(a)

(b)

(c) Look for the x-intercept of the graph

$$ \text{From graph, } x = 2 $$

(a)

(b)

(c) Look for the x-intercepts of the graph

$$ \text{From graph, } x = 0, 3 $$

(a)

(b)

(c) The graph does not meet the x-axis at all!

$$ \text{From graph, there are no solutions} $$

(a)

\begin{align} x^2 & = x + 6 \\ \underbrace{x^2 - x - 6}_\text{Curve} & = 0 \\ \\ \text{From graph, } & x = -2, 3 \phantom{000000} [\text{Look for } x \text{-intercepts}] \end{align}

(b)(i)

\begin{align} x^2 - x - 5 & = 0 \\ x^2 - x - 5 - 1 & = -1 \\ \underbrace{x^2 - x - 6}_\text{Curve} & = -1 \\ \\ \text{Draw } & y = -1 \end{align}

$$ \text{From graph, } x = -1.8, 2.8 $$

(b)(ii)

\begin{align} x^2 - 2x + 1 & = 0 \\ x^2 - 2x + x + 1 & = 0 + x \\ x^2 - x + 1 & = x \\ x^2 - x + 1 - 7 & = x - 7 \\ \underbrace{x^2 - x - 6}_\text{Curve} & = x - 7 \\ \\ \text{Draw } & y = x - 7 \end{align}

x	0	2	4
y	-7	-5	-3

$$ \text{From graph, } x = 1 $$

(a)

x	-2	-1	0	1	2	3	4
y	5	0	-3	-4	-3	0	5

(b)

\begin{align} x^2 & = 2x + 3 \\ \underbrace{x^2 - 2x - 3}_\text{Curve} & = 0 \\ \\ \text{From graph, } & x = -1, 3 \phantom{000000} [\text{Look for } x \text{-intercepts}] \end{align}

(c)(i)

\begin{align} x^2 - 2x & = 6 \\ x^2 - 2x - 3 & = 6 - 3 \\ \underbrace{x^2 - 2x - 3}_\text{Curve} & = 3 \\ \\ \text{Draw } & y = 3 \\ \\ \text{From graph, } & x = -1.65, 3.65 \end{align}

(c)(ii)

(a)

x	-4	-3	-2	-1	0	1	2
y	11	2	-3	-4	-1	6	17

(b)

\begin{align} 2x^2 + 5x & = 1 \\ \underbrace{2x^2 + 5x - 1}_\text{Curve} & = 0 \\ \\ \text{From graph, } & x = -2.7, 0.2 \phantom{000000} [\text{Look for } x \text{-intercepts}] \end{align}

(c)(i)

\begin{align} 2x^2 + 5x - 5 & = 0 \\ 2x^2 + 5x - 5 + 4 & = 0 + 4 \\ \underbrace{2x^2 + 5x - 1}_\text{Curve} & = 4 \\ \\ \text{Draw } & y = 4 \\ \\ \text{From graph, } & x = -3.25, 0.75 \end{align}

(c)(ii)

(a)

x	-2	-1	0	1	2	3
y	20	9	4	5	12	25

(b)

\begin{align} 3x^2 & = 2x - 4 \\ \underbrace{3x^2 - 2x + 4}_\text{Curve} & = 0 \\ \\ \text{From graph, no solutions} & \text{ since there are no } x \text{-intecept(s)} \end{align}

(c)(i)

\begin{align} 3x^2 - 2x - 10 & = 0 \\ 3x^2 - 2x - 10 + 14 & = 0 + 14 \\ \underbrace{3x^2 - 2x + 4}_\text{Curve} & = 14 \\ \\ \text{Draw } & y = 14 \\ \\ \text{From graph, } & x = -1.5, 2.2 \end{align}

(c)(ii)

(a)

x	-4	-3	-2	-1	0	1	2
y	-2	3	6	7	6	3	-2

(b)

\begin{align} x^2 + 2x - 6 & = 0 \\ 0 & = \underbrace{6 - 2x - x^2}_\text{Curve} \\ \\ \text{From graph, } & x = -3.65, 1.65 \phantom{000000} [\text{Look for } x \text{-intercepts}] \end{align}

(c)(i)

\begin{align} x^2 + 2x - 2 & = 0 \\ 0 & = 2 - 2x - x^2 \\ 0 + 4 & = 2 - 2x - x^2 + 4 \\ 4 & = \underbrace{6 - 2x - x^2}_\text{Curve} \\ \\ \text{Draw } & y = 4 \\ \\ \text{From graph, } & x = -2.75, 0.75 \end{align}

(c)(ii)

(a)

x	-2	-1	0	1	2	3	4
y	-22	-7	2	5	2	-7	-22

(b)(i)

\begin{align} 3x^2 - 6x - 2 & = 0 \\ 0 & = \underbrace{-3x^2 + 6x + 2}_\text{Curve} \\ \\ \text{From graph, } & x = -0.3, 2.3 \phantom{000000} [\text{Look for } x \text{-intercepts}] \end{align}

(b)(ii)

\begin{align} 7 + 6x - 3x^2 & = 0 \\ - 3x^2 + 6x + 7 & = 0 \\ - 3x^2 + 6x + 7 - 5 & = 0 - 5 \\ \underbrace{ -3x^2 + 6x + 2 }_\text{Curve} & = - 5 \\ \\ \text{Draw } & y = -5 \\ \\ \text{From graph, } & x = -0.8, 2.8 \end{align}

(c)(i)

(c)(ii)

\begin{align} y & = -4 + 5x \phantom{00} \text{--- (1)} \\ \\ y & = - 3x^2 + 6x + 2 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ -4 + 5x & = -3x^2 + 6x + 2 \\ 0 & = -3x^2 + 6x - 5x + 2 + 4 \\ 0 & = -3x^2 + x + 6 \\ \\ \therefore b & = 1, c = 6 \end{align}

(a)

x	-6	-5	-4	-3	-2	-1	0	1	2
y	6	2.5	0	-1.5	-2	-1.5	0	2.5	6

(b)

\begin{align} {1 \over 2}x^2 + 2x + 1 & = 0 \\ \underbrace{ {1 \over 2}x^2 + 2x }_\text{Curve} & = -1 \\ \\ \text{Draw } & y = - 1 \\ \\ \text{From graph, } & x = -3.4, -0.6 \end{align}

(c)(i)

(c)(ii)

\begin{align} y & = {1 \over 2}x^2 + 2x \phantom{00} \text{--- (1)} \\ \\ y & = 2 - {1 \over 2}x \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ {1 \over 2}x^2 + 2x & = 2 - {1 \over 2}x \\ {1 \over 2}x^2 + 2x + {1 \over 2}x - 2 & = 0 \\ {1 \over 2}x^2 + {5 \over 2}x - 2 & = 0 \\ 2 \left({1 \over 2}x^2 + {5 \over 2}x - 2\right) & = 2(0) \\ x^2 + 5x - 4 & = 0 \end{align}

(a)

x	-3	-2	-1	0	1	2	3	4
y	-16	-9	-4	-1	0	-1	-4	-9

(b)

\begin{align} -2x^2 + 4x & = 2 \\ {1 \over 2}(-2x^2 + 4x) & = {1 \over 2}(2) \\ -x^2 + 2x & = 1 \\ \underbrace{-x^2 + 2x - 1}_\text{Curve} & = 0 \\ \\ \text{From graph, } & x = 1 \phantom{000000} [\text{Look for } x \text{-intercept}] \end{align}

(c) The maximum point of the graph is (1, 0). Thus, k must be smaller than 0,

$$ k < 0 $$

(d) I think this part is more for A Maths

\begin{align} x^2 + 2x + 2 & = mx \\ -(x^2 + 2x + 2) & = - mx \\ - x^2 - 2x - 2 & = - mx \\ - x^2 - 2x - 2 + 4x & = -mx + 4x \\ - x^2 + 2x - 2 & = (4 - m)x \\ - x^2 + 2x - 2 + 1 & = (4 - m)x + 1 \\ \underbrace{-x^2 + 2x - 1}_\text{Curve} & = (4 - m)x + 1 \\ \\ \text{Draw } & y = (4 - m)x + 1 \\ \\ \text{Let } & x = 0, \\ y & = (4 - m)(0) + 1 \\ y & = 0 + 1 \\ y & = 1 \\ \\ \therefore \text{Construct two lines } & \text{from } (0, 1) \text{ that meet curve only once} \\ \\ \\ y & = (4 - m)x + 1 \end{align} \begin{align} \text{Using } & (-1.45, -6), &&& \text{Using } & (1.45, -0.2), \\ -6 & = (4 - m)(-1.45) + 1 &&& -0.2 & = (4 - m)(1.45) + 1 \\ -6 & = -5.8 + 1.45m + 1 &&& -0.2 & = 5.8 - 1.45m + 1 \\ -1.45m & = -5.8 + 1 + 6 &&& 1.45m & = 5.8 + 1 + 0.2 \\ -1.45m & = 1.2 &&& 1.45m & = 7 \\ m & = {1.2 \over -1.45} &&& m & = {7 \over 1.45} \\ m & = -0.82758 &&& m & = 4.8275 \\ m & \approx -0.83 \text{ ( 2 d.p.)} &&& m & \approx 4.83 \text{ (2 d.p.)} \end{align}