y=3}$$
This means that as time passes, the concentration in the tank settles towards, but never quite reaches, $3$ mg/L -- the long-term (steady-state) concentration.
**(c)**
Setting $C(t) = 3.5$:
$$\frac{3t+20}{t+5} = 3.5$$
Multiplying both sides by $(t+5)$:
$$3t+20 = 3.5(t+5) = 3.5t+17.5$$
Collecting terms:
$$20-17.5 = 3.5t-3t$$
$$2.5 = 0.5t$$
$$\boxed{t=5 \text{ minutes}}$$
QUESTION 47
9 marks
Hard
The function $f(x) = \dfrac{px+q}{x-4}$ has a graph that passes through the points $(0,-3)$ and $(2,1)$.
(a) By substituting each point into $f$, form two equations in $p$ and $q$, and hence find the values of $p$ and $q$. [5]
(b) Find the equations of the vertical and horizontal asymptotes of the graph of $f$. [2]
(c) Write down the range of $f$. [2]
Show complete worked solution
**(a)**
Substituting $(0,-3)$ into $f(x) = \dfrac{px+q}{x-4}$:
$$-3 = \frac{p(0)+q}{0-4} = \frac{q}{-4}$$
so
$$q = 12$$
Substituting $(2,1)$:
$$1 = \frac{2p+q}{2-4} = \frac{2p+q}{-2}$$
so
$$2p+q = -2$$
Substituting $q=12$:
$$2p+12=-2 \Rightarrow 2p=-14$$
$$\boxed{p=-7,\ q=12}$$
**(b)**
So $f(x) = \dfrac{-7x+12}{x-4}$.
Vertical asymptote: setting the denominator equal to zero,
$$x-4=0 \Rightarrow x=4$$
Horizontal asymptote: as $x \to \pm\infty$, the ratio of the leading terms dominates,
$$f(x) \to \frac{-7x}{x} = -7$$
$$\boxed{x=4 \text{ and } y=-7}$$
**(c)**
Since $f$ takes every real value except the horizontal asymptote value:
$$\boxed{\text{Range: } y \in \mathbb{R},\ y \neq -7}$$
QUESTION 48
9 marks
Hard
Consider the function $f(x) = \dfrac{2x+4}{x-1}$.
(a) Find the equations of the vertical and horizontal asymptotes of the graph of $f$. [2]
(b) Find the coordinates of the $x$-intercept and the $y$-intercept of the graph of $f$. [2]
(c) Sketch the graph of $f$, showing the asymptotes as dashed lines and labelling the intercepts. [3]
(d) Write down the domain and range of $f$. [2]
Show complete worked solution
**(a)**
Vertical asymptote: setting the denominator equal to zero,
$$x-1=0 \Rightarrow x=1$$
Horizontal asymptote: as $x \to \pm\infty$, the ratio of the leading terms dominates,
$$f(x) \to \frac{2x}{x} = 2$$
$$\boxed{x=1 \text{ and } y=2}$$
**(b)**
The $x$-intercept occurs where the numerator is zero:
$$2x+4=0 \Rightarrow x=-2$$
giving $(-2,0)$.
The $y$-intercept is found by substituting $x=0$:
$$f(0) = \frac{2(0)+4}{0-1} = \frac{4}{-1} = -4$$
giving $(0,-4)$.
$$\boxed{x\text{-intercept } (-2,0); \quad y\text{-intercept } (0,-4)}$$
**(c)**
Using the asymptotes $x=1$ and $y=2$ (both dashed) together with the intercepts $(-2,0)$ and $(0,-4)$, the graph is sketched with one branch in the region $x<1$ (passing through both intercepts, lying below $y=2$, and falling steeply as $x \to 1^-$) and one branch in the region $x>1$ (lying above $y=2$, and rising steeply as $x \to 1^+$). See diagram.
**(d)**
$$\boxed{\text{Domain: } x \in \mathbb{R},\ x \neq 1; \quad \text{Range: } y \in \mathbb{R},\ y \neq 2}$$
QUESTION 49
10 marks
Hard
Consider the function $f(x) = \dfrac{6-2x}{x+3}$.
(a) Show that $f(x)$ can be written in the form $f(x) = -2+\dfrac{12}{x+3}$. [3]
(b) Hence write down the equations of the vertical and horizontal asymptotes of the graph of $f$. [2]
(c) Find the coordinates of the point where the two asymptotes intersect. [2]
(d) Sketch the graph of $f$, showing the asymptotes as dashed lines. [3]
Show complete worked solution
**(a)**
Combining the right-hand side into a single fraction:
$$-2+\frac{12}{x+3} = \frac{-2(x+3)}{x+3}+\frac{12}{x+3} = \frac{-2(x+3)+12}{x+3}$$
Expanding the numerator:
$$-2(x+3)+12 = -2x-6+12 = -2x+6 = 6-2x$$
So:
$$-2+\frac{12}{x+3} = \frac{6-2x}{x+3} = f(x) \checkmark$$
$$\boxed{f(x) = -2+\frac{12}{x+3}}$$
**(b)**
Using the form $f(x) = -2+\dfrac{12}{x+3}$: the vertical asymptote occurs where $x+3=0$, and the horizontal asymptote is the constant term $y=-2$ (since $\dfrac{12}{x+3} \to 0$ as $x \to \pm\infty$).
$$\boxed{x=-3 \text{ and } y=-2}$$
**(c)**
The two asymptotes $x=-3$ and $y=-2$ intersect at:
$$\boxed{(-3,-2)}$$
**(d)**
Using the asymptotes from part (b), together with the $y$-intercept $f(0)=\dfrac{6}{3}=2$ and $x$-intercept where $6-2x=0 \Rightarrow x=3$, the graph is sketched with one branch in the region $x<-3$ and one branch in the region $x>-3$, each approaching both asymptotes. See diagram.
$$\boxed{\text{Sketch as shown, with asymptotes } x=-3,\ y=-2}$$
QUESTION 50
10 marks
Hard
The function $f(x) = \dfrac{ax+8}{x+b}$ has a vertical asymptote at $x=-2$ and its graph passes through the point $(1,2)$.
(a) Find the value of $b$. [2]
(b) Find the value of $a$. [3]
(c) Find the $x$-coordinates of the points where the graph of $f$ intersects the line $y=x-1$. [5]
Show complete worked solution
**(a)**
The vertical asymptote occurs where $x+b=0$. Since this happens at $x=-2$:
$$-2+b=0$$
$$\boxed{b=2}$$
**(b)**
So $f(x) = \dfrac{ax+8}{x+2}$. Substituting the point $(1,2)$:
$$2 = \frac{a(1)+8}{1+2} = \frac{a+8}{3}$$
Solving for $a$:
$$6 = a+8$$
$$\boxed{a=-2}$$
**(c)**
So $f(x) = \dfrac{-2x+8}{x+2}$. Setting $f(x)=x-1$:
$$\frac{-2x+8}{x+2} = x-1$$
Multiplying both sides by $(x+2)$:
$$-2x+8 = (x-1)(x+2)$$
Expanding the right-hand side:
$$(x-1)(x+2) = x^2+2x-x-2 = x^2+x-2$$
So:
$$-2x+8 = x^2+x-2$$
Rearranging so that all terms are on one side:
$$0 = x^2+x-2+2x-8$$
$$x^2+3x-10=0$$
Factorising: we need two numbers that multiply to $-10$ and add to $3$, which are $5$ and $-2$.
$$x^2+3x-10 = (x+5)(x-2)$$
Setting each factor equal to zero:
$$x+5=0 \Rightarrow x=-5 \qquad \text{or} \qquad x-2=0 \Rightarrow x=2$$
Since neither value equals $-2$ (the excluded domain value), both are valid.
$$\boxed{x=-5 \text{ or } x=2}$$
Exponential and Logarithmic Functions 50 questions
QUESTION 1
4 marks
Easy
Consider the function $f(x) = 2^x - 5$.
(a) Write down the equation of the horizontal asymptote of the graph of $f$.
(b) Find the $y$-intercept of the graph of $f$.
(c) Write down the range of $f$.
Show complete worked solution
**(a)**
As $x \to -\infty$, $2^x \to 0$, so:
$$f(x) \to -5$$
$$\boxed{y=-5}$$
**(b)**
Substituting $x=0$:
$$f(0) = 2^0 - 5 = 1-5=-4$$
$$\boxed{(0,-4)}$$
**(c)**
Since $2^x>0$ for all real $x$:
$$f(x)=2^x-5>-5 \text{ for all } x$$
$$\boxed{f(x)>-5}$$
QUESTION 2
3 marks
Easy
Solve for $x$: $\log_2(x+3) = 4$.
Show complete worked solution
Using the method of converting from logarithmic to exponential form, $\log_a(N) = b \iff N = a^b$:
$$\log_2(x+3)=4 \Rightarrow x+3 = 2^4 = 16$$
Solving for $x$:
$$x = 13$$
Checking the domain: we need $x+3>0$, i.e. $x>-3$; $x=13$ satisfies this.
$$\boxed{x=13}$$
QUESTION 3
6 marks
Medium
Solve the equation $3^{2x} - 10(3^x) + 9 = 0$.
Show complete worked solution
Using the substitution $y=3^x$ (noting $y>0$) to reduce the equation to a quadratic in $y$:
$$y^2 - 10y+9=0$$
Factorising:
$$(y-1)(y-9)=0$$
$$y=1 \text{ or } y=9$$
Case 1: solving $3^x=1=3^0$:
$$x=0$$
Case 2: solving $3^x=9=3^2$:
$$x=2$$
$$\boxed{x=0 \text{ or } x=2}$$
QUESTION 4
6 marks
Medium
The number of bacteria in a culture is modelled by $P(t) = 200e^{0.15t}$, where $t$ is the time in hours after the culture is set up.
(a) Find the initial number of bacteria.
(b) Find the number of bacteria after $10$ hours, correct to 3 significant figures.
(c) Find the time taken for the population to reach $1000$, correct to one decimal place.
Show complete worked solution
**(a)**
Substituting $t=0$:
$$P(0) = 200e^0 = 200$$
$$\boxed{P(0)=200}$$
**(b)**
Substituting $t=10$:
$$P(10) = 200e^{0.15(10)} = 200e^{1.5}$$
Evaluating:
$$P(10) = 896.338\ldots$$
$$\boxed{P(10)\approx896}$$
**(c)**
Setting $P(t)=1000$:
$$200e^{0.15t}=1000$$
Dividing both sides by 200:
$$e^{0.15t}=5$$
Taking $\ln$ of both sides:
$$0.15t = \ln5$$
Solving for $t$:
$$t = \frac{\ln5}{0.15} = 10.7296\ldots$$
$$\boxed{t\approx10.7 \text{ hours}}$$
QUESTION 5
6 marks
Hard
Solve the equation $\log_2(x+1) + \log_2(x-1) = 3$.
Show complete worked solution
Using the law $\log_2A+\log_2B=\log_2(AB)$ to combine the logarithms, then converting to exponential form. First, stating the domain:
$$\text{Domain requires } x+1>0 \text{ and } x-1>0, \text{ i.e. } x>1$$
Combining the logarithms:
$$\log_2[(x+1)(x-1)] = 3$$
Converting to exponential form:
$$(x+1)(x-1) = 2^3 = 8$$
Simplifying the left-hand side using the difference of two squares:
$$x^2-1=8$$
$$x^2=9$$
Taking square roots:
$$x=3 \text{ or } x=-3$$
Since the domain requires $x>1$, reject $x=-3$.
$$\boxed{x=3}$$
QUESTION 6
4 marks
Easy
Consider the function $f(x) = 3^x + 2$.
(a) Write down the equation of the horizontal asymptote of the graph of $f$. [1]
(b) Find the $y$-intercept of the graph of $f$. [2]
(c) Write down the range of $f$. [1]
Show complete worked solution
**(a)**
As $x \to -\infty$, $3^x \to 0$, so:
$$f(x) \to 2$$
$$\boxed{y=2}$$
**(b)**
Substituting $x=0$:
$$f(0) = 3^0+2 = 1+2 = 3$$
$$\boxed{(0,3)}$$
**(c)**
Since $3^x>0$ for all real $x$:
$$f(x) = 3^x+2 > 2 \text{ for all } x$$
$$\boxed{f(x)>2}$$
QUESTION 7
3 marks
Easy
The diagram shows the graph of $y=2^x$.
(a) Write down the coordinates of the point where the curve crosses the $y$-axis. [1]
(b) Write down the equation of the horizontal asymptote of the graph. [1]
(c) Write down the domain of the function $y=2^x$. [1]
Show complete worked solution
**(a)**
Reading from the graph, the curve crosses the $y$-axis at:
$$\boxed{(0,1)}$$
**(b)**
As $x\to-\infty$, the curve approaches but never reaches the line $y=0$:
$$\boxed{y=0}$$
**(c)**
The function $2^x$ is defined for every real value of $x$ (there is no restriction on $x$):
$$\boxed{x\in\mathbb{R}}$$
QUESTION 8
3 marks
Easy
Consider the function $g(x) = \log_3 x$.
(a) Write down the domain of $g$. [1]
(b) Write down the range of $g$. [1]
(c) Write down the equation of the vertical asymptote of the graph of $g$. [1]
Show complete worked solution
**(a)**
Logarithms are only defined for a positive argument, so:
$$\boxed{x>0}$$
**(b)**
As $x$ ranges over all positive reals, $\log_3 x$ takes every real value:
$$\boxed{g(x)\in\mathbb{R}}$$
**(c)**
As $x\to0^{+}$, $\log_3x\to-\infty$, so the graph has a vertical asymptote:
$$\boxed{x=0}$$
QUESTION 9
4 marks
Easy
Consider the function $f(x) = e^x - 4$.
(a) Write down the equation of the horizontal asymptote of the graph of $f$. [1]
(b) Find the $y$-intercept of the graph of $f$. [1]
(c) Find the exact $x$-intercept of the graph of $f$, and hence give its value correct to 3 significant figures. [2]
Show complete worked solution
**(a)**
As $x\to-\infty$, $e^x\to0$, so:
$$\boxed{y=-4}$$
**(b)**
Substituting $x=0$:
$$f(0) = e^0-4 = 1-4=-3$$
$$\boxed{(0,-3)}$$
**(c)**
Setting $f(x)=0$:
$$e^x-4=0 \implies e^x=4$$
Taking $\ln$ of both sides:
$$x=\ln4 = 1.3863\ldots$$
$$\boxed{x=\ln4\approx1.39}$$
QUESTION 10
5 marks
Easy
Consider the function $h(x) = \ln(x-2)$.
(a) Write down the domain of $h$. [1]
(b) Write down the equation of the vertical asymptote of the graph of $h$. [1]
(c) Find the exact coordinates of the $x$-intercept of the graph of $h$. [3]
Show complete worked solution
**(a)**
The argument of the logarithm must be positive:
$$x-2>0 \implies x>2$$
$$\boxed{x>2}$$
**(b)**
As $x\to2^{+}$, $\ln(x-2)\to-\infty$, so:
$$\boxed{x=2}$$
**(c)**
Setting $h(x)=0$:
$$\ln(x-2)=0$$
Converting to exponential form, $\ln(N)=0 \iff N=e^0=1$:
$$x-2=1 \implies x=3$$
$$\boxed{(3,0)}$$
QUESTION 11
3 marks
Easy
Describe the single transformation that maps the graph of $y=2^x$ onto the graph of $y=2^{x+3}$, and state the equation of the horizontal asymptote of each graph.
Show complete worked solution
Rewriting $y=2^{x+3}$ as $y=2^{(x-(-3))}$, this is of the form $y=f(x-h)$ with $h=-3$, so this is a horizontal translation.
$$\boxed{\text{Horizontal translation of 3 units to the left, i.e. by the vector } \begin{pmatrix}-3\\0\end{pmatrix}}$$
A horizontal translation does not change the value that the function approaches as $x\to-\infty$, since $2^{x+3}\to0$ just as $2^x\to0$. So the asymptote is unchanged:
$$\boxed{y=0 \text{ for both graphs}}$$
QUESTION 12
3 marks
Easy
The diagram shows the graph of $y=\log_2 x$. The point $(8,k)$ lies on the graph, as shown.
(a) Use the graph to write down an estimate for the value of $k$. [1]
(b) Find the exact value of $k$ algebraically. [2]
Show complete worked solution
**(a)**
Reading from the graph at $x=8$, the curve height is approximately $k\approx3$.
**(b)**
Since $(8,k)$ lies on the graph, $k=\log_2 8$. Using the definition of a logarithm, $\log_2 8 = y \iff 2^y=8$:
$$2^y=8=2^3 \implies y=3$$
$$\boxed{k=3}$$
QUESTION 13
3 marks
Easy
Consider the function $f(x) = -e^x + 5$.
(a) Write down the equation of the horizontal asymptote of the graph of $f$. [1]
(b) Write down the range of $f$. [2]
Show complete worked solution
**(a)**
As $x\to-\infty$, $e^x\to0$, so $f(x)\to-0+5=5$:
$$\boxed{y=5}$$
**(b)**
Since $e^x>0$ for all real $x$, it follows that $-e^x<0$, so:
$$f(x) = -e^x+5 < 5 \text{ for all } x$$
$$\boxed{f(x)<5}$$
QUESTION 14
3 marks
Easy
The diagram shows two curves, $A$ and $B$. One is the graph of $y=2^x$ and the other is the graph of $y=\left(\dfrac{1}{2}\right)^x$.
(a) Determine which curve, $A$ or $B$, represents $y=\left(\dfrac12\right)^x$. [1]
(b) Write down the coordinates of the point where the two curves intersect. [1]
(c) Write down the equation of the horizontal asymptote shared by both curves. [1]
Show complete worked solution
**(a)**
Since $0<\frac12<1$, the function $\left(\frac12\right)^x$ is decreasing, whereas $2^x$ is increasing. Curve $B$ is decreasing, so:
$$\boxed{B \text{ represents } y=\left(\tfrac12\right)^x}$$
**(b)**
Both functions equal 1 when $x=0$, since $2^0=1$ and $\left(\frac12\right)^0=1$:
$$\boxed{(0,1)}$$
**(c)**
Both $2^x\to0$ and $\left(\frac12\right)^x\to0$ in one of the tails, so both curves share:
$$\boxed{y=0}$$
QUESTION 15
4 marks
Easy
Consider the function $f(x) = \log_5(x+4)$.
(a) Write down the equation of the vertical asymptote of the graph of $f$. [1]
(b) Find the exact $x$-intercept of the graph of $f$. [3]
Show complete worked solution
**(a)**
The vertical asymptote occurs where the argument of the logarithm is zero, i.e. $x+4=0$:
$$\boxed{x=-4}$$
**(b)**
Setting $f(x)=0$:
$$\log_5(x+4)=0$$
Converting to exponential form, $\log_5(N)=0 \iff N=5^0=1$:
$$x+4=1 \implies x=-3$$
$$\boxed{(-3,0)}$$
QUESTION 16
4 marks
Easy
Consider the function $f(x) = 10^x - 1$.
(a) Write down the equation of the horizontal asymptote of the graph of $f$. [1]
(b) Write down the range of $f$. [1]
(c) Find the $y$-intercept of the graph of $f$, and state what is special about this point. [2]
Show complete worked solution
**(a)**
As $x\to-\infty$, $10^x\to0$, so:
$$\boxed{y=-1}$$
**(b)**
Since $10^x>0$ for all $x$, $f(x)=10^x-1>-1$:
$$\boxed{f(x)>-1}$$
**(c)**
Substituting $x=0$:
$$f(0) = 10^0-1 = 1-1 = 0$$
So the $y$-intercept is $(0,0)$. Since $f(0)=0$, this point is also the $x$-intercept:
$$\boxed{(0,0) \text{ is both the } x\text{- and } y\text{-intercept}}$$
QUESTION 17
3 marks
Easy
The diagram shows the graph of $f(x)=e^x+k$, for some constant $k$. The horizontal asymptote of the graph is $y=3$, as shown.
(a) Write down the value of $k$. [1]
(b) Hence find the exact $y$-intercept of the graph. [2]
Show complete worked solution
**(a)**
As $x\to-\infty$, $e^x\to0$, so $f(x)\to k$. Since the asymptote is $y=3$:
$$\boxed{k=3}$$
**(b)**
Substituting $x=0$ into $f(x)=e^x+3$:
$$f(0) = e^0+3 = 1+3=4$$
$$\boxed{(0,4)}$$
QUESTION 18
3 marks
Easy
Consider the function $g(x) = \ln(5-x)$.
(a) Write down the domain of $g$. [1]
(b) Write down the equation of the vertical asymptote of the graph of $g$. [1]
(c) Find the exact $y$-intercept of the graph of $g$. [1]
Show complete worked solution
**(a)**
The argument of the logarithm must be positive:
$$5-x>0 \implies x<5$$
$$\boxed{x<5}$$
**(b)**
As $x\to5^{-}$, $5-x\to0^{+}$ and $\ln(5-x)\to-\infty$, so:
$$\boxed{x=5}$$
**(c)**
Substituting $x=0$:
$$g(0) = \ln(5-0) = \ln5$$
$$\boxed{(0,\ln5)}$$
QUESTION 19
5 marks
Easy
Consider the function $h(x) = \log_4(x+7)$.
(a) Write down the domain of $h$. [1]
(b) Write down the equation of the vertical asymptote of the graph of $h$. [1]
(c) Find the $y$-intercept of the graph of $h$, correct to 3 significant figures. [3]
Show complete worked solution
**(a)**
The argument must be positive:
$$x+7>0 \implies x>-7$$
$$\boxed{x>-7}$$
**(b)**
$$\boxed{x=-7}$$
**(c)**
Substituting $x=0$:
$$h(0) = \log_4(7)$$
Using the change of base formula, $\log_4 7 = \dfrac{\ln7}{\ln4}$:
$$h(0) = \frac{1.9459\ldots}{1.3863\ldots} = 1.4036\ldots$$
$$\boxed{(0,1.40)}$$
QUESTION 20
4 marks
Easy
Describe the single transformation that maps the graph of $y=e^x$ onto the graph of $y=e^{-x}$, and state the horizontal asymptote and $y$-intercept of $y=e^{-x}$.
Show complete worked solution
Replacing $x$ with $-x$ reflects the graph in the $y$-axis:
$$\boxed{\text{Reflection in the } y\text{-axis}}$$
A reflection in the $y$-axis does not affect points on the $y$-axis or the horizontal asymptote, since as $x\to+\infty$, $-x\to-\infty$ and $e^{-x}\to0$ in exactly the way $e^x\to0$ as $x\to-\infty$:
$$\boxed{y=0}$$
The $y$-intercept is unchanged, since $e^{-0}=e^0=1$:
$$\boxed{(0,1)}$$
QUESTION 21
4 marks
Easy
The diagram shows the graph of $y=a^x$, for some constant $a>0$. The graph passes through the point $(2,9)$, as shown.
(a) Find the value of $a$. [2]
(b) Write down the $y$-intercept of the graph. [1]
(c) Write down the equation of the horizontal asymptote of the graph. [1]
Show complete worked solution
**(a)**
Since $(2,9)$ lies on the graph:
$$a^2=9$$
Since $a>0$:
$$\boxed{a=3}$$
**(b)**
Since $a^0=1$ for any $a>0$:
$$\boxed{(0,1)}$$
**(c)**
$$\boxed{y=0}$$
QUESTION 22
4 marks
Easy
Consider the function $f(x) = 3 - 2^{-x}$.
(a) Find the equation of the horizontal asymptote of the graph of $f$, giving a reason for your answer. [2]
(b) Write down the range of $f$. [2]
Show complete worked solution
**(a)**
As $x\to+\infty$, $2^{-x}\to0$, so:
$$f(x) \to 3-0=3$$
$$\boxed{y=3}$$
**(b)**
Since $2^{-x}>0$ for all real $x$, $-2^{-x}<0$, so:
$$f(x) = 3-2^{-x}<3 \text{ for all } x$$
$$\boxed{f(x)<3}$$
QUESTION 23
4 marks
Easy
The diagram shows the graphs of $y=\log_2 x$ and $y=\log_2 x - 3$.
(a) Describe the transformation that maps $y=\log_2 x$ onto $y=\log_2 x-3$. [1]
(b) Write down the equation of the vertical asymptote of $y=\log_2 x - 3$. [1]
(c) Find the exact $x$-intercept of $y=\log_2 x - 3$. [2]
Show complete worked solution
**(a)**
$$\boxed{\text{Vertical translation of 3 units down, i.e. by the vector } \begin{pmatrix}0\\-3\end{pmatrix}}$$
**(b)**
A vertical translation does not affect the vertical asymptote, so it remains:
$$\boxed{x=0}$$
**(c)**
Setting $y=0$:
$$\log_2 x - 3=0 \implies \log_2 x = 3$$
$$x=2^3=8$$
$$\boxed{(8,0)}$$
QUESTION 24
5 marks
Medium
The diagram shows the graph of $y=a\cdot2^x+b$, where $a,b\in\mathbb{R}$. The horizontal asymptote is $y=-3$, and the graph passes through the point $(0,1)$, as shown.
(a) Write down the value of $b$. [1]
(b) Find the value of $a$. [2]
(c) Hence find the value of $y$ when $x=2$. [2]
Show complete worked solution
**(a)**
As $x\to-\infty$, $2^x\to0$, so $y\to b$. Since the asymptote is $y=-3$:
$$\boxed{b=-3}$$
**(b)**
Substituting the point $(0,1)$ into $y=a\cdot2^x-3$:
$$1 = a\cdot2^0-3 = a-3$$
$$\boxed{a=4}$$
**(c)**
The equation is $y=4\cdot2^x-3$. Substituting $x=2$:
$$y = 4\cdot2^2-3 = 4(4)-3=16-3$$
$$\boxed{y=13}$$
QUESTION 25
5 marks
Medium
Consider the function $f(x) = e^{x-1}+2$.
(a) Describe the transformation(s) that map the graph of $y=e^x$ onto the graph of $f$. [2]
(b) Write down the equation of the horizontal asymptote of the graph of $f$. [1]
(c) Find the exact $y$-intercept of $f$, and give its value correct to 3 significant figures. [2]
Show complete worked solution
**(a)**
Writing $f(x)=e^{(x-1)}+2$, this is of the form $y=e^{x-h}+k$ with $h=1$, $k=2$:
$$\boxed{\text{Translation by the vector } \begin{pmatrix}1\\2\end{pmatrix} \text{ (1 unit right, 2 units up)}}$$
**(b)**
Since $e^{x-1}\to0$ as $x\to-\infty$, $f(x)\to0+2=2$:
$$\boxed{y=2}$$
**(c)**
Substituting $x=0$:
$$f(0) = e^{-1}+2$$
$$f(0) = 0.36788\ldots+2 = 2.36788\ldots$$
$$\boxed{(0,\,e^{-1}+2) \approx (0,2.37)}$$
QUESTION 26
6 marks
Medium
Consider the functions $f(x)=3^x$ and $g(x)=\log_3 x$.
(a) State the geometric relationship between the graphs of $f$ and $g$. [1]
(b) Write down the equation of the line about which the two graphs are symmetric. [1]
(c) State the equation of the horizontal asymptote of the graph of $f$ and the equation of the vertical asymptote of the graph of $g$, and explain how these two asymptotes are related by the symmetry in part (b). [3]
(d) The point $(0,1)$ lies on the graph of $f$. Write down the coordinates of the corresponding point on the graph of $g$. [1]
Show complete worked solution
**(a)**
$$\boxed{g \text{ is the inverse function of } f, \text{ so the graphs are reflections of each other}}$$
**(b)**
$$\boxed{y=x}$$
**(c)**
The horizontal asymptote of $f$ is $y=0$. The vertical asymptote of $g$ is $x=0$.
Reflecting the horizontal line $y=0$ in the line $y=x$ swaps the roles of $x$ and $y$, turning the horizontal line $y=0$ into the vertical line $x=0$ -- exactly the vertical asymptote of $g$:
$$\boxed{y=0 \text{ (for } f\text{) reflects to } x=0 \text{ (for } g\text{)}}$$
**(d)**
Reflecting $(0,1)$ in $y=x$ swaps the coordinates:
$$\boxed{(1,0)}$$
QUESTION 27
5 marks
Medium
Consider the function $f(x) = \ln x$. The graph of $f$ is transformed to give the graph of $y=\ln x+2$.
(a) Describe the transformation that maps the graph of $f$ onto the graph of $y=\ln x+2$. [1]
(b) State whether the equation of the vertical asymptote changes, giving the new equation. [1]
(c) Find the exact $x$-intercept of $y=\ln x+2$. [3]
Show complete worked solution
**(a)**
$$\boxed{\text{Vertical translation of 2 units up, i.e. by the vector } \begin{pmatrix}0\\2\end{pmatrix}}$$
**(b)**
A vertical translation does not affect a vertical asymptote:
$$\boxed{x=0 \text{ (unchanged)}}$$
**(c)**
Setting $y=0$:
$$\ln x+2=0 \implies \ln x=-2$$
Converting to exponential form:
$$x=e^{-2}$$
$$\boxed{(e^{-2},0) \approx (0.135,0)}$$
QUESTION 28
8 marks
Medium
The temperature, $N$ degrees Celsius, of a cooling liquid $t$ minutes after it starts being timed is modelled by $N(t) = N_0e^{-kt}+c$, where $N_0$, $k$ and $c$ are positive constants. The diagram shows the graph of $N$ against $t$. The horizontal asymptote is $N=20$, and the graph passes through $(0,100)$ and $(4,60)$.
(a) Write down the value of $c$. [1]
(b) Find the value of $N_0$. [2]
(c) Find the value of $k$, correct to 3 significant figures. [3]
(d) Hence find the temperature after 10 minutes, correct to 3 significant figures. [2]
Show complete worked solution
**(a)**
As $t\to\infty$, $e^{-kt}\to0$, so $N(t)\to c$. Since the asymptote is $N=20$:
$$\boxed{c=20}$$
**(b)**
Substituting $t=0$:
$$N(0) = N_0e^0+20 = N_0+20$$
Since $N(0)=100$:
$$N_0+20=100$$
$$\boxed{N_0=80}$$
**(c)**
Substituting $t=4$, $N(4)=60$ into $N(t)=80e^{-kt}+20$:
$$80e^{-4k}+20=60$$
$$80e^{-4k}=40$$
$$e^{-4k}=0.5$$
Taking $\ln$ of both sides:
$$-4k=\ln0.5$$
$$k = \frac{\ln2}{4} = 0.173287\ldots$$
$$\boxed{k\approx0.173}$$
**(d)**
Substituting $t=10$ into $N(t)=80e^{-0.173287t}+20$:
$$N(10) = 80e^{-1.73287}+20$$
$$N(10) = 80(0.176787\ldots)+20 = 14.1430\ldots+20$$
$$\boxed{N(10)\approx34.1^\circ\text{C}}$$
QUESTION 29
5 marks
Medium
Consider the function $f(x) = \log_2(x-1)+3$.
(a) Write down the domain of $f$. [1]
(b) Write down the equation of the vertical asymptote of the graph of $f$. [1]
(c) Find the exact $x$-intercept of the graph of $f$. [3]
Show complete worked solution
**(a)**
$$x-1>0 \implies x>1$$
$$\boxed{x>1}$$
**(b)**
$$\boxed{x=1}$$
**(c)**
Setting $f(x)=0$:
$$\log_2(x-1)+3=0$$
$$\log_2(x-1)=-3$$
Converting to exponential form:
$$x-1=2^{-3}=\frac{1}{8}$$
$$x=1+\frac18=\frac98$$
$$\boxed{\left(\frac98,0\right) = (1.125,0)}$$
QUESTION 30
6 marks
Medium
The diagram shows the graph of $g(x)=p\cdot e^{qx}$, where $p,q\in\mathbb{R}$. The graph passes through the points $(0,4)$ and $(3,32)$, as shown.
(a) Write down the value of $p$. [1]
(b) Find the exact value of $q$, and give its value correct to 3 significant figures. [3]
(c) Hence find the exact value of $g(5)$. [2]
Show complete worked solution
**(a)**
Substituting $x=0$: $g(0)=p\cdot e^0=p$. Since $g(0)=4$:
$$\boxed{p=4}$$
**(b)**
Substituting $x=3$, $g(3)=32$ into $g(x)=4e^{qx}$:
$$4e^{3q}=32$$
$$e^{3q}=8$$
Taking $\ln$ of both sides:
$$3q=\ln8=\ln(2^3)=3\ln2$$
$$\boxed{q=\ln2 \approx 0.693}$$
**(c)**
Substituting $q=\ln2$ and $x=5$:
$$g(5) = 4e^{5\ln2} = 4e^{\ln(2^5)} = 4\times2^5 = 4\times32$$
$$\boxed{g(5)=128}$$
QUESTION 31
5 marks
Medium
Consider the function $h(x) = -\log_3 x$.
(a) Describe the single transformation that maps $y=\log_3 x$ onto the graph of $h$. [1]
(b) Write down the domain and range of $h$. [2]
(c) Write down the equation of the vertical asymptote of the graph of $h$. [1]
(d) Find the exact value of $h(9)$. [1]
Show complete worked solution
**(a)**
$$\boxed{\text{Reflection in the } x\text{-axis}}$$
**(b)**
A reflection in the $x$-axis does not change the domain, and the range of a logarithmic function is still all real numbers:
$$\boxed{\text{Domain: } x>0, \quad \text{Range: } h(x)\in\mathbb{R}}$$
**(c)**
A reflection in the $x$-axis does not affect a vertical asymptote:
$$\boxed{x=0}$$
**(d)**
$$h(9) = -\log_39 = -\log_3(3^2) = -2$$
$$\boxed{h(9)=-2}$$
QUESTION 32
5 marks
Medium
The graph of $f(x)=e^x-k$ crosses the $x$-axis at $x=2$.
(a) Find the exact value of $k$. [2]
(b) Find the exact $y$-intercept of the graph of $f$. [1]
(c) Write down the equation of the horizontal asymptote of the graph of $f$. [1]
(d) Write down the range of $f$. [1]
Show complete worked solution
**(a)**
Since the graph crosses the $x$-axis at $x=2$, $f(2)=0$:
$$e^2-k=0$$
$$\boxed{k=e^2 \approx 7.39}$$
**(b)**
Substituting $x=0$ into $f(x)=e^x-e^2$:
$$f(0) = e^0-e^2 = 1-e^2$$
$$\boxed{(0,\,1-e^2)\approx(0,-6.39)}$$
**(c)**
As $x\to-\infty$, $e^x\to0$, so $f(x)\to-e^2$:
$$\boxed{y=-e^2}$$
**(d)**
$$\boxed{f(x)>-e^2}$$
QUESTION 33
6 marks
Medium
The diagram shows the graph of $y=-2^{x-1}+4$, obtained by transforming the graph of $y=2^x$.
(a) Describe a sequence of transformations that maps $y=2^x$ onto $y=-2^{x-1}+4$. [3]
(b) Write down the equation of the horizontal asymptote of $y=-2^{x-1}+4$. [1]
(c) Find the exact $y$-intercept of $y=-2^{x-1}+4$. [2]
Show complete worked solution
**(a)**
Comparing $y=2^x$ with $y=-2^{x-1}+4$, three transformations are applied:
$$\boxed{\text{(i) Reflection in the } x\text{-axis} \ \ \text{(ii) Translation 1 unit right} \ \ \text{(iii) Translation 4 units up}}$$
**(b)**
The horizontal asymptote of $y=2^x$ is $y=0$. Reflecting in the $x$-axis keeps it at $y=0$; translating up 4 units moves it to $y=4$:
$$\boxed{y=4}$$
**(c)**
Substituting $x=0$:
$$y = -2^{-1}+4 = -0.5+4$$
$$\boxed{\left(0,\,3.5\right)}$$
QUESTION 34
5 marks
Medium
The graph of $y=\log_a x$ passes through the point $(8,3)$.
(a) Find the value of $a$. [2]
(b) Find the value of $x$ for which $y=5$. [2]
(c) Write down the $x$-intercept of the graph of $y=\log_a x$, for any base $a>1$. [1]
Show complete worked solution
**(a)**
Since $(8,3)$ lies on the graph:
$$\log_a 8=3 \implies a^3=8$$
$$\boxed{a=2}$$
**(b)**
Solving $\log_2 x=5$:
$$x=2^5$$
$$\boxed{x=32}$$
**(c)**
For any base $a$, $\log_a1=0$, so the graph always crosses the $x$-axis at:
$$\boxed{(1,0)}$$
QUESTION 35
7 marks
Medium
The diagram shows the graph of $f(x)=5e^{-0.2x}+1$.
(a) Write down the equation of the horizontal asymptote of the graph. [1]
(b) Find the exact $y$-intercept of the graph. [1]
(c) Describe the behaviour of the graph as $x\to+\infty$ and as $x\to-\infty$. [2]
(d) Find the value of $x$ for which $f(x)=3$, correct to 3 significant figures. [3]
Show complete worked solution
**(a)**
As $x\to+\infty$, $e^{-0.2x}\to0$, so $f(x)\to1$:
$$\boxed{y=1}$$
**(b)**
$$f(0) = 5e^0+1 = 5+1$$
$$\boxed{(0,6)}$$
**(c)**
$$\boxed{\text{As } x\to+\infty, f(x)\to1^{+} \text{ (approaches 1 from above); as } x\to-\infty, f(x)\to+\infty}$$
**(d)**
$$5e^{-0.2x}+1=3$$
$$5e^{-0.2x}=2$$
$$e^{-0.2x}=0.4$$
Taking $\ln$ of both sides:
$$-0.2x=\ln0.4=-0.91629\ldots$$
$$x = 4.58145\ldots$$
$$\boxed{x\approx4.58}$$
QUESTION 36
6 marks
Medium
The graph of $y=\ln x$ is reflected in the line $y=x$ to give the graph of $y=e^x$.
(a) State the vertical asymptote of $y=\ln x$, and hence write down the horizontal asymptote of $y=e^x$. [2]
(b) State the $x$-intercept of $y=\ln x$, and hence write down the corresponding point on the graph of $y=e^x$. [2]
(c) State the domain and range of $y=\ln x$, and explain how these relate to the domain and range of $y=e^x$. [2]
Show complete worked solution
**(a)**
The vertical asymptote of $y=\ln x$ is $x=0$. Reflecting a vertical line in $y=x$ produces a horizontal line with the same numerical value:
$$\boxed{x=0 \text{ (for } \ln x\text{)} \ \longrightarrow \ y=0 \text{ (for } e^x\text{)}}$$
**(b)**
The $x$-intercept of $y=\ln x$ is $(1,0)$. Reflecting in $y=x$ swaps the coordinates:
$$\boxed{(1,0) \ \longrightarrow \ (0,1) \text{, the } y\text{-intercept of } e^x}$$
**(c)**
$y=\ln x$ has domain $x>0$ and range $y\in\mathbb{R}$. Since inverse functions swap domain and range:
$$\boxed{y=e^x \text{ has domain } x\in\mathbb{R} \text{ and range } y>0}$$
QUESTION 37
8 marks
Medium
The value, in thousands of dollars, of an investment $t$ years after it is made is modelled by $V(t)=2000(1.05)^t$.
(a) Find $V(0)$, and interpret this value in context. [2]
(b) Explain whether the graph of $V$ has a horizontal asymptote as $t\to+\infty$. [1]
(c) Describe the behaviour of $V(t)$ as $t\to-\infty$, and state the equation of the horizontal asymptote in this direction. [2]
(d) Find the value of $t$ for which $V(t)=3000$, correct to 3 significant figures. [3]
Show complete worked solution
**(a)**
$$V(0) = 2000(1.05)^0 = 2000$$
$$\boxed{V(0)=2000 \text{: the initial value of the investment was } 2000}$$
**(b)**
As $t\to+\infty$, $(1.05)^t\to+\infty$, so $V(t)$ grows without bound:
$$\boxed{\text{No, there is no horizontal asymptote as } t\to+\infty}$$
**(c)**
As $t\to-\infty$, $(1.05)^t\to0$, so $V(t)\to0$:
$$\boxed{V(t)\to0 \text{ as } t\to-\infty\text{; the asymptote is } V=0}$$
**(d)**
$$2000(1.05)^t=3000$$
$$(1.05)^t=1.5$$
Taking $\ln$ of both sides:
$$t\ln(1.05)=\ln1.5$$
$$t = \frac{\ln1.5}{\ln1.05} = \frac{0.405465\ldots}{0.048790\ldots}$$
$$\boxed{t\approx8.31 \text{ years}}$$
QUESTION 38
7 marks
Medium
Consider the function $f(x)=\log_2 x$. It is transformed to give $g(x)=\log_2(x+4)-1$.
(a) Describe the transformation(s) that map $f$ onto $g$. [2]
(b) Write down the equation of the vertical asymptote of $g$. [1]
(c) Write down the domain of $g$. [1]
(d) Find the exact $x$-intercept of $g$. [3]
Show complete worked solution
**(a)**
$$\boxed{\text{Translation by the vector } \begin{pmatrix}-4\\-1\end{pmatrix} \text{ (4 units left, 1 unit down)}}$$
**(b)**
The vertical asymptote of $f$, $x=0$, shifts 4 units left:
$$\boxed{x=-4}$$
**(c)**
$$\boxed{x>-4}$$
**(d)**
Setting $g(x)=0$:
$$\log_2(x+4)-1=0$$
$$\log_2(x+4)=1$$
$$x+4=2^1=2$$
$$\boxed{x=-2, \text{ i.e. } (-2,0)}$$
QUESTION 39
6 marks
Medium
The diagram shows the graphs of $y=2^x$ and $y=4^x$.
(a) Write down the common $y$-intercept of the two graphs. [1]
(b) For $x>0$, state which graph lies above the other, and explain why using the bases of the two functions. [2]
(c) For $x<0$, state which graph lies above the other. [2]
(d) Write down the equation of the horizontal asymptote shared by both graphs. [1]
Show complete worked solution
**(a)**
Since $2^0=4^0=1$:
$$\boxed{(0,1)}$$
**(b)**
Since $4>2>1$, and both functions are increasing for $x>0$, the function with the larger base grows faster. For example, at $x=1$: $4^1=4>2^1=2$:
$$\boxed{y=4^x \text{ lies above } y=2^x \text{ for } x>0}$$
**(c)**
For $x<0$ the inequality reverses, since raising a larger base to a negative power gives a smaller result. For example, at $x=-1$: $4^{-1}=0.25<2^{-1}=0.5$:
$$\boxed{y=2^x \text{ lies above } y=4^x \text{ for } x<0}$$
**(d)**
$$\boxed{y=0}$$
QUESTION 40
7 marks
Medium
The mass, $N$ grams, of a radioactive sample after $t$ years is modelled by $N(t)=N_0\left(\dfrac12\right)^{t/h}$, where $h$ is the half-life in years. The diagram shows the graph of $N$ against $t$. The graph passes through $(0,80)$ and $(6,20)$.
(a) Write down the equation of the horizontal asymptote of the graph, and state what this represents in context. [1]
(b) Using the given points, find the half-life $h$. [3]
(c) Hence find the mass remaining after 10 years, correct to 3 significant figures. [3]
Show complete worked solution
**(a)**
$$\boxed{N=0 \text{: the mass approaches zero but the sample never fully disappears}}$$
**(b)**
From $(0,80)$: $N_0=80$. Substituting the point $(6,20)$ into $N(t)=80\left(\frac12\right)^{t/h}$:
$$80\left(\frac12\right)^{6/h}=20$$
$$\left(\frac12\right)^{6/h}=\frac14=\left(\frac12\right)^2$$
$$\frac6h=2$$
$$\boxed{h=3 \text{ years}}$$
**(c)**
Substituting $t=10$, $h=3$ into $N(t)=80\left(\frac12\right)^{t/3}$:
$$N(10) = 80\left(\frac12\right)^{10/3} = 80\times2^{-10/3}$$
$$2^{10/3} = 10.0794\ldots$$
$$N(10) = \frac{80}{10.0794\ldots} = 7.9369\ldots$$
$$\boxed{N(10)\approx7.94 \text{ grams}}$$
QUESTION 41
8 marks
Medium
Consider the function $f(x)=e^{2x-1}-3$.
(a) Write down the equation of the horizontal asymptote of the graph of $f$. [1]
(b) Find the exact $y$-intercept of the graph of $f$, and give its value correct to 3 significant figures. [2]
(c) Find the exact $x$-intercept of the graph of $f$, and give its value correct to 3 significant figures. [3]
(d) Writing $f(x)=e^{2(x-0.5)}-3$, describe the sequence of transformations that maps $y=e^x$ onto the graph of $f$. [2]
Show complete worked solution
**(a)**
As $x\to-\infty$, $e^{2x-1}\to0$, so $f(x)\to-3$:
$$\boxed{y=-3}$$
**(b)**
$$f(0) = e^{-1}-3 = 0.36788\ldots-3$$
$$\boxed{(0,\,e^{-1}-3)\approx(0,-2.63)}$$
**(c)**
Setting $f(x)=0$:
$$e^{2x-1}-3=0$$
$$e^{2x-1}=3$$
$$2x-1=\ln3$$
$$x = \frac{1+\ln3}{2}$$
$$\boxed{x=\frac{1+\ln3}{2}\approx1.05}$$
**(d)**
$$\boxed{\text{Horizontal stretch, scale factor } \tfrac12, \text{ relative to the } y\text{-axis, followed by a translation of } \begin{pmatrix}0.5\\-3\end{pmatrix}}$$
QUESTION 42
9 marks
Hard
The diagram shows the graph of $f(x)=a\cdot b^x+c$, where $a,b,c\in\mathbb{R}$, $b>0$. The horizontal asymptote is $y=2$, and the graph passes through $(0,5)$ and $(1,8)$.
(a) Write down the value of $c$. [1]
(b) Use the point $(0,5)$ to find the value of $a$. [2]
(c) Use the point $(1,8)$ to find the value of $b$. [2]
(d) Show that the graph of $f$ does not cross the $x$-axis, justifying your answer using the range of $f$. [2]
(e) Find the exact value of $f(3)$. [2]
Show complete worked solution
**(a)**
From the diagram, $f$ is an increasing function, so $b>1$. As $x\to-\infty$, $b^x\to0$, so $f(x)\to c$. Since the asymptote is $y=2$:
$$\boxed{c=2}$$
**(b)**
Substituting $(0,5)$ into $f(x)=a\cdot b^x+2$:
$$5 = a\cdot b^0+2 = a+2$$
$$\boxed{a=3}$$
**(c)**
Substituting $(1,8)$ into $f(x)=3b^x+2$:
$$8 = 3b^1+2$$
$$3b=6$$
$$\boxed{b=2}$$
**(d)**
The function is $f(x)=3\cdot2^x+2$. Since $2^x>0$ for all real $x$:
$$3\cdot2^x>0 \implies f(x) = 3\cdot2^x+2>2$$
Since $f(x)>2>0$ for every real $x$, $f(x)$ is never equal to zero:
$$\boxed{f(x)>2 \text{ for all } x, \text{ so the graph never reaches the } x\text{-axis}}$$
**(e)**
$$f(3) = 3\cdot2^3+2 = 3(8)+2 = 24+2$$
$$\boxed{f(3)=26}$$
QUESTION 43
9 marks
Hard
The diagram shows the graph of $g(x)=p\ln(x+2)+r$, where $p,r\in\mathbb{R}$. The vertical asymptote is $x=-2$, and the graph passes through $(-1,3)$ and $(7,7.39)$, where the second $y$-coordinate is given correct to 3 significant figures.
(a) Use the point $(-1,3)$ to find the value of $r$. [2]
(b) Use the point $(7,7.39)$ to find the value of $p$. [3]
(c) Hence write down the full equation of $g$. [1]
(d) Find the exact $x$-intercept of the graph of $g$, and give its value correct to 3 significant figures. [3]
Show complete worked solution
**(a)**
Substituting $(-1,3)$ into $g(x)=p\ln(x+2)+r$:
$$3 = p\ln(-1+2)+r = p\ln(1)+r$$
Since $\ln1=0$:
$$3 = 0+r$$
$$\boxed{r=3}$$
**(b)**
Substituting $(7,7.39)$ into $g(x)=p\ln(x+2)+3$:
$$7.39 = p\ln(9)+3$$
$$p\ln9=4.39$$
$$p = \frac{4.39}{\ln9} = \frac{4.39}{2.19722\ldots}$$
$$\boxed{p\approx2}$$
**(c)**
$$\boxed{g(x) = 2\ln(x+2)+3}$$
**(d)**
Setting $g(x)=0$:
$$2\ln(x+2)+3=0$$
$$\ln(x+2)=-1.5$$
$$x+2 = e^{-1.5}$$
$$x = e^{-1.5}-2 = 0.22313\ldots-2$$
$$\boxed{x=e^{-1.5}-2\approx-1.78}$$
QUESTION 44
9 marks
Hard
Let $f(x)=5^x$ and $g(x)=\log_5 x$.
(a) Show algebraically that $g$ is the inverse function of $f$. [3]
(b) Write down the equation of the horizontal asymptote of the graph of $f$ and the equation of the vertical asymptote of the graph of $g$, and explain how these two lines are related by a reflection in the line $y=x$. [3]
(c) The point $(2,25)$ lies on the graph of $f$. Write down the coordinates of the corresponding point on the graph of $g$, and verify this algebraically. [3]
Show complete worked solution
**(a)**
Let $y=f(x)=5^x$. To find the inverse, swap $x$ and $y$ and solve for $y$:
$$x=5^y$$
Using the definition of a logarithm, $5^y=x \iff y=\log_5x$:
$$y=\log_5x$$
This is exactly $g(x)$, so:
$$\boxed{g(x)=f^{-1}(x)}$$
**(b)**
The horizontal asymptote of $f$ is $y=0$. The vertical asymptote of $g$ is $x=0$.
Reflecting the line $y=0$ in the line $y=x$ interchanges the roles of $x$ and $y$: the equation $y=0$ becomes $x=0$, which is exactly the asymptote of $g$:
$$\boxed{y=0 \text{ (for } f\text{) reflects onto } x=0 \text{ (for } g\text{), consistent with } g \text{ being the inverse of } f}$$
**(c)**
Since $g=f^{-1}$, the graph of $g$ is the reflection of the graph of $f$ in $y=x$, so coordinates are swapped:
$$\boxed{(25,2)}$$
Verifying: $g(25)=\log_525$. Since $5^2=25$, $\log_525=2$, which matches:
$$\boxed{g(25)=2 \checkmark}$$
QUESTION 45
9 marks
Hard
Consider the function $f(x)=e^x$. The graph of $f$ is transformed to give the graph of $y=3e^{-(x+2)}-1$.
(a) Describe, in order, a sequence of transformations that maps the graph of $y=e^x$ onto the graph of $y=3e^{-(x+2)}-1$. [4]
(b) Write down the equation of the horizontal asymptote of $y=3e^{-(x+2)}-1$. [1]
(c) Find the exact $y$-intercept of $y=3e^{-(x+2)}-1$, and give its value correct to 3 significant figures. [2]
(d) Write down the range of $y=3e^{-(x+2)}-1$. [2]
Show complete worked solution
**(a)**
Starting from $y=e^x$ and building up to $y=3e^{-(x+2)}-1$ step by step:
$$e^x \ \xrightarrow{\text{reflect in } y\text{-axis}}\ e^{-x} \ \xrightarrow{\text{translate left 2}}\ e^{-(x+2)} \ \xrightarrow{\text{stretch vertically, factor 3}}\ 3e^{-(x+2)} \ \xrightarrow{\text{translate down 1}}\ 3e^{-(x+2)}-1$$
$$\boxed{\text{(i) Reflection in the } y\text{-axis} \ \ \text{(ii) Translation 2 units left} \ \ \text{(iii) Vertical stretch, scale factor 3} \ \ \text{(iv) Translation 1 unit down}}$$
**(b)**
As $x\to+\infty$, $-(x+2)\to-\infty$, so $e^{-(x+2)}\to0$ and $y\to0-1=-1$:
$$\boxed{y=-1}$$
**(c)**
Substituting $x=0$:
$$y = 3e^{-2}-1 = 3(0.135335\ldots)-1$$
$$\boxed{(0,\,3e^{-2}-1)\approx(0,-0.594)}$$
**(d)**
Since $e^{-(x+2)}>0$ for all real $x$, $3e^{-(x+2)}>0$, so:
$$y = 3e^{-(x+2)}-1>-1 \text{ for all } x$$
$$\boxed{y>-1}$$
QUESTION 46
10 marks
Hard
The temperature, $T$ degrees Celsius, of a hot drink $t$ minutes after being poured is modelled by $T(t)=T_s+(T_0-T_s)e^{-kt}$, where $T_s$ is the (constant) surrounding room temperature. The diagram shows the graph of $T$ against $t$, with horizontal asymptote $T=20$, $T(0)=90$ and $T(5)=60$.
(a) Use the asymptote to write down the value of $T_s$. [1]
(b) Show that $T_0=90$, and find the value of $k$, correct to 3 significant figures. [4]
(c) Find the time at which the temperature reaches $30^\circ\text{C}$, correct to 3 significant figures. [3]
(d) Explain, with reference to the equation of $T(t)$, why the temperature can never actually equal $T_s$. [2]
Show complete worked solution
**(a)**
As $t\to\infty$, $e^{-kt}\to0$, so $T(t)\to T_s$. Since the asymptote is $T=20$:
$$\boxed{T_s=20}$$
**(b)**
Substituting $t=0$ into $T(t)=20+(T_0-20)e^{-kt}$:
$$T(0) = 20+(T_0-20)e^0 = 20+T_0-20 = T_0$$
Since $T(0)=90$:
$$\boxed{T_0=90}$$
So $T(t)=20+70e^{-kt}$. Substituting $t=5$, $T(5)=60$:
$$20+70e^{-5k}=60$$
$$70e^{-5k}=40$$
$$e^{-5k}=\frac{4}{7}$$
Taking $\ln$ of both sides:
$$-5k=\ln\left(\frac47\right)=-0.55962\ldots$$
$$\boxed{k\approx0.112}$$
**(c)**
Substituting $T(t)=30$ into $T(t)=20+70e^{-0.11192t}$:
$$20+70e^{-0.11192t}=30$$
$$70e^{-0.11192t}=10$$
$$e^{-0.11192t}=\frac17$$
$$-0.11192t=\ln\left(\frac17\right)=-1.94591\ldots$$
$$t = 17.386\ldots$$
$$\boxed{t\approx17.4 \text{ minutes}}$$
**(d)**
$$\boxed{\text{Since } e^{-kt}>0 \text{ for every finite } t, \text{ the term } 70e^{-kt} \text{ is always strictly positive, so } T(t)=20+70e^{-kt}>20 \text{ for all } t; T \text{ only approaches } T_s=20 \text{ as } t\to\infty, \text{ it never reaches it}}$$
QUESTION 47
8 marks
Hard
The diagram shows the graphs of $y=2^x$ and $y=x+2$ for $-4\le x\le4$.
(a) Verify algebraically that $x=2$ is a solution of the equation $2^x=x+2$. [2]
(b) Using the graph, write down the other value of $x$ for which $2^x=x+2$, correct to 1 decimal place. [2]
(c) Hence write down the solution set of the inequality $2^x(d) Explain, with reference to the graph, why $2^x>x+2$ for all $x>2$. [2]
Show complete worked solution
**(a)**
Substituting $x=2$ into both sides:
$$\text{LHS} = 2^2=4, \qquad \text{RHS} = 2+2=4$$
Since LHS = RHS:
$$\boxed{x=2 \text{ is a solution}}$$
**(b)**
Reading from the diagram, the curve $y=2^x$ and the line $y=x+2$ intersect a second time where $x$ is negative, close to where the curve crosses the line near the bottom-left of the diagram:
$$\boxed{x\approx-1.7}$$
**(c)**
$2^x2 \text{ the curve } y=2^x \text{ lies above the line } y=x+2 \text{ in the diagram (exponential growth eventually outpaces linear growth), and the two graphs do not cross again, so } 2^x>x+2 \text{ for all } x>2}$$
QUESTION 48
10 marks
Hard
Consider the function $f(x)=\ln(x^2-4)$.
(a) Find the domain of $f$. [3]
(b) Write down the equations of the two vertical asymptotes of the graph of $f$. [2]
(c) Show that the graph of $f$ is symmetric about the $y$-axis. [2]
(d) Find the exact $x$-intercepts of the graph of $f$. [3]
Show complete worked solution
**(a)**
The argument of the logarithm must be positive:
$$x^2-4>0$$
$$x^2>4$$
$$\boxed{x<-2 \text{ or } x>2}$$
**(b)**
The vertical asymptotes occur at the boundary points of the domain, where $x^2-4\to0^+$:
$$\boxed{x=-2 \text{ and } x=2}$$
**(c)**
Replacing $x$ with $-x$:
$$f(-x) = \ln((-x)^2-4) = \ln(x^2-4) = f(x)$$
Since $f(-x)=f(x)$ for every $x$ in the domain, $f$ is an even function, so its graph is symmetric about the $y$-axis:
$$\boxed{f(-x)=f(x) \implies \text{symmetric about the } y\text{-axis}}$$
**(d)**
Setting $f(x)=0$:
$$\ln(x^2-4)=0$$
$$x^2-4=1$$
$$x^2=5$$
$$x=\pm\sqrt5$$
Both values satisfy the domain found in part (a), since $\sqrt5\approx2.236>2$:
$$\boxed{x=\sqrt5 \text{ and } x=-\sqrt5}$$
QUESTION 49
9 marks
Hard
The number of words, $h(x)$, that a language-learner can recognise after $x$ weeks of study is modelled by $h(x)=A(1-e^{-kx})$, where $A$ and $k$ are positive constants. The diagram shows the graph of $h$, with horizontal asymptote $y=80$ (in hundreds of words) and the graph passing through $(2,40)$.
(a) Write down the value of $A$. [1]
(b) Find the value of $k$, correct to 3 significant figures. [3]
(c) Find the exact number of weeks required for $h(x)=75$. [3]
(d) Explain, in context, the significance of the horizontal asymptote $y=80$. [2]
Show complete worked solution
**(a)**
As $x\to\infty$, $e^{-kx}\to0$, so $h(x)\to A$. Since the asymptote is $y=80$:
$$\boxed{A=80}$$
**(b)**
Substituting $(2,40)$ into $h(x)=80(1-e^{-kx})$:
$$80(1-e^{-2k})=40$$
$$1-e^{-2k}=0.5$$
$$e^{-2k}=0.5$$
$$-2k=\ln0.5$$
$$k = \frac{\ln2}{2} = 0.34657\ldots$$
$$\boxed{k\approx0.347}$$
**(c)**
Setting $h(x)=75$ in $h(x)=80\left(1-e^{-\frac{\ln2}{2}x}\right)$:
$$80\left(1-e^{-\frac{\ln2}{2}x}\right)=75$$
$$1-e^{-\frac{\ln2}{2}x}=\frac{75}{80}=0.9375$$
$$e^{-\frac{\ln2}{2}x}=0.0625=\frac{1}{16}=2^{-4}$$
Taking $\ln$ of both sides:
$$-\frac{\ln2}{2}x = -4\ln2$$
Dividing both sides by $-\frac{\ln2}{2}$:
$$x = \frac{4\ln2}{\tfrac12\ln2}=8$$
$$\boxed{x=8 \text{ weeks}}$$
**(d)**
$$\boxed{\text{The number of recognised words approaches a maximum of 8000 words (80 hundreds) as study time increases, but this limit is never actually attained, no matter how long the learner studies}}$$
QUESTION 50
8 marks
Hard
Consider the graph of $y=\log_2 x$.
(a) Show that $y=\log_2\left(\dfrac8x\right)$ can be written in the form $y=3-\log_2x$. [2]
(b) Describe the transformation(s) that map the graph of $y=\log_2 x$ onto the graph of $y=3-\log_2x$. [2]
(c) Write down the domain and the equation of the vertical asymptote of the graph of $y=3-\log_2x$. [2]
(d) Find the exact $x$-intercept of the graph of $y=3-\log_2x$. [2]
Show complete worked solution
**(a)**
Using the law of logarithms $\log_2\left(\dfrac{A}{B}\right)=\log_2A-\log_2B$:
$$\log_2\left(\frac8x\right) = \log_28-\log_2x$$
Since $8=2^3$, $\log_28=3$:
$$\boxed{\log_2\left(\frac8x\right) = 3-\log_2x}$$
**(b)**
$$\boxed{\text{Reflection in the } x\text{-axis, followed by a translation of 3 units up}}$$
**(c)**
Neither a reflection in the $x$-axis nor a vertical translation affects the domain or the vertical asymptote:
$$\boxed{\text{Domain: } x>0, \quad \text{Vertical asymptote: } x=0}$$
**(d)**
Setting $y=0$:
$$3-\log_2x=0$$
$$\log_2x=3$$
$$\boxed{x=2^3=8, \text{ i.e. } (8,0)}$$
Transformations 50 questions
QUESTION 1
3 marks
Easy
The graph of $y=f(x)$ has a minimum point at $(3,-2)$. Write down the coordinates of the minimum point on the graph of $y=f(x)-4$.
Show complete worked solution
Using the fact that $y=f(x)-4$ is a vertical translation of $y=f(x)$ by $\begin{pmatrix}0\\-4\end{pmatrix}$: the $x$-coordinate is unchanged and 4 is subtracted from the $y$-coordinate.
$$\text{New minimum: } (3,\ -2-4) = (3,-6)$$
$$\boxed{(3,-6)}$$
QUESTION 2
3 marks
Easy
The graph of $y=f(x)$ passes through the point $(5,7)$. Find the coordinates of the corresponding point on the graph of $y=f(x-2)+1$.
Show complete worked solution
Using the fact that $y=f(x-2)+1$ is obtained from $y=f(x)$ by a translation with vector $\begin{pmatrix}2\\1\end{pmatrix}$ (2 units right, 1 unit up):
$$\text{New point: } (5+2,\ 7+1) = (7,8)$$
$$\boxed{(7,8)}$$
QUESTION 3
6 marks
Medium
Describe a sequence of transformations that maps the graph of $y=x^2$ onto the graph of $y=3(x+2)^2-1$.
Show complete worked solution
Rewriting the target function to identify each transformation applied to $y=x^2$, in order.
Starting with $y=x^2$:
Step 1 (horizontal translation by $\begin{pmatrix}-2\\0\end{pmatrix}$):
$$y=(x+2)^2$$
Step 2 (vertical stretch with scale factor 3, relative to the $x$-axis):
$$y=3(x+2)^2$$
Step 3 (vertical translation by $\begin{pmatrix}0\\-1\end{pmatrix}$):
$$y=3(x+2)^2-1$$
$$\boxed{\text{Translation } \begin{pmatrix}-2\\0\end{pmatrix} \text{, then vertical stretch scale factor } 3 \text{, then translation } \begin{pmatrix}0\\-1\end{pmatrix}}$$
QUESTION 4
5 marks
Medium
The point $(4,-8)$ lies on the graph of $y=f(x)$. The graph of $g$ is defined by $g(x) = -f\left(\dfrac{x}{2}\right)$. Find the coordinates of the corresponding point on the graph of $g$.
Show complete worked solution
Using the fact that $y=f\left(\dfrac{x}{2}\right)$ represents a horizontal stretch of $y=f(x)$ with scale factor 2 (each $x$-coordinate doubles), and the negative sign then reflects the result in the $x$-axis (each $y$-coordinate is negated).
Since $(4,-8)$ lies on $f$, i.e. $f(4)=-8$, the horizontal stretch maps this point to the $x$-value $2(4)=8$: the point $(8,-8)$ lies on $y=f(x/2)$.
Reflecting in the $x$-axis:
$$(8,\ -(-8)) = (8,8)$$
$$\boxed{(8,8)}$$
QUESTION 5
7 marks
Hard
The graph of $y=f(x)$ has a local maximum point at $(-1,4)$. The graph of $h$ is defined by $h(x) = -2f(x-3)+1$. Find the coordinates of the corresponding point on the graph of $h$, and state whether it is a local maximum or a local minimum.
Show complete worked solution
Identifying the $x$-value on $h$ corresponding to $x=-1$ on $f$, then applying the vertical transformations to the $y$-value.
Since $h(x)=-2f(x-3)+1$ involves $f(x-3)$, the input to $f$ equals $-1$ when:
$$x-3=-1 \implies x=2$$
At this point, using $f(-1)=4$:
$$h(2) = -2f(-1)+1 = -2(4)+1 = -8+1=-7$$
So the corresponding point is $(2,-7)$.
Determining the nature of the point: horizontal translation and vertical translation do not change the type of turning point, and a stretch with positive scale factor also preserves the type; however, the factor of $-2$ includes a reflection in the $x$-axis, which converts a local maximum into a local minimum.
$$\boxed{(2,-7),\ \text{a local minimum}}$$
QUESTION 6
3 marks
Easy
The graph of $y=f(x)$ has a minimum point at $(3,-2)$, as shown (in blue) on the diagram below, together with a possible image curve. Write down the coordinates of the minimum point on the graph of $y=f(x)-4$.
Show complete worked solution
Using the fact that $y=f(x)-4$ is a vertical translation of $y=f(x)$ by $\begin{pmatrix}0\\-4\end{pmatrix}$: the $x$-coordinate of the minimum point is unchanged, and 4 is subtracted from the $y$-coordinate.
$$\text{New minimum: } (3,\ -2-4) = (3,-6)$$
$$\boxed{(3,-6)}$$
QUESTION 7
3 marks
Easy
The graph of $y=f(x)$ passes through the point $(-3,6)$. Find the coordinates of the corresponding point on the graph of $y=f(x+5)$.
Show complete worked solution
Using the fact that $y=f(x+5)$ is obtained from $y=f(x)$ by a horizontal translation with vector $\begin{pmatrix}-5\\0\end{pmatrix}$ (5 units to the left).
Every point on $y=f(x)$ moves 5 units to the left, so the $x$-coordinate decreases by 5 and the $y$-coordinate is unchanged.
$$\text{New point: } (-3-5,\ 6) = (-8,6)$$
$$\boxed{(-8,6)}$$
QUESTION 8
4 marks
Easy
Describe fully the single transformation that maps the graph of $y=x^2$ onto the graph of $y=(x-3)^2+2$.
Show complete worked solution
Comparing $y=(x-3)^2+2$ with $y=x^2$: the function has been replaced by $f(x-3)+2$, which corresponds to a translation.
The vector of the translation is $\begin{pmatrix}3\\2\end{pmatrix}$ (3 units in the positive $x$-direction and 2 units in the positive $y$-direction), which can be checked using the vertex: $(0,0) \to (3,2)$.
$$\boxed{\text{Translation by the vector } \begin{pmatrix}3\\2\end{pmatrix}}$$
QUESTION 9
3 marks
Easy
The graph of $y=f(x)$ has an $x$-intercept at $x=7$. Write down the $x$-intercept of the graph of $y=f(x-4)$.
Show complete worked solution
Using the fact that $y=f(x-4)$ is a horizontal translation of $y=f(x)$ by $\begin{pmatrix}4\\0\end{pmatrix}$ (4 units to the right): every point on the curve, including the $x$-intercept, moves 4 units to the right.
$$\text{New } x\text{-intercept: } 7+4=11$$
$$\boxed{x=11}$$
QUESTION 10
5 marks
Easy
The diagram below shows the graph of $y=\sin x$ for $0 \leq x \leq 2\pi$. On the same diagram, the image of the curve under a translation is shown. State the vector of this translation, and write down the equation of the image curve.
Show complete worked solution
Comparing the two curves in the diagram: every point on the blue curve $y=\sin x$ has moved vertically upward by 2 units to give the orange curve, while the $x$-coordinate of every point is unchanged.
This is a translation with vector $\begin{pmatrix}0\\2\end{pmatrix}$.
Applying $y \to y+2$ to $y=\sin x$:
$$\boxed{\text{Translation } \begin{pmatrix}0\\2\end{pmatrix}, \quad y=\sin x + 2}$$
QUESTION 11
3 marks
Easy
The graph of $y=f(x)$ has a zero at $x=-2$. Write down a zero of the graph of $y=2f(x)$.
Show complete worked solution
Using the fact that $y=2f(x)$ is a vertical stretch of $y=f(x)$ with scale factor 2, relative to the $x$-axis: each $y$-coordinate is doubled, while each $x$-coordinate is unchanged.
Since $f(-2)=0$, doubling this value gives $2f(-2)=2(0)=0$, so the point $(-2,0)$ still lies on the image curve.
$$\boxed{x=-2}$$
QUESTION 12
3 marks
Easy
Describe fully the single transformation that maps the graph of $y=x^2$ onto the graph of $y=x^2-5$.
Show complete worked solution
Comparing $y=x^2-5$ with $y=x^2$: the function has been replaced by $f(x)-5$.
This corresponds to a vertical translation, where every point on the curve moves 5 units in the negative $y$-direction while the $x$-coordinate is unchanged.
$$\boxed{\text{Translation by the vector } \begin{pmatrix}0\\-5\end{pmatrix}}$$
QUESTION 13
3 marks
Easy
Describe fully the single transformation that maps the graph of $y=x^3$ onto the graph of $y=(x+1)^3$.
Show complete worked solution
Comparing $y=(x+1)^3$ with $y=x^3$: the function has been replaced by $f(x+1)$.
This corresponds to a horizontal translation, where every point on the curve moves 1 unit in the negative $x$-direction while the $y$-coordinate is unchanged.
$$\boxed{\text{Translation by the vector } \begin{pmatrix}-1\\0\end{pmatrix}}$$
QUESTION 14
3 marks
Easy
The graph of $y=f(x)$ passes through the point $(4,9)$, as shown on the diagram below. The graph of $g$ is defined by $g(x)=f(x)+6$. Find the coordinates of the corresponding point on the graph of $g$.
Show complete worked solution
Using the fact that $y=f(x)+6$ is a vertical translation of $y=f(x)$ by $\begin{pmatrix}0\\6\end{pmatrix}$: the $x$-coordinate is unchanged, and 6 is added to the $y$-coordinate.
$$g(4) = f(4)+6 = 9+6 = 15$$
$$\boxed{(4,15)}$$
QUESTION 15
5 marks
Easy
The diagram below shows the graph of $y=\cos x$ for $0 \leq x \leq 2\pi$, together with its image under a reflection. State the equation of the image curve shown, and describe the transformation.
Show complete worked solution
Comparing the two curves in the diagram: the orange curve is the reflection of the blue curve $y=\cos x$ in the $x$-axis, since every $y$-coordinate has been negated while every $x$-coordinate is unchanged.
This transformation, $y \to -y$, applied to $y=\cos x$ gives $y=-\cos x$.
$$\boxed{\text{Reflection in the } x\text{-axis}, \quad y=-\cos x}$$
QUESTION 16
3 marks
Easy
The graph of $y=f(x)$ has a minimum point at $(0,-3)$. The graph of $g$ is defined by $g(x)=f(x-2)$. Find the coordinates of the minimum point on the graph of $g$.
Show complete worked solution
Using the fact that $y=f(x-2)$ is a horizontal translation of $y=f(x)$ by $\begin{pmatrix}2\\0\end{pmatrix}$: the $y$-coordinate of the minimum point is unchanged, and 2 is added to the $x$-coordinate.
$$\text{New minimum: } (0+2,\ -3) = (2,-3)$$
$$\boxed{(2,-3)}$$
QUESTION 17
5 marks
Easy
The graph of $y=\sqrt{x}$ is transformed to the graph of $y=\sqrt{x}+4$.
(a) Describe fully this transformation. [2]
(b) Find the value of $y$ on each curve when $x=9$. [3]
Show complete worked solution
(a) Comparing $y=\sqrt{x}+4$ with $y=\sqrt{x}$: the function has been replaced by $f(x)+4$, so this is a vertical translation of $\begin{pmatrix}0\\4\end{pmatrix}$ (4 units upward).
$$\boxed{\text{(a) Translation by the vector } \begin{pmatrix}0\\4\end{pmatrix}}$$
(b) At $x=9$:
$$y=\sqrt{9}=3 \ \text{ on the original curve}$$
$$y=\sqrt{9}+4=3+4=7 \ \text{ on the image curve}$$
$$\boxed{\text{(b) } y=3 \text{ and } y=7}$$
QUESTION 18
3 marks
Easy
The graph of $y=f(x)$ has a $y$-intercept of 5. Write down the $y$-intercept of the graph of $y=3f(x)$.
Show complete worked solution
Using the fact that $y=3f(x)$ is a vertical stretch of $y=f(x)$ with scale factor 3, relative to the $x$-axis: every $y$-coordinate is multiplied by 3, while the $x$-coordinate is unchanged.
$$\text{New } y\text{-intercept: } 3 \times 5 = 15$$
$$\boxed{y=15}$$
QUESTION 19
4 marks
Easy
Describe fully the single transformation that maps the graph of $y=x^2$ onto the graph of $y=(2x)^2$.
Show complete worked solution
Comparing $y=(2x)^2$ with $y=x^2$: the function has been replaced by $f(2x)$.
This corresponds to a horizontal stretch with scale factor $\dfrac{1}{2}$, relative to the $y$-axis: every $x$-coordinate is multiplied by $\dfrac{1}{2}$, while the $y$-coordinate is unchanged.
$$\boxed{\text{Horizontal stretch, scale factor } \dfrac{1}{2}\text{, relative to the } y\text{-axis}}$$
QUESTION 20
3 marks
Easy
The graph of $y=f(x)$ passes through the point $(6,-2)$. The graph of $h$ is defined by $h(x)=f(-x)$. Find the coordinates of the corresponding point on the graph of $h$.
Show complete worked solution
Using the fact that $y=f(-x)$ is a reflection of $y=f(x)$ in the $y$-axis: each $x$-coordinate is negated, while the $y$-coordinate is unchanged.
$$\text{New point: } (-6,\ -2)$$
$$\boxed{(-6,-2)}$$
QUESTION 21
6 marks
Easy
The diagram below shows part of the graph of $y=f(x)$, passing through the points $A(-2,1)$ and $B(1,4)$.
(a) Describe fully the transformation that maps $y=f(x)$ onto $y=f(x)-3$. [2]
(b) State the coordinates of the image of $A$ under this transformation. [2]
(c) State the coordinates of the image of $B$ under this transformation. [2]
Show complete worked solution
(a) Comparing $y=f(x)-3$ with $y=f(x)$: the function has been replaced by $f(x)-3$, so this is a vertical translation by the vector $\begin{pmatrix}0\\-3\end{pmatrix}$ (3 units downward).
$$\boxed{\text{(a) Translation by the vector } \begin{pmatrix}0\\-3\end{pmatrix}}$$
(b) Applying the translation to $A(-2,1)$: the $x$-coordinate is unchanged and 3 is subtracted from the $y$-coordinate.
$$A \to (-2,\ 1-3) = (-2,-2)$$
$$\boxed{\text{(b) } A^{\prime}(-2,-2)}$$
(c) Applying the same translation to $B(1,4)$:
$$B \to (1,\ 4-3) = (1,1)$$
$$\boxed{\text{(c) } B^{\prime}(1,1)}$$
QUESTION 22
5 marks
Easy
The graph of $y=e^x$ is transformed to the graph of $y=e^x-1$.
(a) Describe fully this transformation. [2]
(b) Find the $y$-intercept of each curve. [3]
Show complete worked solution
(a) Comparing $y=e^x-1$ with $y=e^x$: the function has been replaced by $f(x)-1$, so this is a vertical translation by the vector $\begin{pmatrix}0\\-1\end{pmatrix}$ (1 unit downward).
$$\boxed{\text{(a) Translation by the vector } \begin{pmatrix}0\\-1\end{pmatrix}}$$
(b) At $x=0$:
$$y=e^0=1 \ \text{ on the original curve}$$
$$y=e^0-1=1-1=0 \ \text{ on the image curve}$$
$$\boxed{\text{(b) } y=1 \text{ and } y=0}$$
QUESTION 23
4 marks
Easy
Given that $f(3)=8$ and that $g(x)=\dfrac{1}{2}f(x)$, find the value of $g(3)$.
Show complete worked solution
Using the fact that $y=\dfrac{1}{2}f(x)$ is a vertical stretch of $y=f(x)$ with scale factor $\dfrac{1}{2}$, relative to the $x$-axis: each $y$-coordinate is multiplied by $\dfrac{1}{2}$.
$$g(3) = \dfrac{1}{2}f(3) = \dfrac{1}{2}(8) = 4$$
$$\boxed{g(3)=4}$$
QUESTION 24
6 marks
Medium
Describe a sequence of transformations that maps the graph of $y=x^2$ onto the graph of $y=2(x-1)^2+3$.
Show complete worked solution
Rewriting the target function to identify each transformation applied to $y=x^2$, in order.
Starting with $y=x^2$:
Step 1 (horizontal translation by $\begin{pmatrix}1\\0\end{pmatrix}$):
$$y=(x-1)^2$$
Step 2 (vertical stretch with scale factor 2, relative to the $x$-axis):
$$y=2(x-1)^2$$
Step 3 (vertical translation by $\begin{pmatrix}0\\3\end{pmatrix}$):
$$y=2(x-1)^2+3$$
$$\boxed{\text{Translation } \begin{pmatrix}1\\0\end{pmatrix} \text{, then vertical stretch scale factor } 2 \text{, then translation } \begin{pmatrix}0\\3\end{pmatrix}}$$
QUESTION 25
7 marks
Medium
The diagram below shows the graph of $y=f(x)$, which has a local maximum point at $(1,3)$ and $x$-intercepts at $x=-1$ and $x=3$.
(a) Describe fully the transformation that maps $y=f(x)$ onto $y=-f(x)$. [2]
(b) State the coordinates of the image of the local maximum point. [2]
(c) State the $x$-intercepts of $y=-f(x)$. [3]
Show complete worked solution
(a) The graph of $y=-f(x)$ is obtained from $y=f(x)$ by a reflection in the $x$-axis: every $y$-coordinate is negated, while each $x$-coordinate is unchanged.
$$\boxed{\text{(a) Reflection in the } x\text{-axis}}$$
(b) Applying $y \to -y$ to the local maximum point $(1,3)$:
$$(1,\ -3)$$
Since the reflection turns a local maximum into a local minimum, this becomes a local minimum point.
$$\boxed{\text{(b) } (1,-3), \text{ a local minimum}}$$
(c) A reflection in the $x$-axis leaves $x$-intercepts unchanged, since if $f(x)=0$ then $-f(x)=0$ at the same $x$-values.
$$\boxed{\text{(c) } x=-1 \text{ and } x=3}$$
QUESTION 26
6 marks
Medium
Let $f(x)=x^2-4x$. The graph of $g$ is obtained by translating the graph of $f$ by the vector $\begin{pmatrix}2\\-3\end{pmatrix}$. Find $g(x)$, giving your answer in expanded and simplified form.
Show complete worked solution
Using the fact that a translation by $\begin{pmatrix}2\\-3\end{pmatrix}$ applied to $y=f(x)$ gives $y=f(x-2)-3$.
$$g(x) = f(x-2)-3 = (x-2)^2-4(x-2)-3$$
Expanding $(x-2)^2$:
$$(x-2)^2 = x^2-4x+4$$
Expanding $-4(x-2)$:
$$-4(x-2) = -4x+8$$
Combining all terms:
$$g(x) = x^2-4x+4-4x+8-3$$
Simplifying:
$$g(x) = x^2-8x+9$$
$$\boxed{g(x)=x^2-8x+9}$$
QUESTION 27
5 marks
Medium
The point $(-2,5)$ lies on the graph of $y=f(x)$. The graph of $h$ is defined by $h(x)=f(3x)$. Find the coordinates of the point on the graph of $h$ that corresponds to $(-2,5)$.
Show complete worked solution
Using the fact that $h(x)=f(3x)$ is a horizontal stretch of $y=f(x)$ with scale factor $\dfrac{1}{3}$, relative to the $y$-axis.
The point $(-2,5)$ on $f$ corresponds to the point on $h$ where the input to $f$ equals $-2$:
$$3x=-2 \implies x=-\dfrac{2}{3}$$
The $y$-coordinate is unchanged, since $h\left(-\dfrac{2}{3}\right) = f(3 \times (-\tfrac{2}{3})) = f(-2) = 5$.
$$\boxed{\left(-\dfrac{2}{3},\ 5\right)}$$
QUESTION 28
6 marks
Medium
The diagram below shows the graph of $y=f(x)$, which has $x$-intercepts at $x=-3$ and $x=1$. The graph of $g$ is defined by $g(x)=f(2x)$.
(a) Describe fully the transformation that maps $y=f(x)$ onto $y=g(x)$, and explain its effect on the $x$-coordinates of points on the curve. [3]
(b) Hence write down both $x$-intercepts of $y=g(x)$. [3]
Show complete worked solution
(a) The graph of $y=f(2x)$ is obtained from $y=f(x)$ by a horizontal stretch with scale factor $\dfrac{1}{2}$, relative to the $y$-axis.
Each point on the curve moves toward the $y$-axis: every $x$-coordinate is multiplied by $\dfrac{1}{2}$, while the $y$-coordinate is unchanged.
$$\boxed{\text{(a) Horizontal stretch, scale factor } \dfrac{1}{2}\text{, relative to the } y\text{-axis}}$$
(b) Applying $x \to \dfrac{1}{2}x$ to each $x$-intercept of $f$:
$$x=-3 \to x=\dfrac{1}{2}(-3) = -\dfrac{3}{2}$$
$$x=1 \to x=\dfrac{1}{2}(1) = \dfrac{1}{2}$$
$$\boxed{\text{(b) } x=-\dfrac{3}{2} \text{ and } x=\dfrac{1}{2}}$$
QUESTION 29
5 marks
Medium
The point $(4,-3)$ lies on the graph of $y=f(x)$. The graph of $g$ is defined by $g(x)=-2f(x)+5$. Find the coordinates of the point on the graph of $g$ that corresponds to $(4,-3)$.
Show complete worked solution
Using the fact that $g(x)=-2f(x)+5$ applies a vertical stretch with scale factor 2, then a reflection in the $x$-axis, then a vertical translation of 5 units upward, to the $y$-coordinate, while the $x$-coordinate is unchanged.
$$g(4) = -2f(4)+5 = -2(-3)+5 = 6+5=11$$
$$\boxed{(4,11)}$$
QUESTION 30
7 marks
Medium
The diagram below shows the graph of $y=\cos x$ for $0 \leq x \leq 2\pi$, together with the graph of its image, $y=3\cos x - 1$.
(a) State the two transformations, in the order in which they are applied to $y=\cos x$, that produce the image curve. [2]
(b) Write down the equation of the image curve (already given above; confirm by stating it). [2]
(c) State the amplitude and the range of the image curve. [3]
Show complete worked solution
(a) Comparing $y=3\cos x - 1$ with $y=\cos x$:
Step 1: a vertical stretch with scale factor 3, relative to the $x$-axis, giving $y=3\cos x$.
Step 2: a vertical translation by $\begin{pmatrix}0\\-1\end{pmatrix}$ (1 unit downward), giving $y=3\cos x - 1$.
$$\boxed{\text{(a) Vertical stretch scale factor } 3\text{, then translation } \begin{pmatrix}0\\-1\end{pmatrix}}$$
(b) $$\boxed{\text{(b) } y=3\cos x - 1}$$
(c) The amplitude of $y=\cos x$ is 1. A vertical stretch with scale factor 3 multiplies the amplitude by 3, so the new amplitude is 3. A vertical translation does not change the amplitude.
Since $-1 \leq \cos x \leq 1$, multiplying by 3 gives $-3 \leq 3\cos x \leq 3$, then subtracting 1 gives $-4 \leq 3\cos x - 1 \leq 2$.
$$\boxed{\text{(c) Amplitude } = 3, \text{ range } = [-4,2]}$$
QUESTION 31
6 marks
Medium
The graph of $y=f(x)$ has a horizontal asymptote $y=2$ as $x \to \infty$.
(a) State the equation of the horizontal asymptote of $y=f(x)-5$. [3]
(b) State the equation of the horizontal asymptote of $y=3f(x)+1$. [3]
Show complete worked solution
(a) A vertical translation of $\begin{pmatrix}0\\-5\end{pmatrix}$ moves every point on the curve, including points approaching the asymptote, down by 5 units. The asymptote value is therefore also translated down by 5.
$$y=2-5=-3$$
$$\boxed{\text{(a) } y=-3}$$
(b) The transformation $y \to 3y+1$ (a vertical stretch scale factor 3, then a translation up 1) applies to the asymptote value in the same way as to any other $y$-value approached by the curve.
$$y=3(2)+1=6+1=7$$
$$\boxed{\text{(b) } y=7}$$
QUESTION 32
5 marks
Medium
Given that $f(2)=6$, and that $g(x)=f(x-5)$, find the value of $x$ for which $g(x)=6$.
Show complete worked solution
Using the fact that $g(x)=f(x-5)$: the output of $g$ at $x$ equals the output of $f$ at $x-5$.
We require $g(x)=6$, and we know $f(2)=6$, so we need the input to $f$ to equal 2:
$$x-5=2 \implies x=7$$
Checking: $g(7)=f(7-5)=f(2)=6$. Correct.
$$\boxed{x=7}$$
QUESTION 33
7 marks
Medium
The diagram below shows the graph of $y=f(x)$ where $f(x)=x^3-3x$, which has a local maximum point at $(-1,2)$ and a local minimum point at $(1,-2)$. The graph of $h$ is defined by $h(x)=f(x-1)+2$.
(a) Describe fully, in order, the two transformations that map $y=f(x)$ onto $y=h(x)$. [2]
(b) Find the coordinates of the local maximum point on the graph of $h$. [2]
(c) Find the coordinates of the local minimum point on the graph of $h$, and confirm it is still a local minimum. [3]
Show complete worked solution
(a) Comparing $h(x)=f(x-1)+2$ with $f(x)$:
Step 1: a horizontal translation by $\begin{pmatrix}1\\0\end{pmatrix}$ (1 unit to the right), giving $y=f(x-1)$.
Step 2: a vertical translation by $\begin{pmatrix}0\\2\end{pmatrix}$ (2 units upward), giving $y=f(x-1)+2$.
$$\boxed{\text{(a) Translation } \begin{pmatrix}1\\0\end{pmatrix} \text{, then translation } \begin{pmatrix}0\\2\end{pmatrix}}$$
(b) Applying $\begin{pmatrix}1\\2\end{pmatrix}$ overall to $(-1,2)$:
$$(-1+1,\ 2+2) = (0,4)$$
$$\boxed{\text{(b) } (0,4)}$$
(c) Applying $\begin{pmatrix}1\\2\end{pmatrix}$ to $(1,-2)$:
$$(1+1,\ -2+2) = (2,0)$$
Since both transformations are translations (no reflection or stretch is involved), the type of turning point is preserved, so $(2,0)$ is still a local minimum.
$$\boxed{\text{(c) } (2,0), \text{ a local minimum}}$$
QUESTION 34
6 marks
Medium
The diagram below shows part of the graph of $y=f(x)$, which passes through the point $(2,4)$. The graph of $g$ is defined by $g(x)=f\left(-\dfrac{x}{2}\right)$, and is also shown.
(a) Describe fully, in order, the two transformations that map $y=f(x)$ onto $y=g(x)$. [3]
(b) Find the coordinates of the point on $y=g(x)$ that corresponds to $(2,4)$. [3]
Show complete worked solution
(a) Writing $g(x)=f\left(-\dfrac{x}{2}\right) = f\left(-1 \times \dfrac{x}{2}\right)$:
Step 1: a horizontal stretch with scale factor 2, relative to the $y$-axis, giving $y=f\left(\dfrac{x}{2}\right)$.
Step 2: a reflection in the $y$-axis, giving $y=f\left(-\dfrac{x}{2}\right)$.
(Since both transformations act on the $x$-coordinate only and horizontal stretches/reflections in the $y$-axis commute, this order is equivalent to applying them in the reverse order.)
$$\boxed{\text{(a) Horizontal stretch scale factor } 2 \text{, then reflection in the } y\text{-axis}}$$
(b) The point $(2,4)$ on $f$ corresponds to the point on $g$ where the input to $f$ equals 2:
$$-\dfrac{x}{2}=2 \implies x=-4$$
The $y$-coordinate is unchanged, since $g(-4)=f\left(-\dfrac{-4}{2}\right)=f(2)=4$.
$$\boxed{\text{(b) } (-4,4)}$$
QUESTION 35
6 marks
Medium
Given that $f(5)=-3$, and that $h(x)=f(x-3)+1$, find the value of $x$ for which $h(x)=-2$.
Show complete worked solution
Using the fact that $h(x)=f(x-3)+1$.
We require $h(x)=-2$:
$$f(x-3)+1=-2 \implies f(x-3)=-3$$
Since $f(5)=-3$, the input to $f$ must equal 5:
$$x-3=5 \implies x=8$$
Checking: $h(8)=f(8-3)+1=f(5)+1=-3+1=-2$. Correct.
$$\boxed{x=8}$$
QUESTION 36
6 marks
Medium
Describe a sequence of transformations that maps the graph of $y=x^2$ onto the graph of $y=\left(\dfrac{x}{2}\right)^2-4$.
Show complete worked solution
Rewriting the target function to identify each transformation applied to $y=x^2$, in order.
Starting with $y=x^2$:
Step 1 (horizontal stretch with scale factor 2, relative to the $y$-axis):
$$y=\left(\dfrac{x}{2}\right)^2$$
Step 2 (vertical translation by $\begin{pmatrix}0\\-4\end{pmatrix}$):
$$y=\left(\dfrac{x}{2}\right)^2-4$$
$$\boxed{\text{Horizontal stretch scale factor } 2\text{, then translation } \begin{pmatrix}0\\-4\end{pmatrix}}$$
QUESTION 37
6 marks
Medium
The graph of $y=e^x$ is transformed to the graph of $y=e^{x+3}$.
(a) Describe fully this transformation. [2]
(b) Find the $y$-intercept of the image curve. [2]
(c) Find the value of the image function at $x=-3$. [2]
Show complete worked solution
(a) Comparing $y=e^{x+3}$ with $y=e^x$: the function has been replaced by $f(x+3)$, so this is a horizontal translation by the vector $\begin{pmatrix}-3\\0\end{pmatrix}$ (3 units to the left).
$$\boxed{\text{(a) Translation by the vector } \begin{pmatrix}-3\\0\end{pmatrix}}$$
(b) At $x=0$:
$$y=e^{0+3}=e^3$$
$$\boxed{\text{(b) } y=e^3}$$
(c) At $x=-3$:
$$y=e^{-3+3}=e^0=1$$
$$\boxed{\text{(c) } y=1}$$
QUESTION 38
8 marks
Medium
The diagram below shows the graph of $y=f(x)$, where $f(x)=(x-2)^2-5$, which has a minimum point at $(2,-5)$ and passes through $(0,-1)$. The graph of $j$ is obtained by reflecting the graph of $f$ in the $y$-axis, and then translating the result 3 units upward.
(a) Describe, in order, the two transformations used to obtain $j$, and hence write down $j(x)$ in terms of $f$. [3]
(b) Find the coordinates of the minimum point on the graph of $j$. [3]
(c) Find the $y$-intercept of the graph of $j$. [2]
Show complete worked solution
(a) Step 1: reflection in the $y$-axis, giving $y=f(-x)$.
Step 2: translation by $\begin{pmatrix}0\\3\end{pmatrix}$ (3 units upward), giving $y=f(-x)+3$.
$$\boxed{\text{(a) } j(x)=f(-x)+3}$$
(b) Applying the reflection in the $y$-axis to the minimum point $(2,-5)$: the $x$-coordinate is negated.
$$(2,-5) \to (-2,-5)$$
Applying the translation upward by 3:
$$(-2,-5) \to (-2,\ -5+3) = (-2,-2)$$
$$\boxed{\text{(b) } (-2,-2)}$$
(c) The $y$-intercept of $f$ is $(0,-1)$. Since $x=0$ is unchanged by a reflection in the $y$-axis, this point maps to $(0,-1)$, and then the translation adds 3 to the $y$-coordinate:
$$-1+3=2$$
$$\boxed{\text{(c) } y=2}$$
QUESTION 39
4 marks
Medium
A curve $y=f(x)$ has a line of symmetry $x=-1$. Find the line of symmetry of the graph of $y=f(x+5)$.
Show complete worked solution
Using the fact that $y=f(x+5)$ is a horizontal translation of $y=f(x)$ by $\begin{pmatrix}-5\\0\end{pmatrix}$ (5 units to the left): every point on the curve, including the line of symmetry, shifts 5 units to the left.
$$x=-1-5=-6$$
$$\boxed{x=-6}$$
QUESTION 40
7 marks
Medium
The graph of $y=f(x)$ is transformed to the graph of $y=f(2x-6)$.
(a) By writing $f(2x-6)$ in the form $f(2(x-c))$ for a constant $c$, express this as a composition of two transformations, stated in the correct order. [4]
(b) Verify your answer to part (a) by finding the image of the point $(3, f(3))$ under your two transformations, and checking it satisfies $y=f(2x-6)$ at the corresponding $x$-value. [3]
Show complete worked solution
(a) Factorising: $f(2x-6) = f(2(x-3))$, so $c=3$.
The correct order for a composite horizontal transformation of this form is to apply the stretch first, then the translation:
Step 1: a horizontal stretch with scale factor $\dfrac{1}{2}$, relative to the $y$-axis, applied to $y=f(x)$ gives $y=f(2x)$.
Step 2: a horizontal translation by $\begin{pmatrix}3\\0\end{pmatrix}$ (3 units to the right), replacing $x$ by $x-3$ in $y=f(2x)$, gives $y=f(2(x-3))=f(2x-6)$.
$$\boxed{\text{(a) Horizontal stretch scale factor } \dfrac{1}{2}\text{, then translation } \begin{pmatrix}3\\0\end{pmatrix}}$$
(b) Applying Step 1 to $(3,f(3))$: the $x$-coordinate is multiplied by $\dfrac{1}{2}$, giving $\left(\dfrac{3}{2}, f(3)\right)$.
Applying Step 2: add 3 to the $x$-coordinate, giving $\left(\dfrac{3}{2}+3,\ f(3)\right) = \left(\dfrac{9}{2},\ f(3)\right)$.
Checking directly: at $x=\dfrac{9}{2}$, $f(2x-6) = f\left(2 \times \dfrac{9}{2} - 6\right) = f(9-6) = f(3)$, which matches.
$$\boxed{\text{(b) } \left(\dfrac{9}{2},\ f(3)\right), \text{ confirmed}}$$
QUESTION 41
6 marks
Medium
The point $(-6,10)$ lies on the graph of $y=f(x)$. The graph of $m$ is defined by $m(x)=\dfrac{1}{2}f(3x)$. Find the coordinates of the point on the graph of $m$ that corresponds to $(-6,10)$.
Show complete worked solution
Using the fact that $m(x)=\dfrac{1}{2}f(3x)$ applies a horizontal stretch with scale factor $\dfrac{1}{3}$ to the $x$-coordinate and a vertical stretch with scale factor $\dfrac{1}{2}$ to the $y$-coordinate.
The input to $f$ equals $-6$ when:
$$3x=-6 \implies x=-2$$
The corresponding $y$-coordinate:
$$m(-2) = \dfrac{1}{2}f(3 \times -2) = \dfrac{1}{2}f(-6) = \dfrac{1}{2}(10) = 5$$
$$\boxed{(-2,5)}$$
QUESTION 42
9 marks
Hard
The graph of $y=f(x)$ has a local minimum point at $(2,-5)$, a local maximum point at $(-1,3)$, and passes through the origin $(0,0)$. The function $g$ is defined by $g(x)=-3f(x+2)-4$.
(a) Find the coordinates of the image of the local minimum point and the image of the local maximum point under this transformation. [4]
(b) State, with a reason, whether each image point found in part (a) is now a local maximum or a local minimum of $g$. [3]
(c) Find the coordinates of the point on the graph of $g$ that corresponds to the origin on the graph of $f$. [2]
Show complete worked solution
(a) Writing $g(x)=-3f(x+2)-4$: the input to $f$ equals a given value when $x+2$ equals that value, i.e. $x = (\text{value}) - 2$.
For the local minimum $(2,-5)$: the input to $f$ equals 2 when $x+2=2 \implies x=0$.
$$g(0) = -3f(2)-4 = -3(-5)-4 = 15-4=11$$
Image of local minimum: $(0,11)$.
For the local maximum $(-1,3)$: the input to $f$ equals $-1$ when $x+2=-1 \implies x=-3$.
$$g(-3) = -3f(-1)-4 = -3(3)-4 = -9-4=-13$$
Image of local maximum: $(-3,-13)$.
$$\boxed{\text{(a) } (0,11) \text{ and } (-3,-13)}$$
(b) The horizontal translation ($x \to x+2$) does not affect the type of turning point. The factor of $-3$ consists of a vertical stretch (which preserves the type) combined with a reflection in the $x$-axis (which reverses the type), since the overall multiplier is negative.
Since the multiplier is negative, each turning point changes type: the local minimum of $f$ becomes a local maximum of $g$, and the local maximum of $f$ becomes a local minimum of $g$.
$$\boxed{\text{(b) } (0,11) \text{ is a local maximum; } (-3,-13) \text{ is a local minimum}}$$
(c) The input to $f$ equals 0 when $x+2=0 \implies x=-2$.
$$g(-2) = -3f(0)-4 = -3(0)-4=-4$$
$$\boxed{\text{(c) } (-2,-4)}$$
QUESTION 43
9 marks
Hard
Let $f(x)=x^2-2x$. The graph of $g$ is obtained from the graph of $f$ by: a vertical stretch with scale factor 2, relative to the $x$-axis; then a reflection in the $x$-axis; then a horizontal translation of 3 units to the left.
(a) Find $g(x)$, giving your answer as a simplified polynomial in $x$. [5]
(b) Find the coordinates of the vertex of the graph of $g$. [4]
Show complete worked solution
(a) Applying the transformations to $f(x)$ in order.
Step 1 (vertical stretch scale factor 2):
$$2f(x) = 2x^2-4x$$
Step 2 (reflection in the $x$-axis):
$$-(2x^2-4x) = -2x^2+4x$$
Step 3 (horizontal translation 3 units left, replacing $x$ with $x+3$):
$$g(x) = -2(x+3)^2+4(x+3)$$
Expanding $(x+3)^2 = x^2+6x+9$:
$$-2(x^2+6x+9) = -2x^2-12x-18$$
Expanding $4(x+3) = 4x+12$:
$$g(x) = -2x^2-12x-18+4x+12$$
Combining like terms:
$$\boxed{\text{(a) } g(x) = -2x^2-8x-6}$$
(b) Method: completing the square, or applying the three transformations directly to the vertex of $f$.
The vertex of $f(x)=x^2-2x=(x-1)^2-1$ is $(1,-1)$.
Step 1 (vertical stretch scale factor 2): $(1,-1) \to (1,-2)$.
Step 2 (reflection in the $x$-axis): $(1,-2) \to (1,2)$.
Step 3 (translation 3 units left): $(1,2) \to (-2,2)$.
Checking using the expanded form: $g(-2)=-2(4)-8(-2)-6=-8+16-6=2$, confirming the vertex has $y=2$.
$$\boxed{\text{(b) } (-2,2)}$$
QUESTION 44
13 marks
Hard
The diagram below shows the graph of $y=f(x)$, which has a local maximum point at $(-2,4)$, a local minimum point at $(1,-2)$, and a $y$-intercept at $(0,1)$. The graph of $h$ is defined by $h(x)=f(2x-4)$.
(a) By writing $2x-4$ in the form $2(x-c)$, express $h(x)=f(2x-4)$ as a composition of two transformations applied to $y=f(x)$, stated in the correct order. [4]
(b) Find the coordinates of the point on $y=h(x)$ that corresponds to the local maximum point of $f$. [3]
(c) Find the coordinates of the point on $y=h(x)$ that corresponds to the $y$-intercept of $f$. [3]
(d) Find the coordinates of the point on $y=h(x)$ that corresponds to the local minimum point of $f$. [3]
Show complete worked solution
(a) Factorising: $f(2x-4) = f(2(x-2))$.
Step 1: a horizontal stretch with scale factor $\dfrac{1}{2}$, relative to the $y$-axis, giving $y=f(2x)$.
Step 2: a horizontal translation by $\begin{pmatrix}2\\0\end{pmatrix}$ (2 units to the right), giving $y=f(2(x-2))=f(2x-4)$.
$$\boxed{\text{(a) Horizontal stretch scale factor } \dfrac{1}{2}\text{, then translation } \begin{pmatrix}2\\0\end{pmatrix}}$$
(b) For each feature point $(p,q)$ on $f$, the corresponding point on $h$ satisfies $2x-4=p$, i.e. $x=\dfrac{p+4}{2}$, with the same $y$-value $q$.
For the local maximum $(-2,4)$:
$$x=\dfrac{-2+4}{2}=\dfrac{2}{2}=1$$
$$\boxed{\text{(b) } (1,4)}$$
(c) For the $y$-intercept $(0,1)$:
$$x=\dfrac{0+4}{2}=\dfrac{4}{2}=2$$
$$\boxed{\text{(c) } (2,1)}$$
(d) For the local minimum $(1,-2)$:
$$x=\dfrac{1+4}{2}=\dfrac{5}{2}$$
$$\boxed{\text{(d) } \left(\dfrac{5}{2},-2\right)}$$
QUESTION 45
9 marks
Hard
The graph of $y=\sin x$ is transformed by a horizontal stretch with scale factor 3, relative to the $y$-axis, followed by a translation with vector $\begin{pmatrix}\pi/2\\-1\end{pmatrix}$.
(a) Write down the equation of the new curve. [3]
(b) State the amplitude, the period, and the equation of the principal axis (midline) of the new curve. [4]
(c) Find the range of the new curve. [2]
Show complete worked solution
(a) Step 1 (horizontal stretch scale factor 3, relative to the $y$-axis) applied to $y=\sin x$:
$$y=\sin\left(\dfrac{x}{3}\right)$$
Step 2 (translation $\begin{pmatrix}\pi/2\\-1\end{pmatrix}$), replacing $x$ with $x-\dfrac{\pi}{2}$ and subtracting 1:
$$y=\sin\left(\dfrac{x-\pi/2}{3}\right)-1$$
$$\boxed{\text{(a) } y=\sin\left(\dfrac{x-\pi/2}{3}\right)-1}$$
(b) A vertical stretch has not been applied, so the amplitude of $\sin x$, which is 1, is unchanged.
The period of $y=\sin x$ is $2\pi$. A horizontal stretch with scale factor 3 multiplies the period by 3:
$$\text{Period} = 2\pi \times 3 = 6\pi$$
The translation of $-1$ in the $y$-direction moves the principal axis from $y=0$ to $y=-1$.
$$\boxed{\text{(b) Amplitude} = 1, \text{ period} = 6\pi, \text{ principal axis: } y=-1}$$
(c) Since $-1 \leq \sin\left(\dfrac{x-\pi/2}{3}\right) \leq 1$, subtracting 1 throughout:
$$-2 \leq \sin\left(\dfrac{x-\pi/2}{3}\right)-1 \leq 0$$
$$\boxed{\text{(c) Range} = [-2,0]}$$
QUESTION 46
7 marks
Hard
The point $(a,b)$ lies on the graph of the one-to-one function $y=f(x)$. The graph of $g$ is defined by $g(x)=2f(3x-6)+7$. Express, in terms of $a$ and $b$, the coordinates of the point on the graph of $g$ that corresponds to $(a,b)$ on $f$. Show your reasoning fully.
Show complete worked solution
Writing $g(x)=2f(3x-6)+7 = 2f(3(x-2))+7$ by factorising $3x-6=3(x-2)$.
The input to $f$ equals $a$ (so that $f$ outputs $b$) when:
$$3(x-2)=a$$
Solving for $x$:
$$x-2=\dfrac{a}{3} \implies x=2+\dfrac{a}{3}$$
At this value of $x$, using $f(a)=b$:
$$g(x) = 2f(a)+7 = 2b+7$$
So the point $(a,b)$ on $f$ corresponds to the point on $g$ with $x$-coordinate $2+\dfrac{a}{3}$ and $y$-coordinate $2b+7$.
$$\boxed{\left(2+\dfrac{a}{3},\ 2b+7\right)}$$
QUESTION 47
8 marks
Hard
The graph of $y=f(x)$ passes through the points $A(-3,2)$ and $B(1,-4)$. Under a single transformation of the form $g(x)=pf(x)+q$, the graph of $f$ maps onto the graph of $g$, with $A$ mapping to $A^{\prime}(-3,5)$ and $B$ mapping to $B^{\prime}(1,-19)$.
(a) Using the given information, write down two equations satisfied by $p$ and $q$. [2]
(b) Solve your equations to find the values of $p$ and $q$. [3]
(c) The graph of $f$ also passes through the point $C(4,0)$. Find the coordinates of the image of $C$ under this transformation. [3]
Show complete worked solution
(a) Since $g(x)=pf(x)+q$, and using $f(-3)=2$ with image $y$-coordinate 5, and $f(1)=-4$ with image $y$-coordinate $-19$:
$$p(2)+q=5$$
$$p(-4)+q=-19$$
$$\boxed{\text{(a) } 2p+q=5 \text{ and } -4p+q=-19}$$
(b) Subtracting the second equation from the first:
$$(2p+q)-(-4p+q) = 5-(-19)$$
$$6p=24 \implies p=4$$
Substituting into $2p+q=5$:
$$2(4)+q=5 \implies 8+q=5 \implies q=-3$$
$$\boxed{\text{(b) } p=4, \ q=-3}$$
(c) Using $g(x)=4f(x)-3$, and $f(4)=0$ (since $C(4,0)$ lies on $f$):
$$g(4) = 4f(4)-3 = 4(0)-3=-3$$
$$\boxed{\text{(c) } C^{\prime}(4,-3)}$$
QUESTION 48
9 marks
Hard
The graph of $y=x^3$ is transformed to the graph of $y=-\dfrac{1}{2}(x+1)^3-3$.
(a) Describe fully the sequence of transformations, in the correct order, that maps $y=x^3$ onto this curve. [6]
(b) The point $(2,8)$ lies on $y=x^3$. Using your transformations from part (a), find the coordinates of the corresponding point on the transformed curve. [3]
Show complete worked solution
(a) Rewriting the target function to identify each transformation applied to $y=x^3$, in order.
Starting with $y=x^3$:
Step 1 (horizontal translation by $\begin{pmatrix}-1\\0\end{pmatrix}$):
$$y=(x+1)^3$$
Step 2 (vertical stretch with scale factor $\dfrac{1}{2}$, combined with a reflection in the $x$-axis, i.e. an overall vertical scale factor of $-\dfrac{1}{2}$):
$$y=-\dfrac{1}{2}(x+1)^3$$
Step 3 (vertical translation by $\begin{pmatrix}0\\-3\end{pmatrix}$):
$$y=-\dfrac{1}{2}(x+1)^3-3$$
$$\boxed{\text{(a) Translation } \begin{pmatrix}-1\\0\end{pmatrix}\text{, then vertical stretch scale factor } \dfrac{1}{2}\text{ with reflection in the } x\text{-axis, then translation } \begin{pmatrix}0\\-3\end{pmatrix}}$$
(b) Applying Step 1 to $(2,8)$: the $x$-coordinate decreases by 1.
$$(2-1,\ 8) = (1,8)$$
Applying Step 2: the $y$-coordinate is multiplied by $-\dfrac{1}{2}$.
$$\left(1,\ 8 \times -\dfrac{1}{2}\right) = (1,-4)$$
Applying Step 3: the $y$-coordinate decreases by 3.
$$(1,\ -4-3) = (1,-7)$$
Checking directly: at $x=1$, $y=-\dfrac{1}{2}(1+1)^3-3 = -\dfrac{1}{2}(8)-3=-4-3=-7$. Confirmed.
$$\boxed{\text{(b) } (1,-7)}$$
QUESTION 49
12 marks
Hard
The diagram below shows the graph of $y=f(x)$, a downward-opening curve with a maximum point at $(1,6)$ and $x$-intercepts (roots) at $x=-2$ and $x=4$. The graph of $n$ is defined by $n(x)=f(4-x)$.
(a) By writing $4-x$ in the form $-(x-c)$, express $n(x)=f(4-x)$ as a composition of a reflection and a translation, applied to $y=f(x)$ in the correct order. [4]
(b) Find the coordinates of the maximum point of $n$. [3]
(c) Find both roots of $n$. [3]
(d) Write down the range of $n$. [2]
Show complete worked solution
(a) Writing $4-x = -(x-4)$.
Step 1: a reflection in the $y$-axis, applied to $y=f(x)$, giving $y=f(-x)$.
Step 2: a horizontal translation by $\begin{pmatrix}4\\0\end{pmatrix}$ (4 units to the right), replacing $x$ by $x-4$ in $y=f(-x)$, giving $y=f(-(x-4))=f(4-x)$.
$$\boxed{\text{(a) Reflection in the } y\text{-axis, then translation } \begin{pmatrix}4\\0\end{pmatrix}}$$
(b) Applying Step 1 to the maximum point $(1,6)$: the $x$-coordinate is negated.
$$(1,6) \to (-1,6)$$
Applying Step 2: 4 is added to the $x$-coordinate.
$$(-1,6) \to (3,6)$$
$$\boxed{\text{(b) } (3,6)}$$
(c) Applying Step 1 to the roots $x=-2$ and $x=4$: each $x$-coordinate is negated, giving $x=2$ and $x=-4$.
Applying Step 2: 4 is added to each $x$-coordinate, giving $x=2+4=6$ and $x=-4+4=0$.
$$\boxed{\text{(c) } x=0 \text{ and } x=6}$$
(d) Neither a reflection in the $y$-axis nor a horizontal translation changes any $y$-coordinate on the curve, so the set of $y$-values taken by the curve, i.e. the range, is unchanged from that of $f$. Since $f$ has a maximum value of 6 and opens downward (with no minimum), the range of $f$, and hence of $n$, is $y \leq 6$.
$$\boxed{\text{(d) } y \leq 6}$$
QUESTION 50
8 marks
Hard
The functions $f$ and $g$ satisfy $g(x)=f(px+q)$ for constants $p$ and $q$. The graph of $g$ is obtained from the graph of $f$ by a horizontal stretch with scale factor $\dfrac{1}{4}$, relative to the $y$-axis, followed by a horizontal translation of 5 units to the right.
(a) Find the values of $p$ and $q$. [3]
(b) Hence write down $g(x)$ in terms of $f$. [2]
(c) Given that $f$ has a root at $x=13$, find the corresponding root of $g$. [3]
Show complete worked solution
(a) Step 1 (horizontal stretch scale factor $\dfrac{1}{4}$, relative to the $y$-axis) applied to $y=f(x)$:
$$y=f(4x)$$
Step 2 (horizontal translation of 5 units to the right), replacing $x$ by $x-5$:
$$y=f(4(x-5)) = f(4x-20)$$
Comparing $f(4x-20)$ with $f(px+q)$:
$$\boxed{\text{(a) } p=4, \ q=-20}$$
(b) $$\boxed{\text{(b) } g(x)=f(4x-20)}$$
(c) Since $f$ has a root at $x=13$, i.e. $f(13)=0$, we require the input to $f$ to equal 13:
$$4x-20=13$$
$$4x=33$$
$$x=\dfrac{33}{4}$$
Checking: $g\left(\dfrac{33}{4}\right) = f\left(4 \times \dfrac{33}{4}-20\right) = f(33-20)=f(13)=0$. Confirmed.
$$\boxed{\text{(c) } x=\dfrac{33}{4}}$$
Function Sketching 50 questions
QUESTION 1
3 marks
Easy
The graph of \(y = f(x)\), where \(f(x) = x^2\), is transformed to the graph of \(y = f(x-3) + 2\).
Write down the resulting function in the form \(y = x^2 + bx + c\).
Show complete worked solution
Applying the transformation $f(x-3)+2$ (a horizontal translation of 3, then a vertical translation of 2) to $f(x)=x^2$:
$$y = (x-3)^2 + 2$$
Expanding:
$$y = x^2 - 6x + 9 + 2$$
$$\boxed{y = x^2 - 6x + 11}$$
QUESTION 2
4 marks
Easy
A ball is thrown from the top of a cliff. Its height above the ground, \(h\) metres, after \(t\) seconds, is modelled by
\[ h(t) = -5t^2 + 20t + 25, \quad t \geq 0 \]
(a) Write down the height of the cliff.
(b) Find the time at which the ball hits the ground.
Show complete worked solution
**(a)**
The height of the cliff is the height of the ball at $t=0$:
$$h(0) = -5(0)^2 + 20(0) + 25 = 25$$
$$\boxed{25 \text{ m}}$$
**(b)**
The ball hits the ground when $h(t) = 0$:
$$-5t^2 + 20t + 25 = 0$$
Dividing through by $-5$:
$$t^2 - 4t - 5 = 0$$
Factorising:
$$(t-5)(t+1) = 0$$
$$t = 5 \text{ or } t = -1$$
Rejecting $t=-1$ since $t \geq 0$:
$$\boxed{t = 5 \text{ seconds}}$$
QUESTION 3
4 marks
Medium
The graph of \(y = g(x)\) has a local minimum point at \((4, -3)\).
The graph of \(y=g(x)\) is transformed to the graph of \(y = 2g(x-1) + 5\).
Find the coordinates of the corresponding point on the transformed graph, and state whether it is a local minimum or a local maximum.
Show complete worked solution
The transformation $x \to x-1$ inside the function is a horizontal translation of 1 unit in the positive $x$-direction, so the $x$-coordinate of the image point satisfies:
$$x - 1 = 4$$
$$x = 5$$
The transformation $2g(\cdot)+5$ is a vertical stretch with scale factor 2 followed by a vertical translation of 5, applied to the $y$-coordinate:
$$y = 2(-3) + 5 = -1$$
A positive vertical stretch and a vertical translation do not change the nature of a turning point, so the image is still a local minimum.
$$\boxed{(5, -1), \text{ a local minimum}}$$
QUESTION 4
5 marks
Medium
Let \(f(x) = x^3 - 4x^2 + 2x + 5\) and \(g(x) = -x + 3\).
Using your graphic display calculator, find the coordinates of all points of intersection of the graphs of \(f\) and \(g\), giving your answers correct to three significant figures.
Show complete worked solution
Setting $f(x) = g(x)$ to find the points of intersection:
$$x^3 - 4x^2 + 2x + 5 = -x+3$$
$$x^3 - 4x^2 + 3x + 2 = 0$$
Using the GDC to graph $y = x^3-4x^2+3x+2$ and find the roots (or noting that $x=2$ is an exact root and factorising):
$$x^3-4x^2+3x+2 = (x-2)(x^2-2x-1)$$
Solving $x^2 - 2x - 1 = 0$ using the quadratic formula:
$$x = \frac{2 \pm \sqrt{4+4}}{2} = 1 \pm \sqrt{2}$$
So the three $x$-values of intersection are:
$$x = 2, \quad x = 1+\sqrt{2} \approx 2.41421, \quad x = 1-\sqrt{2} \approx -0.41421$$
Substituting each into $g(x) = -x+3$ to find the corresponding $y$-values:
$$x=2: \ y = 1$$
$$x = 1+\sqrt2: \ y = 3-(1+\sqrt2) = 2-\sqrt2 \approx 0.586$$
$$x = 1-\sqrt2: \ y = 3-(1-\sqrt2) = 2+\sqrt2 \approx 3.41$$
$$\boxed{(2, 1), \ (2.41, 0.586), \ (-0.414, 3.41) \text{ (3 s.f.)}}$$
QUESTION 5
6 marks
Hard
The number of insects in a colony, \(P\), after \(t\) weeks is modelled by
\[ P(t) = \frac{200}{1+9e^{-0.5t}}, \quad t \geq 0 \]
(a) Find \(P(0)\) and interpret this value in context.
(b) Find the value of \(t\) at which the population first reaches 150, giving your answer correct to two decimal places.
(c) With reference to the shape of the graph of \(P\), describe how the growth rate of the population changes as \(t\) increases from 0.
Show complete worked solution
**(a)**
Substituting $t=0$:
$$P(0) = \frac{200}{1+9e^{0}} = \frac{200}{1+9} = \frac{200}{10} = 20$$
$$\boxed{P(0) = 20\text{: the colony starts with 20 insects}}$$
**(b)**
Setting $P(t) = 150$:
$$\frac{200}{1+9e^{-0.5t}} = 150$$
Rearranging:
$$1 + 9e^{-0.5t} = \frac{200}{150} = \frac{4}{3}$$
$$9e^{-0.5t} = \frac{1}{3}$$
$$e^{-0.5t} = \frac{1}{27}$$
Taking $\ln$ of both sides:
$$-0.5t = \ln\left(\frac{1}{27}\right) = -\ln 27$$
Solving for $t$:
$$t = 2\ln 27 \approx 6.5917$$
$$\boxed{t \approx 6.59 \text{ weeks}}$$
**(c)**
The graph of $P$ is S-shaped (sigmoid): it is concave up and increasingly steep for small $t$, meaning the growth rate increases; it is steepest around $P=100$ (the point of inflection); after this the graph becomes concave down and flattens as $P$ approaches the upper horizontal asymptote at $P=200$, meaning the growth rate decreases towards zero.
$$\boxed{\text{Growth rate increases, reaches a maximum around } P=100\text{, then decreases towards 0}}$$
QUESTION 6
3 marks
Easy
Let \(f(x) = (x-3)(x+2)\).
Find the \(x\)-intercepts and the \(y\)-intercept of the graph of \(f\).
Show complete worked solution
Using the fact that $x$-intercepts occur where $f(x)=0$:
$$(x-3)(x+2) = 0 \implies x = 3 \text{ or } x = -2$$
$$\boxed{x\text{-intercepts: } (3,0) \text{ and } (-2, 0)}$$
Using the fact that the $y$-intercept occurs where $x=0$:
$$f(0) = (0-3)(0+2) = (-3)(2) = -6$$
$$\boxed{y\text{-intercept: } (0,-6)}$$
QUESTION 7
3 marks
Easy
Let \(f(x) = \dfrac{3}{x+4} - 2\).
Write down the equations of the vertical and horizontal asymptotes of the graph of \(f\).
Show complete worked solution
Using the fact that the vertical asymptote occurs where the denominator is zero:
$$x + 4 = 0 \implies x = -4$$
$$\boxed{\text{Vertical asymptote: } x = -4}$$
Using the fact that as $x \to \pm\infty$, $\dfrac{3}{x+4} \to 0$, so $f(x) \to -2$:
$$\boxed{\text{Horizontal asymptote: } y = -2}$$
QUESTION 8
4 marks
Medium
Let \(f(x) = -3x^2 + 12x - 7\).
By completing the square, find the coordinates of the turning point of the graph of \(f\), and state whether it is a maximum or a minimum.
Show complete worked solution
Completing the square, first factoring out $-3$ from the terms in $x$:
$$f(x) = -3(x^2 - 4x) - 7$$
Completing the square inside the brackets:
$$x^2 - 4x = (x-2)^2 - 4$$
Substituting back:
$$f(x) = -3[(x-2)^2 - 4] - 7 = -3(x-2)^2 + 12 - 7 = -3(x-2)^2 + 5$$
The vertex form $f(x) = -3(x-2)^2+5$ gives the turning point directly:
$$\boxed{\text{Turning point: } (2, 5)}$$
Since the coefficient of $(x-2)^2$ is $-3 < 0$, the parabola opens downward:
$$\boxed{\text{This is a maximum point}}$$
QUESTION 9
5 marks
Medium
Let \(f(x) = x + \dfrac{4}{x}\), \(x \neq 0\).
(a) Write down the equation of the vertical asymptote of the graph of \(f\).
(b) Using your graphic display calculator, find the coordinates of the local minimum point (for \(x>0\)) and the local maximum point (for \(x<0\)).
Show complete worked solution
**(a)**
The function is undefined at $x=0$, giving a vertical asymptote there:
$$\boxed{x = 0}$$
**(b)**
Graphing $y = x + \dfrac{4}{x}$ on the GDC: for $x>0$, the graph decreases then increases, giving a local minimum; for $x<0$, it increases then decreases, giving a local maximum. Reading from the graph, the turning points occur at $x = 2$ and $x=-2$. Substituting these values:
$$f(2) = 2 + \frac{4}{2} = 2+2=4$$
$$f(-2) = -2 + \frac{4}{-2} = -2-2=-4$$
$$\boxed{\text{Local minimum: } (2,4), \quad \text{Local maximum: } (-2,-4)}$$
QUESTION 10
6 marks
Hard
The graph of \(f(x) = 2x^2 + bx + c\) has axis of symmetry \(x = 3\) and \(y\)-intercept \((0,-5)\).
(a) Find the values of \(b\) and \(c\).
(b) Find the value of \(f(5)\).
(c) Find the coordinates of the vertex of the graph of \(f\).
Show complete worked solution
**(a)**
Using the fact that the axis of symmetry of $ax^2+bx+c$ is $x = -\dfrac{b}{2a}$, with $a=2$:
$$-\frac{b}{2(2)} = 3 \implies -\frac{b}{4}=3 \implies b = -12$$
The $y$-intercept gives $c$ directly, since $f(0)=c$:
$$c = -5$$
$$\boxed{b=-12, \ c=-5}$$
So $f(x) = 2x^2 - 12x - 5$.
**(b)**
Substituting $x=5$:
$$f(5) = 2(25) - 12(5) - 5 = 50-60-5 = -15$$
$$\boxed{f(5) = -15}$$
**(c)**
Since $x=3$ is the axis of symmetry, substituting $x=3$ gives the vertex:
$$f(3) = 2(9) - 12(3) - 5 = 18-36-5=-23$$
$$\boxed{\text{Vertex: } (3, -23)}$$
QUESTION 11
4 marks
Easy
Solve the equation $x^2-4x+1=0$ analytically, giving your answers in the form $a\pm\sqrt{b}$, where $a,b \in \mathbb{Z}$.
Show complete worked solution
Using the quadratic formula, $x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$, with $a=1,\ b=-4,\ c=1$:
$$x = \frac{-(-4)\pm\sqrt{(-4)^2-4(1)(1)}}{2(1)}$$
Substituting and simplifying inside the square root:
$$x = \frac{4\pm\sqrt{16-4}}{2} = \frac{4\pm\sqrt{12}}{2}$$
Simplifying the surd, $\sqrt{12}=2\sqrt3$:
$$x = \frac{4\pm2\sqrt3}{2} = 2\pm\sqrt3$$
$$\boxed{x=2+\sqrt3 \text{ or } x=2-\sqrt3}$$
QUESTION 12
3 marks
Easy
Find, analytically, the coordinates of the point of intersection of the lines $y=2x-1$ and $y=-x+5$.
Show complete worked solution
Setting the expressions for $y$ equal, since both lines pass through the point of intersection:
$$2x-1=-x+5$$
Solving for $x$:
$$3x=6$$
$$x=2$$
Substituting back into $y=2x-1$:
$$y=2(2)-1=3$$
$$\boxed{(2,3)}$$
QUESTION 13
5 marks
Medium
Solve the equation $\dfrac{2}{x} = x-1$ analytically.
Show complete worked solution
Multiplying both sides by $x$ (noting $x \neq 0$) to clear the fraction:
$$2 = x(x-1) = x^2-x$$
Rearranging so one side is zero:
$$x^2-x-2=0$$
Factorising:
$$(x-2)(x+1)=0$$
$$x=2 \text{ or } x=-1$$
Both values satisfy $x \neq 0$, so both are valid solutions.
$$\boxed{x=2 \text{ or } x=-1}$$
QUESTION 14
6 marks
Medium
Let $f(x) = x^3-x^2-5x-3$.
(a) Use your graphic display calculator to find the $x$-intercepts of the graph of $f$ for $-3\le x\le4$.
(b) Show that $x=-1$ is a root of $f(x)=0$, and hence find all solutions of $f(x)=0$ analytically.
Show complete worked solution
**(a)**
Using a GDC to graph $y=x^3-x^2-5x-3$ over $-3 \le x \le 4$: the graph touches the $x$-axis at $x=-1$ (without crossing) and crosses the $x$-axis at $x=3$.
**(b)**
Substituting $x=-1$ into $f(x)$:
$$f(-1) = (-1)^3-(-1)^2-5(-1)-3 = -1-1+5-3=0$$
Since $f(-1)=0$, $x=-1$ is a root, so $(x+1)$ is a factor of $f(x)$. Dividing $f(x)$ by $(x+1)$ (e.g. by synthetic division with root $-1$, using coefficients $1,-1,-5,-3$) gives quotient $x^2-2x-3$:
$$f(x) = (x+1)(x^2-2x-3)$$
Factorising the quadratic factor:
$$x^2-2x-3=(x-3)(x+1)$$
So:
$$f(x) = (x+1)(x-3)(x+1) = (x+1)^2(x-3)$$
Setting $f(x)=0$:
$$x=-1 \text{ (repeated root) or } x=3$$
$$\boxed{x=-1 \text{ (double root)},\ x=3}$$
QUESTION 15
7 marks
Hard
The curves $y=\dfrac{6}{x}$ and $y=x^2-5$ intersect at one point where $x>0$.
(a) Show that the $x$-coordinate of this point satisfies $x^3-5x-6=0$.
(b) Use your graphic display calculator to find the coordinates of this point of intersection, giving your answers correct to 3 significant figures.
Show complete worked solution
**(a)**
At the point of intersection, $\dfrac{6}{x}=x^2-5$. Multiplying both sides by $x$ (valid since $x>0$):
$$6 = x(x^2-5) = x^3-5x$$
Rearranging so one side is zero:
$$x^3-5x-6=0, \text{ as required.}$$
**(b)**
Using a GDC to solve $x^3-5x-6=0$ (or graphing $y=6/x$ and $y=x^2-5$ and finding the intersection for $x>0$):
$$x=2.6893\ldots \approx 2.69$$
Substituting into $y=x^2-5$ to find the corresponding $y$-value:
$$y = (2.6893\ldots)^2-5 = 2.2325\ldots \approx 2.23$$
$$\boxed{(2.69,\ 2.23)}$$
QUESTION 16
6 marks
Easy
Let $f(x) = x^2 - 2x - 8$.
Find the $x$-intercepts and the $y$-intercept of the graph of $f$, and hence, by completing the square, find the coordinates of the vertex. Sketch the graph of $f$, labelling all of these features.
Show complete worked solution
Finding the $x$-intercepts, where $f(x) = 0$:
$$x^2-2x-8=0$$
Factorising:
$$(x-4)(x+2)=0$$
$$x=4 \text{ or } x=-2$$
$$\boxed{x\text{-intercepts: } (4,0) \text{ and } (-2,0)}$$
Finding the $y$-intercept, where $x=0$:
$$f(0) = 0-0-8=-8$$
$$\boxed{y\text{-intercept: } (0,-8)}$$
Completing the square:
$$f(x) = x^2-2x-8 = (x-1)^2-1-8=(x-1)^2-9$$
The vertex form $f(x)=(x-1)^2-9$ gives the turning point directly. Since the coefficient of $(x-1)^2$ is positive, the parabola opens upward, so this is a minimum:
$$\boxed{\text{Vertex (minimum): } (1,-9)}$$
The sketch is an upward-opening parabola through $(-2,0)$ and $(4,0)$, with $y$-intercept $(0,-8)$ and minimum turning point $(1,-9)$, as shown in the diagram.
QUESTION 17
5 marks
Easy
The diagram below shows the graph of a quadratic function $y=f(x)$, which has a maximum point at $(-2,5)$ and $y$-intercept $(0,1)$. The function can be written in the form $f(x) = a(x+2)^2+5$.
(a) Show that $a=-1$. [2]
(b) Hence find the $x$-intercepts of the graph of $f$, giving your answers in the form $p \pm \sqrt{q}$, where $p,q \in \mathbb{Z}$. [3]
Show complete worked solution
**(a)**
Since the graph passes through the $y$-intercept $(0,1)$, substituting $x=0, f(x)=1$ into $f(x)=a(x+2)^2+5$:
$$1 = a(0+2)^2+5$$
$$1 = 4a+5$$
$$4a=-4$$
$$a=-1, \text{ as required.}$$
**(b)**
So $f(x) = -(x+2)^2+5$. Setting $f(x)=0$ to find the $x$-intercepts:
$$-(x+2)^2+5=0$$
$$(x+2)^2=5$$
Taking the square root of both sides:
$$x+2 = \pm\sqrt5$$
$$x=-2\pm\sqrt5$$
$$\boxed{x=-2+\sqrt5 \text{ or } x=-2-\sqrt5}$$
QUESTION 18
6 marks
Easy
Let $f(x) = (x-1)(x+2)(x-4)$.
(a) Write down the $x$-intercepts of the graph of $f$. [2]
(b) Find the $y$-intercept of the graph of $f$. [1]
(c) Describe the behaviour of the graph of $f$ as $x \to \infty$ and as $x \to -\infty$. [1]
(d) Sketch the graph of $f$, showing the intercepts found above. [2]
Show complete worked solution
**(a)**
Since $f(x)$ is given in factored form, the $x$-intercepts occur where each factor is zero:
$$\boxed{x\text{-intercepts: } (1,0),\ (-2,0),\ (4,0)}$$
**(b)**
Substituting $x=0$:
$$f(0) = (0-1)(0+2)(0-4) = (-1)(2)(-4)=8$$
$$\boxed{y\text{-intercept: } (0,8)}$$
**(c)**
Expanding, $f(x)$ has leading term $x^3$, so it is a cubic with a positive leading coefficient. For such a cubic:
$$\boxed{\text{As } x\to\infty,\ f(x)\to\infty; \text{ as } x\to-\infty,\ f(x)\to-\infty}$$
**(d)**
The sketch crosses the $x$-axis at $x=-2, 1, 4$, passes through $(0,8)$, falls to $-\infty$ on the left and rises to $\infty$ on the right, with a local maximum between $x=-2$ and $x=1$ and a local minimum between $x=1$ and $x=4$, as shown in the diagram.
QUESTION 19
6 marks
Easy
Let $f(x) = \dfrac{3}{x-2}+1$.
(a) Write down the equations of the vertical and horizontal asymptotes of the graph of $f$. [2]
(b) Find the $x$-intercept of the graph of $f$. [1]
(c) Find the $y$-intercept of the graph of $f$. [1]
(d) Sketch the graph of $f$, showing the asymptotes and both intercepts. [2]
Show complete worked solution
**(a)**
The vertical asymptote occurs where the denominator is zero:
$$x-2=0 \implies x=2$$
As $x\to\pm\infty$, $\dfrac{3}{x-2}\to0$, so $f(x)\to1$:
$$\boxed{\text{Vertical asymptote: } x=2, \quad \text{Horizontal asymptote: } y=1}$$
**(b)**
Setting $f(x)=0$:
$$\frac{3}{x-2}+1=0 \implies \frac{3}{x-2}=-1 \implies x-2=-3 \implies x=-1$$
$$\boxed{x\text{-intercept: } (-1,0)}$$
**(c)**
Substituting $x=0$:
$$f(0) = \frac{3}{-2}+1 = -1.5+1=-0.5$$
$$\boxed{y\text{-intercept: } (0,-0.5)}$$
**(d)**
The sketch shows two branches approaching the dashed asymptotes $x=2$ and $y=1$, passing through $(-1,0)$ and $(0,-0.5)$, as shown in the diagram.
QUESTION 20
5 marks
Easy
Let $f(x) = 2^x - 8$.
(a) Find the $y$-intercept of the graph of $f$. [1]
(b) Write down the equation of the horizontal asymptote of the graph of $f$. [1]
(c) Find the $x$-intercept of the graph of $f$. [2]
(d) Sketch the graph of $f$, showing the asymptote and both intercepts. [1]
Show complete worked solution
**(a)**
Substituting $x=0$:
$$f(0) = 2^0-8 = 1-8=-7$$
$$\boxed{y\text{-intercept: } (0,-7)}$$
**(b)**
As $x\to-\infty$, $2^x\to0$, so $f(x)\to-8$:
$$\boxed{\text{Horizontal asymptote: } y=-8}$$
**(c)**
Setting $f(x)=0$:
$$2^x-8=0 \implies 2^x=8=2^3$$
Since the bases are equal, the exponents are equal:
$$\boxed{x\text{-intercept: } (3,0)}$$
**(d)**
The sketch is an increasing exponential curve, approaching the dashed horizontal asymptote $y=-8$ as $x\to-\infty$, passing through $(0,-7)$ and $(3,0)$, and increasing steeply for $x>3$, as shown in the diagram.
QUESTION 21
6 marks
Easy
Let $f(x) = \log_2(x-1)$.
(a) Write down the domain of $f$, and hence state the equation of the vertical asymptote of the graph of $f$. [2]
(b) Find the $x$-intercept of the graph of $f$. [2]
(c) Sketch the graph of $f$, showing the asymptote and the $x$-intercept. [2]
Show complete worked solution
**(a)**
The expression $\log_2(x-1)$ is defined only when $x-1>0$:
$$\boxed{\text{Domain: } x>1}$$
Since $f$ is undefined at $x=1$ and $f(x) \to -\infty$ as $x \to 1^+$:
$$\boxed{\text{Vertical asymptote: } x=1}$$
**(b)**
Setting $f(x)=0$:
$$\log_2(x-1)=0$$
Using the definition of a logarithm, $\log_2(x-1)=0$ means $2^0=x-1$:
$$x-1=1 \implies x=2$$
$$\boxed{x\text{-intercept: } (2,0)}$$
**(c)**
The sketch shows an increasing logarithmic curve defined only for $x>1$, approaching the dashed vertical asymptote $x=1$ as $x\to1^+$, passing through $(2,0)$, and continuing to increase slowly for larger $x$, as shown in the diagram.
QUESTION 22
5 marks
Easy
Let $f(x) = 2(x-3)^2-8$.
(a) Write down the coordinates of the vertex of the graph of $f$, and state the equation of the axis of symmetry. [1]
(b) Find the $x$-intercepts of the graph of $f$. [2]
(c) Find the $y$-intercept of the graph of $f$. [1]
(d) Sketch the graph of $f$, showing the vertex and both intercepts. [1]
Show complete worked solution
**(a)**
Since $f$ is given in vertex form $a(x-h)^2+k$, the vertex is read off directly:
$$\boxed{\text{Vertex: } (3,-8), \quad \text{Axis of symmetry: } x=3}$$
**(b)**
Setting $f(x)=0$:
$$2(x-3)^2-8=0$$
$$(x-3)^2=4$$
$$x-3=\pm2$$
$$\boxed{x\text{-intercepts: } (1,0) \text{ and } (5,0)}$$
**(c)**
Substituting $x=0$:
$$f(0) = 2(9)-8=10$$
$$\boxed{y\text{-intercept: } (0,10)}$$
**(d)**
Since the coefficient of $(x-3)^2$ is positive, the parabola opens upward, with minimum at $(3,-8)$, passing through $(1,0)$, $(5,0)$ and $(0,10)$, as shown in the diagram.
QUESTION 23
4 marks
Easy
The diagram below shows the graph of a cubic function $y=f(x)$ with leading coefficient $1$, which crosses the $x$-axis at $x=-3$, $x=1$ and $x=2$, and at no other points.
(a) Write down $f(x)$ in fully factorised form. [1]
(b) Expand $f(x)$, and hence find the $y$-intercept of the graph of $f$. [3]
Show complete worked solution
**(a)**
Since the graph crosses the $x$-axis at $x=-3, 1, 2$, and the leading coefficient is $1$:
$$\boxed{f(x) = (x+3)(x-1)(x-2)}$$
**(b)**
Expanding $(x-1)(x-2)$ first:
$$(x-1)(x-2) = x^2-3x+2$$
Multiplying by $(x+3)$:
$$f(x) = (x+3)(x^2-3x+2) = x^3-3x^2+2x+3x^2-9x+6$$
Collecting like terms:
$$f(x) = x^3-7x+6$$
The $y$-intercept occurs where $x=0$, which is simply the constant term:
$$\boxed{y\text{-intercept: } (0,6)}$$
QUESTION 24
5 marks
Easy
Let $f(x) = -\dfrac{2}{x+1}-3$.
(a) Write down the equations of the vertical and horizontal asymptotes of the graph of $f$. [2]
(b) Find the $x$-intercept of the graph of $f$. [2]
(c) Find the $y$-intercept of the graph of $f$. [1]
Show complete worked solution
**(a)**
The vertical asymptote occurs where the denominator is zero:
$$x+1=0 \implies x=-1$$
As $x\to\pm\infty$, $-\dfrac{2}{x+1}\to0$, so $f(x)\to-3$:
$$\boxed{\text{Vertical asymptote: } x=-1, \quad \text{Horizontal asymptote: } y=-3}$$
**(b)**
Setting $f(x)=0$:
$$-\frac{2}{x+1}-3=0$$
$$-\frac{2}{x+1}=3$$
$$x+1 = -\frac{2}{3}$$
$$x=-\frac{5}{3}$$
$$\boxed{x\text{-intercept: } \left(-\tfrac{5}{3},0\right)}$$
**(c)**
Substituting $x=0$:
$$f(0) = -\frac{2}{1}-3=-2-3=-5$$
$$\boxed{y\text{-intercept: } (0,-5)}$$
QUESTION 25
4 marks
Easy
Let $f(x) = 3^{-x}+1$.
(a) Find the $y$-intercept of the graph of $f$. [1]
(b) Write down the equation of the horizontal asymptote of the graph of $f$. [1]
(c) Explain why the graph of $f$ has no $x$-intercept. [1]
(d) Sketch the graph of $f$, showing the asymptote and the $y$-intercept. [1]
Show complete worked solution
**(a)**
Substituting $x=0$:
$$f(0) = 3^0+1=1+1=2$$
$$\boxed{y\text{-intercept: } (0,2)}$$
**(b)**
As $x\to\infty$, $3^{-x}\to0$, so $f(x)\to1$:
$$\boxed{\text{Horizontal asymptote: } y=1}$$
**(c)**
Since $3^{-x}>0$ for all real $x$, it follows that $f(x)=3^{-x}+1>1>0$ for all $x$. Since $f(x)$ is never zero:
$$\boxed{\text{The graph of } f \text{ never crosses the } x\text{-axis, so there is no } x\text{-intercept.}}$$
**(d)**
The sketch is a decreasing exponential curve, approaching the dashed horizontal asymptote $y=1$ as $x\to\infty$, passing through $(0,2)$, and increasing without bound as $x\to-\infty$, always staying above the line $y=1$, as shown in the diagram.
QUESTION 26
5 marks
Easy
Let $f(x) = \ln(x) - 2$.
(a) Write down the domain of $f$, and hence state the equation of the vertical asymptote of the graph of $f$. [2]
(b) Find the exact value of the $x$-intercept of the graph of $f$. [2]
(c) Sketch the graph of $f$, showing the asymptote and the $x$-intercept. [1]
Show complete worked solution
**(a)**
The expression $\ln(x)$ is defined only for $x>0$:
$$\boxed{\text{Domain: } x>0}$$
Since $f$ is undefined at $x=0$ and $f(x)\to-\infty$ as $x\to0^+$:
$$\boxed{\text{Vertical asymptote: } x=0}$$
**(b)**
Setting $f(x)=0$:
$$\ln(x)-2=0$$
$$\ln(x)=2$$
Using the definition of $\ln$ as $\log_e$, this means $x=e^2$:
$$\boxed{x\text{-intercept: } (e^2,0)}$$
**(c)**
The sketch shows an increasing logarithmic curve, approaching the dashed vertical asymptote $x=0$ (the $y$-axis) as $x\to0^+$, and passing through $(e^2,0) \approx (7.39,0)$, as shown in the diagram.
QUESTION 27
6 marks
Easy
The graph of $f(x)=a(x+1)(x-5)$ has $x$-intercepts at $x=-1$ and $x=5$, and $y$-intercept $(0,10)$.
(a) Find the value of $a$, and hence find $f(x)$ in expanded form. [2]
(b) Using the axis of symmetry, find the coordinates of the vertex of the graph of $f$, and state whether it is a maximum or a minimum. [3]
(c) Sketch the graph of $f$, showing the intercepts and the vertex. [1]
Show complete worked solution
**(a)**
Substituting the $y$-intercept $(0,10)$ into $f(x)=a(x+1)(x-5)$:
$$10 = a(0+1)(0-5) = -5a$$
$$a=-2$$
So $f(x) = -2(x+1)(x-5)$. Expanding:
$$f(x) = -2(x^2-4x-5) = -2x^2+8x+10$$
$$\boxed{a=-2, \quad f(x)=-2x^2+8x+10}$$
**(b)**
The axis of symmetry lies midway between the $x$-intercepts $-1$ and $5$:
$$x = \frac{-1+5}{2}=2$$
Substituting $x=2$ into $f(x)$:
$$f(2) = -2(2+1)(2-5) = -2(3)(-3)=18$$
Since the coefficient of $x^2$ is $-2<0$, the parabola opens downward:
$$\boxed{\text{Vertex: } (2,18), \text{ a maximum}}$$
**(c)**
The sketch is a downward-opening parabola through $(-1,0)$, $(5,0)$ and $(0,10)$, with maximum turning point $(2,18)$, as shown in the diagram.
QUESTION 28
5 marks
Easy
Let $f(x) = \dfrac{4}{x}$, $x \neq 0$.
(a) Write down the equations of the vertical and horizontal asymptotes of the graph of $f$. [1]
(b) Show algebraically that $f$ is an odd function, and state what this means for the symmetry of its graph. [2]
(c) Sketch the graph of $f$ for $-6 \le x \le 6$, $x \neq 0$, showing both branches. [2]
Show complete worked solution
**(a)**
The function is undefined at $x=0$, giving a vertical asymptote there. As $x\to\pm\infty$, $\dfrac{4}{x}\to0$:
$$\boxed{\text{Vertical asymptote: } x=0, \quad \text{Horizontal asymptote: } y=0}$$
**(b)**
Finding $f(-x)$:
$$f(-x) = \frac{4}{-x} = -\frac{4}{x} = -f(x)$$
Since $f(-x)=-f(x)$ for all $x\neq0$:
$$\boxed{f \text{ is odd}}$$
$$\boxed{\text{The graph of } f \text{ has rotational (point) symmetry of order 2 about the origin}}$$
**(c)**
Using this symmetry, the branch for $x<0$ is the $180^{\circ}$ rotation of the branch for $x>0$ about the origin. For $x>0$, the curve passes through $(1,4)$ and $(4,1)$, decreasing towards the asymptotes; for $x<0$, it passes through $(-1,-4)$ and $(-4,-1)$, as shown in the diagram.
QUESTION 29
5 marks
Easy
The diagram below shows the graph of $y=f(x)$, where $f(x) = ab^x+1$ and $b>0$. The graph has horizontal asymptote $y=1$, $y$-intercept $(0,4)$, and passes through the point $(2,13)$.
(a) Show that $a=3$. [2]
(b) Find the value of $b$. [2]
(c) Write down $f(x)$ in the form $ab^x+1$. [1]
Show complete worked solution
**(a)**
Substituting the $y$-intercept $(0,4)$ into $f(x)=ab^x+1$:
$$4 = ab^0+1 = a(1)+1=a+1$$
$$a=3, \text{ as required.}$$
**(b)**
So $f(x)=3b^x+1$. Substituting the point $(2,13)$:
$$13 = 3b^2+1$$
$$3b^2=12$$
$$b^2=4$$
Since $b>0$:
$$\boxed{b=2}$$
**(c)**
$$\boxed{f(x) = 3(2^x)+1}$$
QUESTION 30
9 marks
Medium
Let $f(x) = x^3-3x^2-9x+5$.
(a) Using differentiation, find the coordinates of the local maximum and local minimum points of the graph of $f$. [5]
(b) Find the $y$-intercept of the graph of $f$. [1]
(c) Sketch the graph of $f$, showing the turning points, the $y$-intercept, and the behaviour of the graph as $x \to \pm\infty$. [3]
Show complete worked solution
**(a)**
Differentiating $f(x)$:
$$f'(x) = 3x^2-6x-9$$
Setting $f'(x)=0$ to find the turning points:
$$3x^2-6x-9=0$$
Dividing by 3:
$$x^2-2x-3=0$$
Factorising:
$$(x-3)(x+1)=0$$
$$x=3 \text{ or } x=-1$$
Substituting each value back into $f(x)$:
$$f(-1) = (-1)^3-3(-1)^2-9(-1)+5 = -1-3+9+5=10$$
$$f(3) = 3^3-3(3)^2-9(3)+5 = 27-27-27+5=-22$$
Since $f'(x)$ changes from positive to negative at $x=-1$ (a downward-opening region of the cubic) and from negative to positive at $x=3$:
$$\boxed{\text{Local maximum: } (-1,10), \quad \text{Local minimum: } (3,-22)}$$
**(b)**
Substituting $x=0$:
$$f(0) = 0-0-0+5=5$$
$$\boxed{y\text{-intercept: } (0,5)}$$
**(c)**
Since the leading coefficient is positive, $f(x)\to-\infty$ as $x\to-\infty$ and $f(x)\to\infty$ as $x\to\infty$. The sketch rises to the local maximum $(-1,10)$, falls through $(0,5)$ to the local minimum $(3,-22)$, then rises again, as shown in the diagram.
QUESTION 31
8 marks
Medium
Let $f(x) = \dfrac{2x-1}{x+3}$.
(a) Write down the equations of the vertical and horizontal asymptotes of the graph of $f$. [2]
(b) Find the $x$-intercept and the $y$-intercept of the graph of $f$. [2]
(c) Sketch the graph of $f$, showing the asymptotes and both intercepts. [4]
Show complete worked solution
**(a)**
The vertical asymptote occurs where the denominator is zero:
$$x+3=0 \implies x=-3$$
For the horizontal asymptote, consider the behaviour as $x\to\pm\infty$. Dividing numerator and denominator by $x$:
$$f(x) = \frac{2-\frac1x}{1+\frac3x} \to \frac{2}{1}=2 \text{ as } x\to\pm\infty$$
$$\boxed{\text{Vertical asymptote: } x=-3, \quad \text{Horizontal asymptote: } y=2}$$
**(b)**
Setting $f(x)=0$, the $x$-intercept occurs where the numerator is zero:
$$2x-1=0 \implies x=\frac12$$
$$\boxed{x\text{-intercept: } \left(\tfrac12,0\right)}$$
Substituting $x=0$:
$$f(0) = \frac{-1}{3} = -\frac13$$
$$\boxed{y\text{-intercept: } \left(0,-\tfrac13\right)}$$
**(c)**
The sketch has two branches separated by the vertical asymptote $x=-3$: for $x<-3$, the curve approaches $y=2$ from above as $x\to-\infty$ and falls steeply near $x=-3$; for $x>-3$, the curve rises from near $x=-3$, passes through $\left(0,-\tfrac13\right)$ and $\left(\tfrac12,0\right)$, and approaches $y=2$ from below as $x\to\infty$, as shown in the diagram.
QUESTION 32
6 marks
Medium
The number of bacteria in a culture, $N$, is decreasing according to the model
$$N(t) = 100e^{-0.2t}+20, \quad t\ge0$$
where $t$ is the time in hours.
(a) Find $N(0)$, and write down the value that $N(t)$ approaches as $t\to\infty$. [2]
(b) Find the exact time at which $N=70$, giving your answer in the form $a\ln2$, where $a \in \mathbb{Q}$. [4]
Show complete worked solution
**(a)**
Substituting $t=0$:
$$N(0) = 100e^0+20 = 100+20=120$$
As $t\to\infty$, $e^{-0.2t}\to0$, so $N(t)$ approaches the horizontal asymptote:
$$\boxed{N(0)=120, \quad N(t) \to 20 \text{ as } t\to\infty}$$
**(b)**
Setting $N(t)=70$:
$$100e^{-0.2t}+20=70$$
$$100e^{-0.2t}=50$$
$$e^{-0.2t}=0.5$$
Taking $\ln$ of both sides:
$$-0.2t = \ln(0.5) = -\ln2$$
$$t = \frac{\ln2}{0.2} = 5\ln2$$
$$\boxed{t=5\ln2 \text{ hours} \approx 3.47 \text{ hours}}$$
QUESTION 33
10 marks
Medium
Let $f(x) = x^3-4x$.
(a) Show algebraically that $f$ is an odd function. [2]
(b) Find the $x$-intercepts of the graph of $f$. [2]
(c) Using differentiation, find the exact coordinates of the local maximum and local minimum points of the graph of $f$, using the symmetry from part (a) to deduce the second turning point without repeating the calculation. [4]
(d) Sketch the graph of $f$, showing the intercepts and turning points. [2]
Show complete worked solution
**(a)**
Finding $f(-x)$:
$$f(-x) = (-x)^3-4(-x) = -x^3+4x = -(x^3-4x) = -f(x)$$
$$\boxed{\text{Since } f(-x)=-f(x) \text{ for all } x, \ f \text{ is odd}}$$
**(b)**
Setting $f(x)=0$:
$$x^3-4x=0$$
$$x(x^2-4)=0$$
$$x(x-2)(x+2)=0$$
$$\boxed{x\text{-intercepts: } (0,0),\ (2,0),\ (-2,0)}$$
**(c)**
Differentiating:
$$f'(x) = 3x^2-4$$
Setting $f'(x)=0$:
$$3x^2=4 \implies x^2=\frac43 \implies x=\pm\frac{2}{\sqrt3}=\pm\frac{2\sqrt3}{3}$$
Substituting $x=\dfrac{2\sqrt3}{3}$ into $f(x)$, using $x^2=\dfrac43$:
$$f(x) = x^3-4x = x(x^2-4) = \frac{2\sqrt3}{3}\left(\frac43-4\right) = \frac{2\sqrt3}{3}\left(-\frac83\right)=-\frac{16\sqrt3}{9}$$
Since $f'(x)$ changes from positive to negative to positive as $x$ increases through these points (the coefficient of $x^3$ is positive), $x=\dfrac{2\sqrt3}{3}$ gives a local minimum. Because $f$ is odd (part (a)), its graph has point symmetry about the origin, so the local maximum is the reflection of the local minimum through the origin, with coordinates negated:
$$\boxed{\text{Local minimum: } \left(\tfrac{2\sqrt3}{3},-\tfrac{16\sqrt3}{9}\right), \quad \text{Local maximum: } \left(-\tfrac{2\sqrt3}{3},\tfrac{16\sqrt3}{9}\right)}$$
**(d)**
The sketch, symmetric about the origin, rises from $-\infty$ through the local maximum near $(-1.15,3.08)$, falls through the intercepts $(-2,0)$, $(0,0)$, $(2,0)$ to the local minimum near $(1.15,-3.08)$, then rises to $\infty$, as shown in the diagram.
QUESTION 34
8 marks
Medium
Let $f(x) = 2\ln(x+1)-1$.
(a) Write down the domain of $f$, and hence state the equation of the vertical asymptote of the graph of $f$. [2]
(b) Find the exact value of the $x$-intercept of the graph of $f$. [3]
(c) Find the $y$-intercept of the graph of $f$. [1]
(d) Sketch the graph of $f$, showing the asymptote and both intercepts. [2]
Show complete worked solution
**(a)**
The expression $\ln(x+1)$ is defined only when $x+1>0$:
$$\boxed{\text{Domain: } x>-1}$$
$$\boxed{\text{Vertical asymptote: } x=-1}$$
**(b)**
Setting $f(x)=0$:
$$2\ln(x+1)-1=0$$
$$\ln(x+1)=\frac12$$
Using the definition of $\ln$:
$$x+1 = e^{1/2}$$
$$x = \sqrt{e}-1$$
$$\boxed{x\text{-intercept: } (\sqrt{e}-1,\ 0) \approx (0.649,0)}$$
**(c)**
Substituting $x=0$:
$$f(0) = 2\ln(1)-1 = 2(0)-1=-1$$
$$\boxed{y\text{-intercept: } (0,-1)}$$
**(d)**
The sketch shows an increasing logarithmic curve, approaching the dashed vertical asymptote $x=-1$ as $x\to-1^+$, passing through $(0,-1)$ and $(\sqrt{e}-1,0)$, and continuing to increase slowly for larger $x$, as shown in the diagram.
QUESTION 35
8 marks
Medium
The graph of $f(x) = a(x+2)(x-6)$ has $x$-intercepts at $x=-2$ and $x=6$, and $y$-intercept $(0,-18)$.
(a) Find the value of $a$. [3]
(b) Find the coordinates of the vertex of the graph of $f$, and state whether it is a maximum or a minimum. [3]
(c) Sketch the graph of $f$, showing the intercepts and the vertex. [2]
Show complete worked solution
**(a)**
Substituting the $y$-intercept $(0,-18)$ into $f(x) = a(x+2)(x-6)$:
$$-18 = a(0+2)(0-6) = -12a$$
$$\boxed{a=1.5}$$
**(b)**
So $f(x) = 1.5(x+2)(x-6)$. The axis of symmetry lies midway between the $x$-intercepts $-2$ and $6$:
$$x = \frac{-2+6}{2}=2$$
Substituting $x=2$:
$$f(2) = 1.5(2+2)(2-6) = 1.5(4)(-4)=-24$$
Since $a=1.5>0$, the parabola opens upward:
$$\boxed{\text{Vertex: } (2,-24), \text{ a minimum}}$$
**(c)**
The sketch is an upward-opening parabola through $(-2,0)$, $(6,0)$ and $(0,-18)$, with minimum turning point $(2,-24)$, as shown in the diagram.
QUESTION 36
7 marks
Medium
The graph of $y=f(x)$, a rational function of the form $f(x) = \dfrac{k}{x-1}-2$, has vertical asymptote $x=1$, horizontal asymptote $y=-2$, and passes through the point $(3,-3)$.
(a) Find the value of $k$, and hence write down $f(x)$. [4]
(b) Find the $x$-intercept of the graph of $f$, and comment on its relationship to the $y$-intercept. [3]
Show complete worked solution
**(a)**
Substituting the point $(3,-3)$ into $f(x)=\dfrac{k}{x-1}-2$:
$$-3 = \frac{k}{3-1}-2$$
$$-3 = \frac{k}{2}-2$$
$$-1 = \frac{k}{2}$$
$$\boxed{k=-2, \quad f(x) = \frac{-2}{x-1}-2}$$
**(b)**
Setting $f(x)=0$:
$$\frac{-2}{x-1}-2=0$$
$$\frac{-2}{x-1}=2$$
$$x-1=-1$$
$$x=0$$
$$\boxed{x\text{-intercept: } (0,0)}$$
The $y$-intercept is $f(0)$, which by the calculation above is also $0$, since $x=0$ satisfies $f(x)=0$:
$$\boxed{\text{The } x\text{-intercept and the } y\text{-intercept coincide at the single point } (0,0), \text{ so the graph passes through the origin.}}$$
QUESTION 37
10 marks
Medium
Let $f(x) = (x-2)^2(x+1)$.
(a) Write down the $x$-intercepts of the graph of $f$, stating the multiplicity of each root and what this means for the shape of the graph at that point. [3]
(b) Find the $y$-intercept of the graph of $f$. [1]
(c) Using differentiation, find the coordinates of the local maximum and local minimum points of the graph of $f$. [4]
(d) Sketch the graph of $f$, showing the intercepts and turning points. [2]
Show complete worked solution
**(a)**
From the factorised form, the $x$-intercepts occur where each factor is zero:
$$\boxed{x=2 \text{ (repeated root, multiplicity 2)}, \quad x=-1 \text{ (multiplicity 1)}}$$
$$\boxed{\text{At } x=2 \text{ the graph touches the } x\text{-axis (does not cross); at } x=-1 \text{ it crosses the } x\text{-axis.}}$$
**(b)**
Substituting $x=0$:
$$f(0) = (0-2)^2(0+1) = (4)(1)=4$$
$$\boxed{y\text{-intercept: } (0,4)}$$
**(c)**
Expanding $f(x)$ first:
$$f(x) = (x^2-4x+4)(x+1) = x^3-3x^2+4$$
Differentiating:
$$f'(x) = 3x^2-6x$$
Setting $f'(x)=0$:
$$3x(x-2)=0$$
$$x=0 \text{ or } x=2$$
Substituting back into $f(x)=x^3-3x^2+4$:
$$f(0)=4, \qquad f(2) = 8-12+4=0$$
Since $f'(x)$ changes from positive to negative at $x=0$ and from negative to positive at $x=2$:
$$\boxed{\text{Local maximum: } (0,4), \quad \text{Local minimum: } (2,0)}$$
Note that the local minimum $(2,0)$ coincides with the repeated root found in part (a), confirming that the graph touches the $x$-axis there.
**(d)**
The sketch rises from $-\infty$, crosses the $x$-axis at $(-1,0)$, rises to the local maximum $(0,4)$, falls to touch the $x$-axis at the local minimum $(2,0)$, then rises to $\infty$, as shown in the diagram.
QUESTION 38
7 marks
Medium
Let $f(x) = -e^{x-2}+5$.
(a) Find the $y$-intercept of the graph of $f$, correct to 3 significant figures. [1]
(b) Describe the behaviour of the graph of $f$ as $x\to-\infty$ and as $x\to\infty$, and hence write down the equation of the horizontal asymptote of the graph of $f$. [3]
(c) Find the exact $x$-intercept of the graph of $f$. [3]
Show complete worked solution
**(a)**
Substituting $x=0$:
$$f(0) = -e^{-2}+5 \approx -0.1353+5=4.8647$$
$$\boxed{y\text{-intercept} \approx (0,4.86)}$$
**(b)**
As $x\to-\infty$, $e^{x-2}\to0$, so $f(x)\to0+5=5$. As $x\to\infty$, $e^{x-2}\to\infty$, so $-e^{x-2}\to-\infty$ and $f(x)\to-\infty$:
$$\boxed{\text{As } x\to-\infty, f(x)\to5; \text{ as } x\to\infty, f(x)\to-\infty}$$
$$\boxed{\text{Horizontal asymptote: } y=5}$$
**(c)**
Setting $f(x)=0$:
$$-e^{x-2}+5=0$$
$$e^{x-2}=5$$
Taking $\ln$ of both sides:
$$x-2=\ln5$$
$$\boxed{x = 2+\ln5 \ (\approx 3.61,\ 0)}$$
QUESTION 39
7 marks
Medium
Let $f(x) = \dfrac{3x+6}{x-1}$.
(a) Write down the equations of the vertical and horizontal asymptotes of the graph of $f$. [2]
(b) Find the $x$-intercept and the $y$-intercept of the graph of $f$. [2]
(c) Sketch the graph of $f$, showing the asymptotes and both intercepts. [3]
Show complete worked solution
**(a)**
The vertical asymptote occurs where the denominator is zero:
$$x-1=0 \implies x=1$$
Dividing numerator and denominator by $x$ to find the behaviour as $x\to\pm\infty$:
$$f(x) = \frac{3+\frac6x}{1-\frac1x} \to \frac{3}{1}=3$$
$$\boxed{\text{Vertical asymptote: } x=1, \quad \text{Horizontal asymptote: } y=3}$$
**(b)**
Setting the numerator equal to zero:
$$3x+6=0 \implies x=-2$$
$$\boxed{x\text{-intercept: } (-2,0)}$$
Substituting $x=0$:
$$f(0) = \frac{6}{-1}=-6$$
$$\boxed{y\text{-intercept: } (0,-6)}$$
**(c)**
The sketch has two branches either side of the vertical asymptote $x=1$: for $x<1$, the curve passes through $(-2,0)$ and $(0,-6)$, falling steeply as $x\to1^-$, and approaching $y=3$ from below as $x\to-\infty$; for $x>1$, the curve falls steeply from near $x=1^+$ and approaches $y=3$ from above as $x\to\infty$, as shown in the diagram.
QUESTION 40
8 marks
Medium
Let $f(x) = \dfrac{x^2}{x^2-4}$.
(a) Show algebraically that $f$ is an even function. [2]
(b) Write down the equations of the vertical and horizontal asymptotes of the graph of $f$. [2]
(c) Find the coordinates of any point where the graph of $f$ intersects the axes. [1]
(d) Using the symmetry from part (a), sketch the graph of $f$. [3]
Show complete worked solution
**(a)**
Finding $f(-x)$:
$$f(-x) = \frac{(-x)^2}{(-x)^2-4} = \frac{x^2}{x^2-4} = f(x)$$
$$\boxed{\text{Since } f(-x)=f(x) \text{ for all } x \text{ in the domain, } f \text{ is even}}$$
**(b)**
The vertical asymptotes occur where the denominator is zero:
$$x^2-4=0 \implies x=\pm2$$
Dividing numerator and denominator by $x^2$:
$$f(x) = \frac{1}{1-\frac{4}{x^2}} \to \frac11=1 \text{ as } x\to\pm\infty$$
$$\boxed{\text{Vertical asymptotes: } x=2 \text{ and } x=-2, \quad \text{Horizontal asymptote: } y=1}$$
**(c)**
Substituting $x=0$:
$$f(0) = \frac{0}{-4}=0$$
$$\boxed{\text{The graph passes through the origin } (0,0), \text{ which is both the } x\text{- and } y\text{-intercept.}}$$
**(d)**
Since $f$ is even, its graph is symmetric about the $y$-axis, so the branch for $x<0$ is the mirror image of the branch for $x>0$. There are three sections: for $-22$ (and by symmetry $x<-2$), the curve falls from $+\infty$ near the asymptote and approaches $y=1$ from above, as shown in the diagram.
QUESTION 41
8 marks
Medium
Let $f(x) = -x^3+3x^2+9x-2$.
(a) Using differentiation, find the coordinates of the local maximum and local minimum points of the graph of $f$, justifying which is which. [5]
(b) Find the $y$-intercept of the graph of $f$. [1]
(c) Sketch the graph of $f$, showing the turning points, the $y$-intercept, and describing the behaviour of the graph as $x\to\pm\infty$. [2]
Show complete worked solution
**(a)**
Differentiating $f(x)$:
$$f'(x) = -3x^2+6x+9$$
Setting $f'(x)=0$:
$$-3x^2+6x+9=0$$
Dividing by $-3$:
$$x^2-2x-3=0$$
$$(x-3)(x+1)=0$$
$$x=3 \text{ or } x=-1$$
Substituting back into $f(x)$:
$$f(-1) = -(-1)^3+3(-1)^2+9(-1)-2 = 1+3-9-2=-7$$
$$f(3) = -(27)+3(9)+9(3)-2 = -27+27+27-2=25$$
To determine the nature of each turning point, differentiate again:
$$f''(x) = -6x+6$$
At $x=-1$: $f''(-1) = 6+6=12>0$, so this is a local minimum. At $x=3$: $f''(3)=-18+6=-12<0$, so this is a local maximum:
$$\boxed{\text{Local minimum: } (-1,-7), \quad \text{Local maximum: } (3,25)}$$
**(b)**
Substituting $x=0$:
$$f(0) = -0+0+0-2=-2$$
$$\boxed{y\text{-intercept: } (0,-2)}$$
**(c)**
Since the leading coefficient of $f$ is negative:
$$\boxed{\text{As } x\to-\infty, f(x)\to\infty; \text{ as } x\to\infty, f(x)\to-\infty}$$
The sketch falls from $+\infty$ to the local minimum $(-1,-7)$, rises through $(0,-2)$ to the local maximum $(3,25)$, then falls to $-\infty$, as shown in the diagram.
QUESTION 42
8 marks
Medium
Let $f(x) = -\log_2(x)+4$.
(a) Write down the domain of $f$, and hence state the equation of the vertical asymptote of the graph of $f$. [2]
(b) Find the $x$-intercept of the graph of $f$. [3]
(c) Describe the behaviour of the graph of $f$ as $x\to0^+$ and as $x\to\infty$, and hence sketch the graph of $f$. [3]
Show complete worked solution
**(a)**
The expression $\log_2(x)$ is defined only for $x>0$:
$$\boxed{\text{Domain: } x>0, \quad \text{Vertical asymptote: } x=0}$$
**(b)**
Setting $f(x)=0$:
$$-\log_2(x)+4=0$$
$$\log_2(x)=4$$
Using the definition of a logarithm:
$$x=2^4=16$$
$$\boxed{x\text{-intercept: } (16,0)}$$
**(c)**
As $x\to0^+$, $\log_2(x)\to-\infty$, so $-\log_2(x)\to\infty$ and $f(x)\to\infty$. As $x\to\infty$, $\log_2(x)\to\infty$, so $-\log_2(x)\to-\infty$ and $f(x)\to-\infty$:
$$\boxed{\text{As } x\to0^+, f(x)\to\infty; \text{ as } x\to\infty, f(x)\to-\infty}$$
The graph is a decreasing logarithmic curve (the reflection of $y=\log_2x$ in the $x$-axis, translated up by 4), rising without bound near the $y$-axis, passing through $(16,0)$, and continuing to decrease slowly for larger $x$, as shown in the diagram.
QUESTION 43
9 marks
Medium
A company's monthly profit, $P$, in thousands of dollars, when it sells $x$ hundred units of a product, is modelled by
$$P(x) = -2x^2+40x-150$$
(a) Find the break-even points (the values of $x$ for which $P(x)=0$). [3]
(b) By completing the square, find the number of units that should be sold to maximise profit, and find the maximum profit. [4]
(c) Write down the range of $P$. [2]
Show complete worked solution
**(a)**
Setting $P(x)=0$:
$$-2x^2+40x-150=0$$
Dividing by $-2$:
$$x^2-20x+75=0$$
Factorising:
$$(x-5)(x-15)=0$$
$$\boxed{x=5 \text{ or } x=15 \text{ (i.e. 500 or 1500 units)}}$$
**(b)**
Completing the square, first factoring out $-2$ from the terms in $x$:
$$P(x) = -2(x^2-20x)-150$$
Completing the square inside the brackets:
$$x^2-20x = (x-10)^2-100$$
Substituting back:
$$P(x) = -2[(x-10)^2-100]-150 = -2(x-10)^2+200-150 = -2(x-10)^2+50$$
The vertex form gives the maximum directly, since the coefficient of $(x-10)^2$ is negative:
$$\boxed{\text{Sell } x=10 \text{ (i.e. 1000 units) for a maximum monthly profit of 50 thousand dollars}}$$
**(c)**
Since the maximum value of $P$ is $50$ (attained at $x=10$) and the parabola opens downward, $P$ takes all values at or below this maximum:
$$\boxed{\text{Range: } P(x) \le 50, \text{ i.e. } P \in (-\infty,50]}$$
QUESTION 44
10 marks
Hard
Let $f(x) = \dfrac{4-2x}{x+3}$.
(a) By performing algebraic division, express $f(x)$ in the form $q+\dfrac{k}{x+3}$, where $q,k \in \mathbb{Z}$. [3]
(b) Hence write down the equations of the vertical and horizontal asymptotes of the graph of $f$. [2]
(c) Find the $x$-intercept and the $y$-intercept of the graph of $f$. [2]
(d) Sketch the graph of $f$, showing the asymptotes and both intercepts. [3]
Show complete worked solution
**(a)**
Rewriting the numerator $4-2x$ in terms of $(x+3)$:
$$4-2x = -2(x+3)+6+4 = -2(x+3)+10$$
Checking: $-2(x+3)=-2x-6$, so $-2(x+3)+10=-2x-6+10=-2x+4=4-2x$. This confirms the rewriting. So:
$$f(x) = \frac{-2(x+3)+10}{x+3} = -2+\frac{10}{x+3}$$
$$\boxed{f(x) = -2+\frac{10}{x+3}, \quad q=-2,\ k=10}$$
**(b)**
From this form, the vertical asymptote occurs where $x+3=0$, and as $x\to\pm\infty$, $\dfrac{10}{x+3}\to0$ so $f(x)\to-2$:
$$\boxed{\text{Vertical asymptote: } x=-3, \quad \text{Horizontal asymptote: } y=-2}$$
**(c)**
Setting the numerator of the original form equal to zero:
$$4-2x=0 \implies x=2$$
$$\boxed{x\text{-intercept: } (2,0)}$$
Substituting $x=0$:
$$f(0) = \frac{4}{3}$$
$$\boxed{y\text{-intercept: } \left(0,\tfrac43\right)}$$
**(d)**
The sketch has two branches either side of the vertical asymptote $x=-3$: for $x<-3$, the curve approaches $y=-2$ from below as $x\to-\infty$ and falls steeply near $x=-3$; for $x>-3$, the curve falls from near $x=-3^+$, passes through $\left(0,\tfrac43\right)$ and $(2,0)$, and approaches $y=-2$ from above as $x\to\infty$, as shown in the diagram.
QUESTION 45
10 marks
Hard
Let $f(x) = x^3-6x^2+9x$.
(a) Factorise $f(x)$ fully, and hence write down the $x$-intercepts of the graph of $f$, stating the multiplicity of each root. [3]
(b) Using differentiation, find the coordinates of the local maximum and local minimum points of the graph of $f$. [4]
(c) Sketch the graph of $f$, showing all intercepts, both turning points, and the behaviour of the graph as $x\to\pm\infty$. [3]
Show complete worked solution
**(a)**
Factoring out $x$:
$$f(x) = x(x^2-6x+9)$$
The quadratic factor is a perfect square:
$$x^2-6x+9=(x-3)^2$$
So:
$$\boxed{f(x) = x(x-3)^2}$$
$$\boxed{x\text{-intercepts: } (0,0) \text{ (multiplicity 1)}, \quad (3,0) \text{ (multiplicity 2)}}$$
**(b)**
Differentiating $f(x) = x^3-6x^2+9x$:
$$f'(x) = 3x^2-12x+9$$
Setting $f'(x)=0$:
$$3x^2-12x+9=0$$
Dividing by 3:
$$x^2-4x+3=0$$
$$(x-1)(x-3)=0$$
$$x=1 \text{ or } x=3$$
Substituting back into $f(x)$:
$$f(1) = 1-6+9=4$$
$$f(3) = 27-54+27=0$$
Since $f'(x)$ changes from positive to negative at $x=1$ and from negative to positive at $x=3$:
$$\boxed{\text{Local maximum: } (1,4), \quad \text{Local minimum: } (3,0)}$$
This local minimum coincides with the repeated root at $x=3$ found in part (a), consistent with the graph touching the $x$-axis there.
**(c)**
Since the leading coefficient is positive, $f(x)\to-\infty$ as $x\to-\infty$ and $f(x)\to\infty$ as $x\to\infty$. The sketch rises from $-\infty$, crosses the $x$-axis at the origin, rises to the local maximum $(1,4)$, falls to touch the $x$-axis at the local minimum $(3,0)$, then rises to $\infty$, as shown in the diagram.
QUESTION 46
9 marks
Hard
The graph of $f(x) = Ae^{kx}$, where $A,k>0$, passes through the points $(0,6)$ and $(3,48)$.
(a) Find the exact value of $A$, and show that $k=\ln2$. [4]
(b) Hence show that $f(x)$ can be written as $6(2^x)$, and write down the equation of the horizontal asymptote of the graph of $f$. [2]
(c) Solve the equation $f(x)=192$. [3]
Show complete worked solution
**(a)**
Substituting the point $(0,6)$ into $f(x)=Ae^{kx}$:
$$6 = Ae^0 = A$$
$$\boxed{A=6}$$
So $f(x)=6e^{kx}$. Substituting the point $(3,48)$:
$$48 = 6e^{3k}$$
$$e^{3k}=8$$
Taking $\ln$ of both sides:
$$3k = \ln8 = \ln(2^3)=3\ln2$$
$$k=\ln2, \text{ as required.}$$
**(b)**
Substituting $k=\ln2$:
$$f(x) = 6e^{(\ln2)x} = 6(e^{\ln2})^x = 6(2^x), \text{ as required.}$$
Since $2^x>0$ for all $x$ and $2^x\to0$ as $x\to-\infty$, $f(x)\to0$ as $x\to-\infty$:
$$\boxed{\text{Horizontal asymptote: } y=0}$$
**(c)**
Setting $f(x)=192$:
$$6(2^x) = 192$$
$$2^x = 32=2^5$$
Since the bases are equal:
$$\boxed{x=5}$$
QUESTION 47
12 marks
Hard
The graph of $f(x) = \dfrac{ax+b}{x+2}$ has horizontal asymptote $y=3$ and $x$-intercept $(4,0)$.
(a) Find the values of $a$ and $b$. [4]
(b) Find the $y$-intercept of the graph of $f$. [1]
(c) The two asymptotes of the graph of $f$ intersect at the point $(-2,3)$. By considering the function $h(x) = f(x-2)-3$, show that $h$ is an odd function, and explain what this shows about the symmetry of the graph of $f$. [4]
(d) Sketch the graph of $f$, marking both asymptotes, both intercepts, and the point $(-2,3)$. [3]
Show complete worked solution
**(a)**
As $x\to\pm\infty$, dividing numerator and denominator by $x$ gives $f(x)\to a$, so the horizontal asymptote $y=3$ gives:
$$a=3$$
So $f(x)=\dfrac{3x+b}{x+2}$. The $x$-intercept occurs where the numerator is zero, so substituting $x=4$:
$$3(4)+b=0$$
$$12+b=0$$
$$b=-12$$
$$\boxed{a=3,\ b=-12, \quad f(x)=\frac{3x-12}{x+2}}$$
**(b)**
Substituting $x=0$:
$$f(0) = \frac{-12}{2}=-6$$
$$\boxed{y\text{-intercept: } (0,-6)}$$
**(c)**
Simplifying $f(x-2)$ by substituting $x-2$ for $x$:
$$f(x-2) = \frac{3(x-2)-12}{(x-2)+2} = \frac{3x-6-12}{x} = \frac{3x-18}{x} = 3-\frac{18}{x}$$
So:
$$h(x) = f(x-2)-3 = 3-\frac{18}{x}-3 = -\frac{18}{x}$$
Finding $h(-x)$:
$$h(-x) = -\frac{18}{-x} = \frac{18}{x} = -\left(-\frac{18}{x}\right)=-h(x)$$
$$\boxed{\text{Since } h(-x)=-h(x) \text{ for all } x\neq0,\ h \text{ is odd.}}$$
The transformation $h(x)=f(x-2)-3$ translates the graph of $f$ by $2$ units in the negative $x$-direction and $3$ units in the negative $y$-direction, moving the intersection point of the asymptotes, $(-2,3)$, to the origin. Since $h$ is odd, its graph has point symmetry about the origin, so:
$$\boxed{\text{The graph of } f \text{ has point (rotational) symmetry about } (-2,3), \text{ the intersection of its asymptotes.}}$$
**(d)**
The sketch has vertical asymptote $x=-2$ and horizontal asymptote $y=3$ (dashed), intersecting at the centre of symmetry $(-2,3)$; the curve passes through $(0,-6)$ and $(4,0)$, and the branch for $x<-2$ is the $180^{\circ}$ rotation of the branch for $x>-2$ about $(-2,3)$, as shown in the diagram.
QUESTION 48
12 marks
Hard
Let $f(x) = x^3-7x+6$.
(a) Show that $x=1$ is a root of $f(x)=0$, and hence factorise $f(x)$ fully to find all three $x$-intercepts of the graph of $f$. [4]
(b) Using differentiation, find the exact coordinates of the local maximum and local minimum points of the graph of $f$. [5]
(c) Sketch the graph of $f$, showing the $x$-intercepts, the $y$-intercept, and the turning points (with coordinates given correct to 3 significant figures). [3]
Show complete worked solution
**(a)**
Substituting $x=1$:
$$f(1) = 1-7+6=0$$
Since $f(1)=0$, $x=1$ is a root, so $(x-1)$ is a factor of $f(x)$. Dividing $f(x)$ by $(x-1)$ (e.g. by synthetic division with root $1$, using coefficients $1,0,-7,6$) gives quotient $x^2+x-6$:
$$f(x) = (x-1)(x^2+x-6)$$
Factorising the quadratic factor:
$$x^2+x-6=(x+3)(x-2)$$
So:
$$\boxed{f(x) = (x-1)(x+3)(x-2)}$$
$$\boxed{x\text{-intercepts: } (1,0),\ (-3,0),\ (2,0)}$$
**(b)**
Differentiating $f(x)=x^3-7x+6$:
$$f'(x) = 3x^2-7$$
Setting $f'(x)=0$:
$$3x^2=7 \implies x^2=\frac73 \implies x=\pm\sqrt{\frac73}=\pm\frac{\sqrt{21}}{3}$$
Substituting $x=\dfrac{\sqrt{21}}{3}$ into $f(x)$, using $x^2=\dfrac73$ so that $x^3=x\cdot x^2=\dfrac{7x}{3}$:
$$f(x) = x^3-7x+6 = \frac{7x}{3}-7x+6 = -\frac{14x}{3}+6$$
$$f\left(\frac{\sqrt{21}}{3}\right) = 6-\frac{14}{3}\cdot\frac{\sqrt{21}}{3} = 6-\frac{14\sqrt{21}}{9}$$
Since $f'(x)$ changes from positive to negative to positive as $x$ increases through $-\dfrac{\sqrt{21}}{3}$ and $\dfrac{\sqrt{21}}{3}$ (the coefficient of $x^3$ is positive), $x=\dfrac{\sqrt{21}}{3}$ gives a local minimum. Replacing $x$ with $-\dfrac{\sqrt{21}}{3}$ in the expression $6-\dfrac{14x}{3}$ gives the local maximum value:
$$\boxed{\text{Local minimum: } \left(\tfrac{\sqrt{21}}{3},\ 6-\tfrac{14\sqrt{21}}{9}\right) \approx (1.53,-1.13)}$$
$$\boxed{\text{Local maximum: } \left(-\tfrac{\sqrt{21}}{3},\ 6+\tfrac{14\sqrt{21}}{9}\right) \approx (-1.53,13.1)}$$
**(c)**
The sketch rises from $-\infty$, crosses the $x$-axis at $(-3,0)$, rises to the local maximum near $(-1.53,13.1)$, falls through the $y$-intercept $(0,6)$ and the $x$-intercepts $(1,0)$ and $(2,0)$ to the local minimum near $(1.53,-1.13)$, then rises to $\infty$, as shown in the diagram.
QUESTION 49
11 marks
Hard
The temperature of a cup of coffee, $T$ degrees Celsius, $t$ minutes after it is poured, is modelled by
$$T(t) = 18+72e^{-0.3t}, \quad t\ge0$$
(a) Find $T(0)$, and interpret this value in context. [2]
(b) Write down the equation of the horizontal asymptote of the graph of $T$, and interpret it in context. [2]
(c) Find the time at which the temperature first reaches $30$ degrees Celsius, giving your answer correct to 3 significant figures. [4]
(d) Sketch the graph of $T$ against $t$ for $t\ge0$, showing the $y$-intercept and the horizontal asymptote. [3]
Show complete worked solution
**(a)**
Substituting $t=0$:
$$T(0) = 18+72e^0 = 18+72=90$$
$$\boxed{T(0)=90\text{: the coffee is poured at a temperature of } 90\text{ degrees Celsius}}$$
**(b)**
As $t\to\infty$, $e^{-0.3t}\to0$, so $T(t)\to18$:
$$\boxed{\text{Horizontal asymptote: } T=18}$$
$$\boxed{\text{This represents the surrounding room temperature, which the coffee cools towards but never reaches.}}$$
**(c)**
Setting $T(t)=30$:
$$18+72e^{-0.3t}=30$$
$$72e^{-0.3t}=12$$
$$e^{-0.3t}=\frac{1}{6}$$
Taking $\ln$ of both sides:
$$-0.3t = \ln\left(\frac16\right)=-\ln6$$
$$t = \frac{\ln6}{0.3}$$
$$\boxed{t \approx 5.97 \text{ minutes}}$$
**(d)**
The sketch is a decreasing exponential curve starting at $(0,90)$, falling steeply at first and then more slowly, approaching the dashed horizontal asymptote $T=18$ as $t\to\infty$, as shown in the diagram.
QUESTION 50
12 marks
Hard
Let $f(x) = \ln(4-x)$.
(a) State the domain of $f$, and hence write down the equation of the vertical asymptote of the graph of $f$. [3]
(b) Find the exact $x$-intercept of the graph of $f$, and find the $y$-intercept correct to 3 significant figures. [3]
(c) Describe the sequence of transformations that maps the graph of $y=\ln x$ onto the graph of $f$. [3]
(d) Sketch the graph of $f$, showing the asymptote and both intercepts. [3]
Show complete worked solution
**(a)**
The expression $\ln(4-x)$ is defined only when $4-x>0$:
$$\boxed{\text{Domain: } x<4}$$
Since $f$ is undefined at $x=4$ and $f(x)\to-\infty$ as $x\to4^-$:
$$\boxed{\text{Vertical asymptote: } x=4}$$
**(b)**
Setting $f(x)=0$:
$$\ln(4-x)=0$$
$$4-x=1$$
$$\boxed{x\text{-intercept: } (3,0)}$$
Substituting $x=0$:
$$f(0) = \ln(4) \approx 1.39$$
$$\boxed{y\text{-intercept} \approx (0,1.39)}$$
**(c)**
Writing $f(x)$ as $\ln(-(x-4))$, this is the composition of two transformations of $y=\ln x$: first a reflection in the $y$-axis, giving $y=\ln(-x)$, then a horizontal translation of $4$ units in the positive $x$-direction (replacing $x$ with $x-4$), giving $y=\ln(-(x-4))=\ln(4-x)$:
$$\boxed{\text{Reflection in the } y\text{-axis, followed by a horizontal translation of 4 units in the positive } x\text{-direction.}}$$
**(d)**
The sketch shows a decreasing logarithmic curve, defined for $x<4$, rising without bound as $x\to4^-$ (approaching the dashed vertical asymptote $x=4$), passing through $(3,0)$ and $(0,1.39)$, and decreasing slowly as $x\to-\infty$, as shown in the diagram.