(i)

\begin{align*} y & = (x + 1)(2 - x) \\ \\ \text{Let } & x = 0, \\ y & = (0 + 1)(2 - 0) \\ y & = 2 \\ \\ \implies & B(0, 2) \\ \\ \text{Let } & y = 0, \\ 0 & = (x + 1)(2 - x) \end{align*} \begin{align*} x + 1 & = 0 && \text{ or } & 2 - x & = 0 \\ x & = -1 &&& -x & = -2 \\ & &&& x & = 2 \\ \\ \implies & A(-1, 0) &&& \implies & C(2, 0) \end{align*}

(ii)

\begin{align*} \text{Line of symmetry: } x & = {-1 + 2 \over 2} \\ x & = 0.5 \end{align*}

(i)

\begin{align*} y & = x^2 + hx - 5 \\ \\ \text{Let } & x = 0, \\ y & = (0)^2 + h(0) - 5 \\ y & = -5 \\ \\ \implies & A(0, -5) \end{align*}

(ii)

\begin{align*} y & = x^2 + hx - 5 \\ \\ \text{Using } & (1, -2), \\ -2 & = (1)^2 + h(1) - 5 \\ -2 & = 1 + h - 5 \\ -h & = 1 - 5 + 2 \\ -h & = -2 \\ h & = 2 \end{align*}

\begin{align*} \text{When } & y = 0, \end{align*} \begin{align*} x & = 1 && \text{ or } & x & = 4 \\ x - 1 & =0 &&& x - 4 & =0 \end{align*} \begin{align*} y & = (x - 1)(x - 4) \\ y & = x^2 - 4x - x + 4 \\ y & = x^2 - 5x + 4 \\ \\ \therefore p & = -5, q = 4 \end{align*}

(a)

\begin{align*} y & = x^3 + 4 \end{align*}

(b)

\begin{align*} y & = 4^x \end{align*}

(c)

\begin{align*} y & = {4 \over x} \end{align*}

\begin{align*} y & = x^{-2} \\ \\ n & = -2 \end{align*}

\begin{align*} y & = ka^x \\ \\ \text{Using } & (0, 3), \\ 3 & = k a^0 \\ 3 & = k(1) \\ 3 & = k \\ \\ y & = 3 a^x \\ \\ \text{Using } & (4, 48), \\ 48 & = 3 a^4 \\ {48 \over 3} & = a^4 \\ 16 & = a^4 \\ \pm \sqrt{16} & = a \end{align*} \begin{align*} a & = 2 && \text{ or } & a & = -2 \text{ (Reject)} \end{align*} $$ \therefore a = 2, k = 3 $$

(a)

\begin{align*} y & = x^2 + 2 \\ \\ \text{Let } & x = 0,\\ y & = (0)^2 + 2 \\ y & = 2 \phantom{000000} [y \text{-intercept}] \end{align*}

(ii)

\begin{align*} x + y & = 2 \\ y & = -x + 2 \phantom{000000} [y = mx + c] \\ \\ y \text{-inter} & \text{cept is } 2 \\ \\ \text{Let } & y = 0, \\ 0 & = -x + 2 \\ x & = 2 \phantom{00000000000} [x \text{-intercept}] \end{align*}

\begin{align*} y & = 9 - 4x^2 \\ \\ \text{Let } & x = 0, \\ y & = 9 - 4(0)^2 \\ y & = 9 \\ \\ \implies & P(0, 9) \\ \\ \text{Let } & y = 0, \\ 0 & = 9 - 4x^2 \\ 4x^2 & = 9 \\ x^2 & = {9 \over 4} \\ x & = \pm \sqrt{9 \over 4} \\ x & = \pm 1.5 \\ \\ \implies & Q(-1.5, 0), R(1.5, 0) \end{align*}

(a)

\begin{align*} y & = (x + 1)(x - 5) \phantom{000000} [\text{Minimum curve since } x^2 \text{ is positive}] \\ \\ \text{Let } & x = 0, \\ y & = (0 + 1)(0 - 5) \\ y & = -5 \phantom{0000000000000000}[ y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = (x + 1)(x - 5) \end{align*} \begin{align*} x + 1 & = 0 && \text{ or } & x - 5 & = 0 \\ x & = -1 &&& x & = 5 \phantom{0000000} [x \text{-intercepts}] \end{align*}
\begin{align*} \text{Line of symmetry: } x & = {-1 + 5 \over 2} \\ x & = 2 \\ \\ \text{Substitute } & x = 2 \text{ into eqn of curve,} \\ y & = (2 + 1)(2 - 5) \\ y & = -9 \\ \\ \text{Turning point: } & (2, -9) \end{align*}

(b)

$$ x = 2 $$

(c)

$$ (2, -9) $$

(i)

\begin{align*} y & = -(x + 3)^2 + 1 \\ \\ \text{Let } & y = 0, \\ 0 & = -(x + 3)^2 + 1 \\ (x + 3)^2 & = 1 \\ x + 3 & = \pm \sqrt{1} \end{align*} \begin{align*} x + 3 & = 1 && \text{ or } & x + 3 & = - 1 \\ x & = 1 - 3 &&& x & = -1 - 3 \\ x & = -2 &&& x & = -4 \\ & (-2, 0) &&& & (-4, 0) \end{align*}
\begin{align*} y & = -(x + 3)^2 + 1 \\ \\ \text{Let } & x = 0, \\ y & = -(0 + 3)^2 + 1 \\ y & = -8 \\ \\ & (0, -8) \end{align*}

(ii)

\begin{align*} y & = -(x + 3)^2 + 1 \\ \\ \text{Max. } & \text{point: } (-3, 1) \end{align*}

(iii)

\begin{align*} y & = -(x + 3)^2 + 1 \\ \\ \text{Shape: } & \text{Maximum curve } (\cap) \\ \\ y \text{-intercept: } & y = -8 \\ \\ x \text{-intercepts: } & x = -2, -4 \\ \\ \text{Turning point: } & (-3, 1) \end{align*}

(iv)

$$ x = - 3 $$

(i)

\begin{align*} x^2 - 6x + 4 & = x^2 - 6x + \left(6 \over 2\right)^2 - \left(6 \over 2\right)^2 + 4 \phantom{000000} [\text{Complete the square}] \\ & = x^2 - 6x + 3^2 - 9 + 4 \\ & = (x - 3)(x - 3) - 5 \\ & = (x - 3)^2 - 5 \end{align*}

(ii)

\begin{align*} y & = x^2 - 6x + 4 \\ y & = (x - 3)^2 - 5 \\ \\ TP: & (3, -5) \end{align*}

(iii)

\begin{align*} y & = (x - 3)^2 - 5 \phantom{000000} [\text{Minimum curve } \cup] \\ \\ TP: & (3, -5) \\ \\ \text{Let } & x = 0, \\ y & = (0 - 3)^2 - 5 \\ y & = 4 \phantom{000000000000000.} [y \text{-intercept}] \end{align*}

(iv)

$$ x = 3 $$

(i)

\begin{align*} x^2 + x + 3 & = x^2 + x + \left(1 \over 2\right)^2 - \left(1 \over 2\right)^2 + 3 \phantom{000000} [\text{Complete the square}] \\ & = \left(x + {1 \over 2}\right)^2 - {1 \over 4} + 3 \\ & = \left(x + {1 \over 2}\right)^2 + {11 \over 4} \end{align*}

(ii)

\begin{align*} y & = x^2 + x + 3 \\ & = \left(x + {1 \over 2}\right)^2 + {11 \over 4} \\ & = (x + 0.5)^2 + 2.75 \\ \\ \text{TP: } & (-0.5, 2.75) \end{align*}

(iii)

\begin{align*} y & = (x + 0.5)^2 + 2.75 \phantom{000} [\text{Minimum curve } \cup] \\ \\ \text{TP: } & (-0.5, 2.75) \\ \\ \text{Let } & x = 0, \\ y & = (0 + 0.5)^2 + 2.75 \\ y & = 3 \phantom{00000000000000000} [y \text{-intercept}] \end{align*}

(iv)

$$ x = -0.5 $$

(a)

(b)

(c)(ii)

\begin{align*} \text{Area of trapezium} & = {1 \over 2} \times \text{Sum of parallel sides} \times \text{Height} \\ & = {1 \over 2} \times (2 + 1) \times 2 \\ & = 3 \text{ units}^2 \end{align*}

(a)

$x$	$-1.5$	$-1$	$0$	$1$	$2$	$2.5$
$y$	$-4.646$	$-4.5$	$-4$	$-3$	$-1$	$0.657$

(b)

$$ x = 1.8 $$

(c)

\begin{align*} \text{For all real} & \text{ values of } x, \\ 2^x & > 0 \\ 2^x - 5 & > - 5 \\ \\ \therefore \text{Curve of } y = 2^x - 5 & \text{ always lies above } y = -5 \end{align*}

(a)

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y$	$10$	$0$	$-6$	$-8$	$-6$	$0$	$10$

(b)

\begin{align*} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ & = {2 - (-10) \over 4 - 1} \\ & = 3 \end{align*}

(c)(i)

(c)(ii)

$$ x = -0.3 \text{ or } 3.3 $$

(c)(iii)

$$ -0.3 \le x \le 3.3 $$

(a)

$x$	$0.5$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$y$	$8$	$4$	$3.5$	$4.67$	$6.25$	$8$	$9.83$	$11.7$

(b)

\begin{align*} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ & = {11 - 1.2 \over 7.4 - 0.6} \\ & = 1.4411 \\ & \approx 1.44 \end{align*}

(c)(i)

\begin{align*} {5 \over x} + 2x - 8 & = 0 \\ {5 \over x} + 2x - 8 + 5 & = 0 + 5 \\ \underbrace{ {5 \over x} + 2x - 3}_\text{Curve} & = 5 \\ \\ \text{Draw } & y = 5 \\ \\ \therefore x & = 0.8 \text{ or } 3.2 \end{align*}

(c)(ii)

(a)

(b)

Bottom half of the graph (more relevant for subsequent parts):

(c)

\begin{align*} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ & = { 0 - (-3) \over 3 - (-1) } \\ & = 0.75 \end{align*}

(d)

\begin{align*} \underbrace{ x^2(x - 2) }_\text{Curve} & = -1 \\ \\ \text{Draw } & y = -1 \\ \\ \text{From graph, } x & = -0.6 \text{ or } 1 \text{ or } 1.6 \end{align*}

(e)

(f)(i)

$$ x = -1, 1, 2 $$

(f)(ii)

\begin{align*} y & = x^2(x - 2) \phantom{0} \text{--- (1)} \\ \\ y & = x - 2 \phantom{0} \text{--- (2)} \\ \\ \text{Sub } & \text{(1) into (2),} \\ x^2(x - 2) & = x - 2 \\ x^3 - 2x & = x - 2 \\ x^3 - 2x - x + 2& = 0 \\ \\ A = -2, B& = -1, C = 2 \end{align*}

(a)

(b)(i)

\begin{align*} x^3 + 3x & = 8 \\ x^3 + 3x - 3 & = 8 - 3 \\ \underbrace{x^3 +3 x - 3}_\text{Curve} & = 5 \\ \\ \text{Draw } & y = 5 \\ \\ \text{From graph, } x & = -4.7 \text{ or } 1.7 \end{align*}

(b)(ii)

(b)(iii)

(c)

\begin{align*} \text{Line from (b)(iii) } y & = x \text{ has gradient} = 1 \phantom{000000} [y = mx + c] \\ \\ \therefore \text{Tangent must be } & \text{parallel to } y = x \\ \\ & \therefore (-1, -5) \end{align*}

(a)

\begin{align*} \text{Area of square} & = x \times x \\ & = x^2 \\ \\ \text{Volume} & = x^2 \times h \\ 35 & = hx^2 \\ {35 \over x^2 } & = h \\ \\ \text{Height} & = {35 \over x^2} \text{ m} \end{align*}

(b)

\begin{align*} A & = x^2 + 4(x \times h) \\ & = x^2 + 4xh \\ & = x^2 + 4x \left(35 \over x^2\right) \phantom{000000} [\text{From (a)}] \\ & = x^2 + \left(4x \over 1\right)\left(35 \over x^2\right) \\ & = x^2 + {140x \over x^2} \\ & = x^2 + {140 \over x} \phantom{0} \text{ (Shown)} \end{align*}

(c)

(d)

Bottom half of the graph:

(e)(i)

$$ x = 3.075 \text{ m} \text{ or } 5.4 \text{ m} $$

(e)(ii)

$$ \text{Minimum value of } A = 51 $$

(e)(iii)

\begin{align*} \text{From graph, } x & = 4.1 \\ \\ h & = {35 \over x^2} \\ & = {35 \over 4.1^2} \\ & = 2.08209 \\ & \approx 2.08 \text{ m} \end{align*}

First part:

\begin{align*} \text{Initial } (t = 2) \text{ speed of car} & = 0 \text{ m/s} \implies (2, 0) \\ t = 3, \text{ Speed of car} & = 3 \text{ m/s} \implies (3, 3) \\ t = 4, \text{ Speed of car} & = 6 \text{ m/s} \implies (4, 6) \end{align*}

(a)

$$ \text{From graph, } t = 6.4 \text{ seconds} $$

(b) On the speed-time graph, gradient of curve = acceleration of object

\begin{align*} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ & = {14.1 - 9 \over 7.5 - 0} \\ & = 0.68 \\ \\ \text{Acceleration} & = 0.68 \text{ m/s}^2 \end{align*}

(a)

\begin{align*} \text{Base of box} & = (16 - 2x) \text{ cm} \\ \\ \text{Height of box} & = x \text{ cm} \\ \\ \text{Volume} & = (16 - 2x) \times (16 - 2x) \times x \\ & = x(16 - 2x)^2 \\ & = x[ (16)^2 - 2(16)(2x) + (2x)^2 ] \phantom{000000} [ (a - b)^2 = a^2 - 2ab + b^2 ] \\ & = x (256 - 64x + 4x^2) \\ & = 256x - 64x^2 + 4x^3 \\ V & = ( 4x^3 - 64x^2 + 256x) \text{ cm}^2 \phantom{0} \text{ (Shown)} \end{align*}

(b)

(c)

(d)(i)

$$ V = 304 \phantom{000000} [y \text{-coordinate of maximum point}] $$

(d)(ii)

\begin{align*} \text{Base of box} & = 16 - 2x \\ & = 16 - 2(2.65) \phantom{000000} [x \text{-coordinate of maximum point}] \\ & = 10.7 \\ \\ \text{Base area} & = 10.7 \times 10.7 \\ & = 114.49 \text{ cm}^2 \end{align*}

(e)

\begin{align*} \text{Base of box} & = (16 - 2x) \text{ cm} \\ \\ \text{If } x > 8, \text{ base } & \text{of box will have negative value} \end{align*}

(a)

\begin{align*} \text{Radius of semicircle} & = {1 \over 2}x \text{ cm} \\ \\ \text{Circumference of semicircle} & = {1 \over 2} \times 2 \pi r \\ & = \pi r \\ & = \pi \left({1 \over 2}x\right) \\ & = {1 \over 2} \pi x \text{ cm} \\ \\ \text{Perimeter} & = {1 \over 2} \pi x + 2y + x \\ 12 & = {1 \over 2} \pi x + 2y + x \\ -2y & = {1 \over 2} \pi x + x - 12 \\ 2y & = 12 - {1 \over 2} \pi x - x \\ y & = {1 \over 2} \left(12 - {1 \over 2} \pi x - x \right) \\ y & = 6 - {1 \over 4} \pi x - {1 \over 2} x \end{align*}

(b)

\begin{align*} \text{Area of semicircle} & = {1 \over 2} \pi r^2 \\ & = {1 \over 2} \pi \left({1 \over 2}x\right)^2 \\ & = {1 \over 2} \pi \left({1 \over 4}x^2\right) \phantom{000000} \left[ {1 \over 2} x \times {1 \over 2}x = {1 \over 4} x^2 \right] \\ & = {1 \over 8} \pi x^2 \\ \\ \text{Area of rectangle} & = xy \\ & = x \left(6 - {1 \over 4} \pi x - {1 \over 2}x \right) \\ & = 6x - {1 \over 4} \pi x^2 - {1 \over 2}x^2 \\ \\ A & = {1 \over 8} \pi x^2 + 6x - {1 \over 4} \pi x^2 - {1 \over 2}x^2 \\ & = 6x - {1 \over 2}x^2 + {1 \over 8} \pi x^2 - {1 \over 4} \pi x^2 \\ & = \left( 6x - {1 \over 2}x^2 - {1 \over 8} \pi x^2 \right) \text{ m}^2 \phantom{0} \text{ (Shown)} \end{align*}

(c)

(d)

(e)(i)

$$ A = 10.1 \text{ m}^2 \phantom{000000} [y \text{-coordinate of maximum point}] $$

(e)(ii)

\begin{align*} \text{Arc length} & = \text{Cicumference of semicircle} \\ & = {1 \over 2} \pi x \\ & = {1 \over 2} \pi (3.35) \phantom{000000} [x \text{-coordinate of maximum point}] \\ & = 5.2621 \\ & \approx 5.26 \text{ cm} \end{align*}