Functions I

(a) Method: use the gradient formula through the two given points on the line, $(0, 800)$ and $(1200, 1400)$. $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ Substituting the two points: $$m = \frac{1400 - 800}{1200 - 0}$$ Simplifying: $$m = \frac{600}{1200} = 0.5$$ $$\boxed{m = 0.5}$$ (b) Since the base station $(0, 800)$ lies on the line at $x = 0$, this point is the $y$-intercept, so $c = 800$. Using the gradient found in (a): $$y = mx + c$$ Substituting $m = 0.5$ and $c = 800$: $$\boxed{y = 0.5x + 800}$$ (c) The gradient represents the rate of change of elevation $y$ with respect to horizontal distance $x$, i.e. how many metres the cable-car rises for every metre it moves horizontally. $$\boxed{\text{The gradient } 0.5 \text{ means the cable-car rises } 0.5\text{ m in elevation for every } 1\text{ m of horizontal distance travelled.}}$$

(a) Method: rearrange the equation of the existing road into gradient-intercept form $y = mx + c$ to read off the gradient. $$3x - 2y + 6 = 0$$ Isolating $y$: $$2y = 3x + 6$$ Dividing through by 2: $$y = 1.5x + 3$$ $$\boxed{\text{gradient of existing road} = 1.5}$$ (b) Method: parallel lines share the same gradient, so $L_1$ has gradient $1.5$; use the point-gradient form with $A(2,5)$. $$y - y_1 = m(x - x_1)$$ Substituting $m = 1.5$ and $(x_1, y_1) = (2, 5)$: $$y - 5 = 1.5(x - 2)$$ Expanding: $$y = 1.5x - 3 + 5$$ Simplifying: $$\boxed{y = 1.5x + 2}$$

(a) Method: rearrange $2x + 5y - 20 = 0$ into gradient-intercept form. $$2x + 5y - 20 = 0$$ $$5y = -2x + 20$$ $$y = -0.4x + 4$$ $$\boxed{\text{gradient of drainage pipe} = -0.4 \left(= -\dfrac{2}{5}\right)}$$ (b) Method: for perpendicular lines $m_1 \times m_2 = -1$, so the connecting pipe's gradient is the negative reciprocal of $-0.4$. $$m_2 = -\frac{1}{m_1}$$ Substituting $m_1 = -0.4$: $$m_2 = -\frac{1}{-0.4} = 2.5$$ Using the point-gradient form with the junction point $(10, 6)$: $$y - 6 = 2.5(x - 10)$$ Expanding: $$y = 2.5x - 25 + 6 = 2.5x - 19$$ Rearranging into the form $ax + by + d = 0$ and multiplying through by 2 to obtain integer coefficients: $$2.5x - y - 19 = 0 \implies 5x - 2y - 38 = 0$$ $$\boxed{5x - 2y - 38 = 0}$$

(a) Method: use the gradient formula with the two given data points, then substitute one point to find $c$. $$m = \frac{d_2 - d_1}{x_2 - x_1}$$ Substituting $(20, 4.5)$ and $(80, 13.5)$: $$m = \frac{13.5 - 4.5}{80 - 20}$$ Simplifying: $$m = \frac{9}{60} = 0.15$$ Using the point $(20, 4.5)$ in $d = mx + c$: $$4.5 = 0.15(20) + c$$ $$4.5 = 3 + c$$ $$c = 1.5$$ $$\boxed{d(x) = 0.15x + 1.5}$$ (b) Method: substitute $x = 150$ into the model. $$d(150) = 0.15(150) + 1.5$$ $$d(150) = 22.5 + 1.5$$ $$\boxed{d(150) = 24 \text{ m}}$$ (c) Method: set $d(x) = 20$ and solve for $x$. $$20 = 0.15x + 1.5$$ $$18.5 = 0.15x$$ $$x = \frac{18.5}{0.15} = 123.33\ldots$$ $$\boxed{x \approx 123 \text{ m (3 s.f.)}}$$

(a) Method: use the gradient formula with $A(2,3)$ and $B(10,7)$, then the point-gradient form. $$m_{AB} = \frac{y_2 - y_1}{x_2 - x_1}$$ Substituting $A(2,3)$ and $B(10,7)$: $$m_{AB} = \frac{7 - 3}{10 - 2} = \frac{4}{8} = 0.5$$ Using point $A(2,3)$: $$y - 3 = 0.5(x - 2)$$ $$y = 0.5x - 1 + 3$$ $$\boxed{y = 0.5x + 2}$$ (b) Method: since the power line is perpendicular to $AB$, its gradient is the negative reciprocal of $0.5$. $$m_{CD} = -\frac{1}{0.5} = -2$$ Using point $C(20, 2)$: $$y - 2 = -2(x - 20)$$ $$y = -2x + 40 + 2$$ $$\boxed{y = -2x + 42}$$ (c) Method: solve the two line equations simultaneously. $$0.5x + 2 = -2x + 42$$ $$2.5x = 40$$ $$x = 16$$ Substituting back into either line equation: $$y = 0.5(16) + 2 = 10$$ $$\boxed{D = (16, 10)}$$ (d) Method: use the distance formula between $C(20, 2)$ and $D(16, 10)$. $$CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ Substituting: $$CD = \sqrt{(20 - 16)^2 + (2 - 10)^2}$$ $$CD = \sqrt{16 + 64} = \sqrt{80}$$ $$CD = 8.9442\ldots$$ $$\boxed{CD \approx 8.94 \text{ km (3 s.f.)}}$$

(a) Method: substitute $x = 4$ into $f(x) = \sqrt{x+5}$. $$f(4) = \sqrt{4+5}$$ $$f(4) = \sqrt{9}$$ $$\boxed{f(4) = 3}$$ (b) Since the domain is $x \geq -5$, the smallest possible value of $x+5$ is $0$, occurring at $x=-5$: $$f(-5) = \sqrt{0} = 0$$ As $x$ increases without bound, $x+5$ also increases without bound, so $f(x)$ increases without bound. Since the square root function only ever returns non-negative outputs, $f(x)$ takes every value from $0$ upward. $$\boxed{\text{Range: } f(x) \geq 0}$$

(a) Method: read the restriction directly from the context - the vending machine sells whole cups, with a minimum of $1$ and a maximum of $10$ per transaction. $$\boxed{\text{Domain: } x \in \{1, 2, 3, \ldots, 10\}}$$ (b) Method: substitute $x = 6$ into $C(x) = 2.50x$. $$C(6) = 2.50 \times 6$$ $$\boxed{C(6) = \$15.00}$$, the cost of buying $6$ cups of hot chocolate. (c) Method: since $C$ is an increasing function of $x$ on a discrete domain, find the range by evaluating $C$ at the least and greatest values of $x$. $$C(1) = 2.50 \times 1 = 2.50$$ $$C(10) = 2.50 \times 10 = 25.00$$ $$\boxed{\text{Range: } C(x) \in \{2.50, 5.00, 7.50, \ldots, 25.00\}}$$

(a) Method: use the two given data points to find the gradient $m$, then substitute a point to find $c$. $$m = \frac{C_2 - C_1}{t_2 - t_1}$$ Substituting $(2, 130)$ and $(5, 220)$: $$m = \frac{220 - 130}{5 - 2} = \frac{90}{3} = 30$$ Using $(2, 130)$ in $C(t) = mt + c$: $$130 = 30(2) + c$$ $$130 = 60 + c$$ $$c = 70$$ $$\boxed{C(t) = 30t + 70}$$ (b) Since $t$ represents a whole number of completed months of membership and cannot be negative: $$\boxed{\text{Domain: } t \in \mathbb{Z}, \ t \geq 0}$$ (c) Method: substitute $t = 12$ into the model. $$C(12) = 30(12) + 70$$ $$C(12) = 360 + 70$$ $$\boxed{C(12) = \$430}$$, the total amount paid after $12$ months of membership (the \$70 joining fee plus $12$ months at \$30 per month).

(a) Method: substitute the domain endpoints $x = 0$ and $x = 5$ into $T(x) = x^2 - 4x + 1$. $$T(0) = 0^2 - 4(0) + 1 = 1$$ $$T(5) = 5^2 - 4(5) + 1 = 25 - 20 + 1 = 6$$ $$\boxed{T(0) = 1, \ T(5) = 6}$$ (b) Method: since $T$ is a quadratic with a positive leading coefficient (an upward-opening parabola), its minimum occurs at the vertex, $x = -\dfrac{b}{2a}$. $$x = -\frac{b}{2a}$$ Substituting $a = 1$, $b = -4$: $$x = -\frac{-4}{2(1)} = 2$$ Checking $x = 2$ lies within the domain $[0,5]$: it does. Substituting $x = 2$ into $T(x)$: $$T(2) = 2^2 - 4(2) + 1 = 4 - 8 + 1 = -3$$ $$\boxed{\text{Minimum } T = -3 \text{ at } x = 2}$$ (c) Method: since the vertex found in (b) lies inside the domain, the minimum of the range is $T = -3$; the maximum of the range is the larger of the two endpoint values from (a). $$\max(T(0), T(5)) = \max(1, 6) = 6$$ $$\boxed{\text{Range: } -3 \leq T(x) \leq 6}$$

(a) Method: compute the gradient between each consecutive pair of data points and check that it is constant. Between $t=0$ and $t=3$: $$m = \frac{11 - 5}{3 - 0} = \frac{6}{3} = 2$$ Between $t=3$ and $t=6$: $$m = \frac{17 - 11}{6 - 3} = \frac{6}{3} = 2$$ Since the gradient is the same ($m = 2$) between each pair of points, the data is consistent with a linear model. Using $(0, 5)$ in $N(t) = mt + c$: $$5 = 2(0) + c \implies c = 5$$ $$\boxed{N(t) = 2t + 5}$$ (b) Since the culture is monitored for the first $10$ hours only: $$\boxed{\text{Domain: } 0 \leq t \leq 10}$$ Since $N$ is an increasing function of $t$, the range is found by evaluating $N$ at the domain endpoints. $$N(0) = 2(0) + 5 = 5$$ $$N(10) = 2(10) + 5 = 25$$ $$\boxed{\text{Range: } 5 \leq N(t) \leq 25}$$ (c) Method: set $N(t) = 50$ and solve for $t$. $$2t + 5 = 50$$ $$2t = 45$$ $$\boxed{t = 22.5 \text{ hours}}$$ (d) The value $t = 22.5$ hours lies far outside the domain over which data was actually collected ($0 \leq t \leq 10$), so this is a significant extrapolation. Bacterial growth typically slows as resources become limited, so the true relationship is unlikely to remain linear that far beyond the observed data. $$\boxed{\text{The prediction } t = 22.5 \text{ hours should NOT be trusted with confidence.}}$$

(a) Method: identify the fixed fee as the $y$-intercept and the per-kilometre charge as the gradient. $$C(d) = (\text{rate per km}) \times d + (\text{fixed fee})$$ Substituting the rate $\$2.50$ per km and the fixed fee $\$3$: $$\boxed{C(d) = 2.50d + 3}$$ (b) Method: substitute $d = 12$ into $C(d)$. $$C(12) = 2.50(12) + 3$$ $$C(12) = 30 + 3$$ $$\boxed{C(12) = \$33}$$

(a) Method: a $10\%$ annual increase corresponds to multiplying the previous year's population by $1 + 0.10 = 1.10$ each year. $$\boxed{r = 1.1}$$ (b) Method: substitute $P_0 = 200$ and $r = 1.1$ into $P(t) = P_0 r^t$. $$\boxed{P(t) = 200(1.1)^t}$$ (c) Method: substitute $t = 5$ into $P(t)$. $$P(5) = 200(1.1)^5$$ $$P(5) = 200 \times 1.61051$$ $$P(5) = 322.102$$ $$\boxed{P(5) \approx 322 \text{ deer (3 s.f.)}}$$

(a) Method: substitute each data point into $h(t) = at^2 + bt + c$ to form a system of equations, then solve. From $t = 0$: $h(0) = c = 2$, so $c = 2$. From $t = 1$: $h(1) = a + b + c = 22$, so $a + b = 20$. From $t = 2$: $h(2) = 4a + 2b + c = 34$, so $4a + 2b = 32$, i.e. $2a + b = 16$. Subtracting the first of these two equations from the second: $$(2a + b) - (a + b) = 16 - 20$$ $$a = -4$$ Substituting back: $$b = 20 - (-4) = 24$$ $$\boxed{a = -4, \ b = 24, \ c = 2}$$, so $h(t) = -4t^2 + 24t + 2$ (b) Method: substitute $t = 3$ into the model. $$h(3) = -4(3)^2 + 24(3) + 2$$ $$h(3) = -36 + 72 + 2$$ $$\boxed{h(3) = 38 \text{ m}}$$

(a) Method: substitute the known values into $m(t) = m_0 e^{-kt}$ and solve for $k$ using logarithms. $$55 = 80e^{-5k}$$ $$e^{-5k} = \frac{55}{80} = 0.6875$$ Taking the natural logarithm of both sides: $$-5k = \ln(0.6875) = -0.374693\ldots$$ $$k = 0.0749386\ldots$$ $$\boxed{k \approx 0.0749 \text{ (3 s.f.)}}$$ (b) Method: substitute $t = 20$, using the unrounded value of $k$ from (a). $$m(20) = 80e^{-0.0749386\ldots \times 20}$$ $$m(20) = 80e^{-1.49877\ldots}$$ $$m(20) = 80 \times 0.223368\ldots$$ $$m(20) = 17.8695\ldots$$ $$\boxed{m(20) \approx 17.9 \text{ mg (3 s.f.)}}$$ (c) Method: the half-life is the time taken for the mass to halve, so solve $0.5 = e^{-kt}$. $$0.5 = e^{-kt}$$ Taking the natural logarithm of both sides: $$-kt = \ln(0.5)$$ $$t = \frac{\ln 2}{k}$$ Substituting the unrounded value of $k$: $$t = \frac{0.693147\ldots}{0.0749386\ldots}$$ $$t = 9.24952\ldots$$ $$\boxed{\text{Half-life} \approx 9.25 \text{ days (3 s.f.)}}$$

(a) Method: subtract the cost function from the revenue function. $$P(x) = R(x) - C(x)$$ Substituting $R(x) = 6x$ and $C(x) = 0.5x^2 - 4x + 30$: $$P(x) = 6x - (0.5x^2 - 4x + 30)$$ Expanding and simplifying: $$P(x) = 6x - 0.5x^2 + 4x - 30$$ $$\boxed{P(x) = -0.5x^2 + 10x - 30}$$ (b) Method: solve $P(x) = 0$ using the quadratic formula. $$-0.5x^2 + 10x - 30 = 0$$ Dividing through by $-0.5$: $$x^2 - 20x + 60 = 0$$ Applying the quadratic formula: $$x = \frac{20 \pm \sqrt{20^2 - 4(1)(60)}}{2}$$ Simplifying inside the square root: $$x = \frac{20 \pm \sqrt{160}}{2}$$ $$\sqrt{160} = 12.6491\ldots$$ $$x = \frac{20 - 12.6491\ldots}{2} = 3.6755\ldots \quad \text{or} \quad x = \frac{20 + 12.6491\ldots}{2} = 16.3246\ldots$$ $$\boxed{x \approx 3.68 \text{ and } x \approx 16.3 \text{ (3 s.f.)}}$$ (c) Method: the maximum profit occurs at the vertex of the concave-down parabola $P(x)$, at $x = -\dfrac{b}{2a}$. $$x = -\frac{b}{2a}$$ Substituting $a = -0.5$, $b = 10$: $$x = -\frac{10}{2(-0.5)} = 10$$ Substituting $x = 10$ into $P(x)$: $$P(10) = -0.5(10)^2 + 10(10) - 30$$ $$P(10) = -50 + 100 - 30$$ $$P(10) = 20$$ $$\boxed{\text{Maximum profit} = \$2000 \text{ (i.e. 20 hundred dollars), at } x = 10}$$ (d) Since $P(x)$ is a concave-down (downward-opening) parabola that equals zero at $x \approx 3.68$ and $x \approx 16.3$ from part (b), profit is positive strictly between these two roots. $$\boxed{3.68 < x < 16.3 \text{ (3 s.f.)}}$$