(a)

\begin{align} f(x) & = |3x - 1| \\ \\ \\ f(-2) & = |3(-2) - 1| \\ & = |-7| \\ & = 7 \\ \\ \\ f(2) & = |3(2) - 1| \\ & = |5| \\ & = 5 \\ \\ \\ \text{When } & f(x) = 3, \\ 3 & = |3x - 1| \end{align} \begin{align} 3 & = 3x - 1 \phantom{000} &\text{or}\phantom{00000} -3 & = 3x - 1 \\ 3 + 1 & = 3x & -3 + 1 & = 3x \\ 4 & = 3x & -2 & = 3x \\ {4 \over 3} & = x & -{2 \over 3} & = x \end{align}
$$ 3 = |3x - 1| $$ \begin{align} \text{When } & x = {4 \over 3}, \phantom{000}&\text{or}\phantom{00000} \text{When } & x = -{2 \over 3}, \\ \\ \text{R.H.S} & = \left| 3\left(4 \over 3\right) - 1 \right| & \text{R.H.S} & = \left| 3\left(-{2 \over 3}\right) - 1 \right| \\ & = |3| & & = |-3| \\ & = 3 & & = 3 \\ & = \text{L.H.S} & & = \text{L.H.S} \\ \\ \\ \therefore x & = {4 \over 3}, -{2 \over 3} \end{align}

(b)

\begin{align} g(x) & = |x^2 - 2x| \\ \\ g(1) & = |(1)^2 - 2(1)| \\ & = | - 1 | \\ & = 1 \\ \\ \\ g(3) & = |(3)^2 - 2(3)| \\ & = |3| \\ & = 3 \end{align}

(a)

$$ |2x - 3| = x $$ \begin{align} 2x - 3 & = x \phantom{000}&\text{or}\phantom{00000} 2x - 3 & = -x \\ 2x - x & = 3 & 2x + x & = 3 \\ x & = 3 & 3x & = 3 \\ & & x & = {3 \over 3} \\ & & x & = 1 \end{align}
$$ |2x - 3| = x $$ \begin{align} \text{When } & x = 3, \phantom{000}&\text{or}\phantom{00000} \text{When } & x = 1, \\ \\ \text{L.H.S} & = |2(3) - 3| & \text{L.H.S} & = |2(1) - 3| \\ & = |3| & & = |-1| \\ & = 3 & & = 1 \\ \\ \text{R.H.S} & = 3 & \text{R.H.S} & = 1 \\ & = \text{L.H.S} & & = \text{L.H.S} \\ \\ \\ \therefore x & = 3, 1 \end{align}

(b)

$$ |x + 1| = 2x - 3 $$ \begin{align} x + 1 & = 2x - 3 \phantom{000}&\text{or}\phantom{00000} x + 1 & = -2x + 3 \\ x - 2x & = -3 - 1 & x + 2x & = 3 - 1 \\ -x & = -4 & 3x & = 2 \\ x & = 4 & x & = {2 \over 3} \end{align}
$$ |x + 1| = 2x - 3 $$ \begin{align} \text{When } & x = 4, \phantom{000}&\text{or}\phantom{00000} \text{When } & x = {2 \over 3}, \\ \\ \text{L.H.S} & = |4 + 1| & \text{L.H.S} & = \left| {2 \over 3} + 1 \right| \\ & = |5| & & = \left| {5 \over 3} \right| \\ & = 5 & & = {5 \over 3} \\ \\ \text{R.H.S} & = 2(4) - 3 & \text{R.H.S} & = 2 \left(2 \over 3\right) - 3 \\ & = 5 & & = -{5 \over 3} \\ & = \text{L.H.S} & & \ne \text{R.H.S} \\ \\ \\ \therefore x & = 4 \end{align}

(c)

$$ |x^2 + 6| = 5x $$ \begin{align} x^2 + 6 & = 5x \phantom{000}&\text{or}\phantom{00000} x^2 + 6 & = -5x \\ x^2 - 5x + 6 & = 0 & x^2 + 5x + 6 & = 0 \\ (x - 2)(x - 3) & = 0 \\ & & b^2 - 4ac & = (5)^2 - 4(1)(6) \\ x = 2 \text{ or } & x = 3 & & = -4 < 0 \\ & & \therefore \text{Equation } & \text{has no real roots}. \end{align}
$$ |x^2 + 6| = 5x $$ \begin{align} \text{When } & x = 2, \phantom{000}&\text{or}\phantom{00000} \text{When } & x = 3, \\ \\ \text{L.H.S} & = |(2)^2 + 6| & \text{L.H.S} & = |(3)^2 + 6| \\ & = |10| & & = |15| \\ & = 10 & & = 15 \\ \\ \text{R.H.S} & = 5(2) & \text{R.H.S} & = 5(3) \\ & = 10 & & = 15 \\ & = \text{L.H.S} & & = \text{L.H.S} \\ \\ \therefore x & = 2, 3 \end{align}

(d)

$$ |\sqrt[3] {x} - 1| = 3 $$ \begin{align} \sqrt[3]{x} - 1 & = 3 \phantom{000}&\text{or}\phantom{00000} \sqrt[3]{x} - 1 & = - 3 \\ \sqrt[3]{x} & = 3 + 1 & \sqrt[3]{x} & = -3 + 1 \\ & = 4 & & = - 2 \\ x & = (4)^3 & x & = (-2)^3 \\ & = 64 & & = -8 \end{align}
\begin{align} \text{When } & x = 64, \phantom{000}&\text{or}\phantom{00000} \text{When } & x = -8, \\ \\ \text{L.H.S} & = |\sqrt[3]{64} - 1| & \text{L.H.S} & = |\sqrt[3]{-8} - 1| \\ & = |3| & & = | - 3| \\ & = 3 & & = 3 \\ & = \text{R.H.S} & & = \text{R.H.S} \\ \\ \\ \therefore x & = 64, -8 \end{align}

(a)

\begin{align} y & = - 3x - 2 \phantom{00000} [\text{Straight line } y = mx + c] \\ \\ \text{When } & x = 0, \\ y & = -3(0) - 2 \\ & = -2 \\ \\ \implies & y \text{-intercept is } -2 \\ \\ \text{When } & y = 0, \\ 0 & = -3x - 2 \\ 3x & = -2 \\ x & = -{2 \over 3} \\ \\ \implies & x \text{-intercept is } -{2 \over 3} \\ \\ \\ \text{When } & x = -2, \\ y & = -3(-2) -2 \\ & = 4 \\ \\ \text{Left } & \text{limit is } (-2, 4) \\ \\ \text{When } & x = 1, \\ y & = -3(1) - 2 \\ & = -5 \\ \\ \text{Right } & \text{limit is } (1, -5) \end{align}

Reflect the portion of the graph below the $x$-axis to obtain the graph of $y = |-3x - 2|$.

(b)

\begin{align} y & = 9 - x^2 \\ & = -x^2 + 9 \\ \\ \text{Coefficient of } x^2 \text{ is negative} & \implies \text{Maximum curve } (\cap) \\ \\ \text{When } & x = 0, \\ y & = -(0)^2 + 9 \\ & = 9 \\ \\ \implies & y \text{-intercept is } 9 \\ \\ \text{When } & y = 0, \\ 0 & = -x^2 + 9 \\ x^2 & = 9 \\ x & = \pm \sqrt{9} \\ & = \pm 3 \\ \\ \implies & x \text{-intercepts are } 3, -3 \\ \\ x-\text{coordinate of turning point} & = {3 + (-3) \over 2} \\ & = 0 \\ \\ \text{When } & x = 0, \\ y & = -(0)^2 + 9 \\ & = 9 \\ \\ \implies & \text{Turning point is } (0, 9) \end{align}

Reflect the portion of the graph below the $x$-axis to obtain the graph of $y = |9 - x^2|$.

(c)

\begin{align} y & = 4x^2 - 1 \\ \\ \text{Coefficient of } x^2 \text{ is positive} & \implies \text{Minimum curve } (\cup) \\ \\ \text{When } & x = 0, \\ y & = 4(0)^2 - 1 \\ & = -1 \\ \\ \implies & y \text{-intercept is } -1 \\ \\ \text{When } & y = 0, \\ 0 & = 4x^2 - 1 \\ 1 & = 4x^2 \\ {1 \over 4} & = x^2 \\ \pm \sqrt{1 \over 4} & = x \\ \pm 0.5 & = x \\ \\ \implies & x \text{-intercepts are } 0.5, -0.5 \\ \\ x-\text{coordinate of turning point} & = {0.5 + (-0.5) \over 2} \\ & = 0 \\ \\ \text{When } & x = 0, \\ y & = 4(0)^2 - 1 \\ & = -1 \\ \\ \implies & \text{Turning point is } (0, -1) \\ \\ \text{When } & x = -1, \\ y & = 4(-1)^2 - 1 \\ & = 3 \\ \\ \text{Left } & \text{limit is } (-1, 3) \\ \\ \text{When } & x = 1, \\ y & = 4(1)^2 - 1 \\ & = 3 \\ \\ \text{Right } & \text{limit is } (1, 3) \end{align}

Reflect the portion of the graph below the $x$-axis to obtain the graph of $y = |4x^2 - 1|$.

(d)

\begin{align} y & = |(2x + 1)(x - 2)| \\ y & = |2x^2 - 4x + x - 2| \\ y & = |2x^2 - 3x - 2| \\ \\ y & = 2x^2 - 3x - 2 \\ \\ \text{Coefficient of } x^2 \text{ is positive} & \implies \text{Minimum curve } (\cup) \\ \\ \text{When } & x = 0, \\ y & = 2(0)^2 - 3(0) - 2 \\ & = - 2 \\ \\ \implies & y \text{-intercept is } -2 \\ \\ \text{When } & y = 0, \\ 0 & = 2x^2 - 3x - 2 \\ 0 & = (2x + 1)(x - 2) \\ \\ x = - 0.5 \text{ or } x = 2 \\ \\ \implies & x \text{-intercepts are } -0.5, 2 \\ \\ x-\text{coordinate of turning point} & = {2 + (-0.5) \over 2} \\ & = 0.75 \\ \\ \text{When } & x = 0.75, \\ y & = 2(0.75)^2 - 3(0.75) - 2 \\ & = -3.125 \\ \\ \implies & \text{Turning point is } (0.75, -3.125) \end{align}

Reflect the portion of the graph below the $x$-axis to obtain the graph of $y = |(2x + 1)(x - 2)|$.

\begin{align} y & = |x - 2| - 5 \\ \\ \text{When } & y = 0, \\ 0 & = |x - 2| - 5 \\ 5 & = |x - 2| \end{align} \begin{align} 5 & = x - 2 \phantom{000}&\text{or}\phantom{00000} -5 & = x - 2 \\ 5 + 2 & = x & -5 + 2 & = x \\ 7 & = x & -3 & = x \end{align}
\begin{align} x \text{-intercepts are } 7, - 3 \\ \\ \therefore \phantom{.} P(7, 0) \text{ and } Q(-3,0) \end{align}

(i)

$$ y = 2x^{1 \over 3} $$

(The shape resembles the graph in the first row of the table on page 103.)

\begin{align} y & = {1 \over 16\sqrt{x}} \\ & = {1 \over 16} \left(1 \over \sqrt{x}\right) \\ & = {1 \over 16} x^{-{1 \over 2}} \end{align}

(The shape resembles the graph in the third row of the table on page 103.)

(ii)

\begin{align} y & = 2x^{1 \over 3} \phantom{000} \text{ --- (1)} \\ \\ y & = {1 \over 16\sqrt{x}} \\ & = {1 \over 16x^{1 \over 2}} \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2x^{1 \over 3} & = {1 \over 16x^{1 \over 2}} \\ 2x^{1 \over 3} (x^{1 \over 2}) & = {1 \over 16} \\ 2 x^{{1 \over 3} + {1 \over 2}} & = {1 \over 16} \\ 2 x^{5 \over 6} & = {1 \over 16} \\ x^{5 \over 6} & = {1 \over 32} \\ (x^{5 \over 6})^{1 \over 5} & = \left(1 \over 32\right)^{1 \over 5} \\ x^{1 \over 6} & = \sqrt[5]{1 \over 32} \\ & = {1 \over 2} \\ (x^{1 \over 6})^6 & = \left(1 \over 2\right)^6 \\ x & = {1 \over 64} \\ \\ \text{Substitute } & x = {1 \over 64} \text{ into (1),} \\ y & = 2 \left(1 \over 64\right)^{1 \over 3} \\ & = 2 \sqrt[3]{1 \over 64} \\ & = 2 \left(1 \over 4\right) \\ & = {1 \over 2} \\ \\ \therefore \text{Point } & \text{of intersection is } \left({1 \over 64}, {1 \over 2}\right) \end{align}

(i)(a)

\begin{align} \text{When } & y = 4, \\ 4 & = 2x^{1 \over 4} \\ {4 \over 2} & = x^{1 \over 4} \\ 2 & = x^{1 \over 4} \\ 2^4 & = (x^{1 \over 4})^4 \\ 16 & = x \end{align}

(i)(b)

\begin{align} \text{When } & x = 25, \\ y & = 2(25)^{1 \over 4} \\ & = 4.4721 \\ & \approx 4.5 \end{align}

(ii)

$$ y = 2x^{1 \over 4} $$

(The shape resembles the graph in the first row of the table on page 103.)

(iii)

\begin{align} y & = x - 16 \phantom{00000} [\text{Straight line } y = mx + c] \\ \\ \text{When } & x = 0, \\ y & = (0) - 16 \\ & = -16 \\ \\ \implies & y \text{-intercept is } -16 \\ \\ \text{When } & y = 0, \\ 0 & = x - 16 \\ 16 & = x \\ \\ \implies & x \text{-intercept is } 16 \end{align}

Reflect the portion of the line below the $x$-axis to obtain the graph of $y = |x - 16|$.

(iv)

\begin{align} & \text{From (iii), graphs meet at two points} \\ \\ & \therefore \text{2 solutions} \end{align}

(a)

$$ |2x + 4| = x^2 + 1 $$ \begin{align} 2x + 4 & = x^2 + 1 \phantom{000}&\text{or}\phantom{00000} 2x + 4 & = -(x^2 + 1) \\ 0 & = x^2 - 2x + 1 - 4 & 2x + 4 & = -x^2 - 1 \\ 0 & = x^2 - 2x - 3 & 0 & = -x^2 - 2x - 1 - 4 \\ 0 & = (x + 1)(x - 3) & 0 & = - x^2 - 2x - 5 \\ & & 0 & = x^2 + 2x + 5 \\ x & = -1 \text{ or } x = 3 \\ & & b^2 - 4ac & = (2)^2 - 4(1)(5) \\ & & & = -16 < 0 \phantom{000} \text{ [No real roots]} \end{align}
$$ |2x + 4| = x^2 + 1| $$ \begin{align} \text{When } & x = -1, \phantom{000}&\text{or}\phantom{00000} \text{When } & x = 3, \\ \\ \text{L.H.S} & = |2(-1) + 4| & \text{L.H.S} & = |2(3) + 4| \\ & = |2| & & = |10| \\ & = 2 & & = 10 \\ \\ \text{R.H.S} & = (-1)^2 + 1 & \text{R.H.S} & = (3)^2 + 1 \\ & = 2 & & = 10 \\ & = \text{L.H.S} & & = \text{L.H.S} \\ \\ \\ \therefore x & = -1, 3 \end{align}

(b)

$$ |x^2 - 1| = x^2 $$ \begin{align} x^2 - 1 & = x^2 \phantom{000}&\text{or}\phantom{00000} x^2 - 1 & = -x^2 \\ -1 & = x^2 - x^2 & x^2 + x^2 & = 1 \\ -1 & = 0 \text{ (Reject)} & 2x^2 & = 1 \\ & & x^2 & = {1 \over 2} \\ & & x & = \pm \sqrt{1 \over 2} \\ &&& = \pm {\sqrt{1} \over \sqrt{2}} \\ &&& = \pm {1 \over \sqrt{2}} \end{align}
$$ |x^2 - 1| = x^2 $$ \begin{align} \text{When } & x = {1 \over \sqrt{2}}, \phantom{000}&\text{or}\phantom{00000} \text{When } & x = -{1 \over \sqrt{2}}, \\ \\ \text{L.H.S} & = \left| \left(1 \over \sqrt{2}\right)^2 - 1 \right| & \text{L.H.S} & = \left| \left(-{1 \over \sqrt{2}}\right)^2 - 1 \right| \\ & = \left| -{1 \over 2} \right| & & = \left| - {1 \over 2} \right| \\ & = {1 \over 2} & & = {1 \over 2} \\ \\ \text{R.H.S} & = \left(1 \over \sqrt{2}\right)^2 & \text{R.H.S} & = \left(-{1 \over \sqrt{2}}\right)^2 \\ & = {1 \over 2} & & = {1 \over 2} \\ & = \text{L.H.S} & & = \text{L.H.S} \\ \\ \\ \therefore x & = \pm {1 \over \sqrt{2}} \end{align}

(c)

$$ |2x^{1 \over 4} - 3| = 5 $$ \begin{align} 2x^{1 \over 4} - 3 & = 5 \phantom{000}&\text{or}\phantom{00000} 2x^{1 \over 4} - 3 & = -5 \\ 2x^{1 \over 4} & = 5 + 3 & 2x^{1 \over 4} & = -5 + 3 \\ & = 8 & & = -2 \\ x^{1 \over 4} & = {8 \over 2} & x^{1 \over 4} & = {-2 \over 2} \\ & = 4 & & = -1 \\ \\ \sqrt[4]{x} & = 4 & \sqrt[4]{x} & = -1 \\ x & = 4^4 & x & = (-1)^4 \\ & = 256 & & = 1 \end{align}
$$ |2x^{1 \over 4} - 3| = 5 $$ \begin{align} \text{When } & x = 256, \phantom{000}&\text{or}\phantom{00000} \text{When } & x = 1, \\ \\ \text{L.H.S} & = |2(256)^{1 \over 4} - 3| & \text{L.H.S} & = |2(1)^{1 \over 4} - 3| \\ & = |5| & & = |-1| \\ & = 5 & & = 1 \\ & = \text{R.H.S} & & \ne \text{R.H.S} \\ \\ \\ \therefore x & = 256 \end{align}

(d)

$$ |\sqrt{x} - 2| = x - 4 $$ \begin{align} \sqrt{x} - 2 & = x - 4 \phantom{000}&\text{or}\phantom{00000} \sqrt{x} - 2 & = -(x - 4) \\ 0 & = x - \sqrt{x} -4 + 2 & \sqrt{x} - 2 & = -x + 4 \\ 0 & = x - \sqrt{x} - 2 & 0 & = -x - \sqrt{x} + 4 + 2 \\ & & 0 & = -x -\sqrt{x} + 6 \\ & & 0 & = x + \sqrt{x} - 6 \\ 0 & = (\sqrt{x})^2 - \sqrt{x} - 2 & 0 & = (\sqrt{x})^2 + \sqrt{x} - 6 \\ 0 & = (\sqrt{x} - 2)(\sqrt{x} + 1) & 0 & = (\sqrt{x} + 3)(\sqrt{x} - 2) \\ \\ \sqrt{x} & = 2 \text{ or } \sqrt{x} = -1 \text{ (Reject)} & \sqrt{x} & = -3 \text{ (Reject)} \text{ or } \sqrt{x} = 2 \\ \\ \sqrt{x} & = 2 \\ x & = 2^2 \\ & = 4 \end{align}
$$ |\sqrt{x} - 2| = x - 4 $$ \begin{align} \text{When } & x = 4, \\ \\ \text{L.H.S} & = |\sqrt{4} - 2| \\ & = |0| \\ & = 0 \\ \\ \text{R.H.S} & = (4) - 4 \\ & = 0 \\ & = \text{L.H.S} \\ \\ \\ \therefore x & = 4 \end{align}

(i)

\begin{align} y & = 2x - 5 \phantom{000000} [\text{Straight line } y = mx + c] \\ \\ \text{When } & x = 0, \\ y & = 2(0) - 5 \\ & = -5 \\ \\ \implies & y\text{-intercept is } - 5 \\ \\ \text{When } & y = 0, \\ 0 & = 2x - 5 \\ -2x & = -5 \\ x & = {-5 \over -2} \\ & = 2.5 \\ \\ \implies & x\text{-intercept is } 2.5 \end{align}

Reflect the portion of the graph below the $x$-axis to obtain the graph of $y = |2x - 5|$.

\begin{align} y & = 4 - x \\ y & = -x + 4 \phantom{00000} [\text{Straight line } y = mx + c] \\ \\ \text{When } & x = 0, \\ y & = 4 - 0 \\ & = 4 \\ \\ \implies & y\text{-intercept is } 4 \\ \\ \text{When } & y = 0, \\ 0 & = 4 - x \\ x & = 4 \\ \\ \implies & x\text{-intercept is } 4 \end{align}

(ii)

\begin{align} y & = |2x - 5| \phantom{000} \text{ --- (1)} \\ \\ y & = 4 - x \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ |2x - 5| & = 4 - x \end{align} \begin{align} 2x - 5 & = 4 -x \phantom{000}&\text{or}\phantom{00000} 2x - 5 & = -(4 - x) \\ 2x + x & = 4 + 5 & 2x - 5 & = -4 + x \\ 3x & = 9 & 2x - x & = -4 + 5 \\ x & = {9 \over 3} & x & = 1 \\ & = 3 \\ \\ \text{Substitute } & \text{into (2),} \\ y & = 4 - (3) & y & = 4 - (1) \\ & = 1 & & = 3 \\ \\ \\ \therefore \text{Coordinates } & \text{are } (3, 1) \text{ and } (1, 3) \end{align}

(a)

\begin{align} 2x + y & = 3 \\ y & = 3 - 2x \phantom{000} \text{ --- (1)} \\ \\ y & = |2x - 1| \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 3 - 2x & = |2x - 1| \end{align} \begin{align} 3 - 2x & = 2x - 1 \phantom{000}&\text{or}\phantom{00000} -(3 - 2x) & = 2x - 1 \\ 3 + 1 & = 2x + 2x & -3 + 2x & = 2x - 1 \\ 4 & = 4x & 2x - 2x & = -1 + 3 \\ {4 \over 4} & = x & 0 & = 2 \text{ (Reject)} \\ 1 & = x \\ \\ \text{Substitute } & x = 1 \text{ into (1),} \\ y & = 3 - 2(1) \\ & = 1 \\ \\ \text{Substitute } & x = 1 \text{ into (2),} \\ y & = |2(1) - 1| \\ & = |1| \\ & = 1 \\ \\ \\ \therefore x & = 1, y = 1 \end{align}

(b)

\begin{align} 2x + 3y & = 19 \\ 3y & = 19 - 2x \\ y & = {19 \over 3} - {2 \over 3}x \phantom{000} \text{ --- (1)} \\ \\ |x - y| & = 3 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ \left| x - \left({19 \over 3} - {2 \over 3}x \right) \right| & = 3 \\ \left| x - {19 \over 3} + {2 \over 3}x \right| & = 3 \\ \left| {5 \over 3}x - {19 \over 3} \right| & = 3 \end{align} \begin{align} {5 \over 3}x - {19 \over 3} & = 3 \phantom{000}&\text{or}\phantom{00000} {5 \over 3}x - {19 \over 3} & = -3 \\ {5 \over 3}x & = 3 + {19 \over 3} & {5 \over 3}x & = -3 + {19 \over 3} \\ & = {28 \over 3} & & = {10 \over 3} \\ x & = {28 \over 3} \div {5 \over 3} & x & = {10 \over 3} \div {5 \over 3} \\ & = {28 \over 5} & & = 2 \\ \\ \text{Substitute } & \text{into (1),} \\ y & = {19 \over 3} - {2 \over 3}\left(28 \over 5\right) & y & = {19 \over 3} - {2 \over 3}(2) \\ & = {13 \over 5} & & = 5 \\ \\ \\ \therefore x & = {28 \over 5}, y = {13 \over 5} \text{ or } x = 2, y = 5 \end{align}

(i)

$$ |x - 36.9| \ge 0.9 $$

(ii)

\begin{align} y & = x - 36.9 \\ \\ \text{When } & x = 0, \\ y & = 0 - 36.9 \\ & = -36.9 \\ \\ \implies & y\text{-intercept is } -36.9 \\ \\ \text{When } & y = 0, \\ 0 & = x - 36.9 \\ -x & = -36.9 \\ x & = 36.9 \\ \\ \implies & x\text{-intercept is } 36.9 \end{align}

Reflect the portion of the graph below the $x$-axis to obtain the graph of $y = |x - 36.9|$.

(i)

$$ |x^2 - 10x + 8| = 8 $$ \begin{align} x^2 - 10x + 8 & = 8 \phantom{000}&\text{or}\phantom{00000} x^2 - 10x + 8 & = - 8 \\ x^2 - 10x + 8 - 8 & = 0 & x^2 - 10x + 8 + 8 & = 0 \\ x^2 - 10x & = 0 & x^2 - 10x + 16 & = 0\\ x(x - 10) & = 0 & (x - 2)(x - 8) & = 0 \\ \\ x = 0 \text{ or } x & = 10 & x = 2 \text{ or } x & = 8 \end{align}
$$ |x^2 - 10x + 8| = 8 $$ \begin{align} \text{When } & x = 0, \phantom{000}&\text{or}\phantom{00000} \text{When } & x = 10, \\ \\ \text{L.H.S} & = |0^2 - 10(0) + 8| & \text{L.H.S} & = |10^2 - 10(10) + 8| \\ & = |8| & & = |8| \\ & = 8 & & = 8 \\ & = \text{R.H.S} & & = \text{R.H.S} \\ \\ \\ \text{When } & x = 2, \phantom{000}&\text{or}\phantom{00000} \text{When } & x = 8, \\ \\ \text{L.H.S} & = |2^2 - 10(2) + 8| & \text{L.H.S} & = |8^2 - 10(8) + 8| \\ & = |-8| & & = |-8| \\ & = 8 & & = 8 \\ & = \text{R.H.S} & & = \text{R.H.S} \\ \\ \\ \therefore x & = 0, 2, 8, 10 \end{align}

(ii)

\begin{align} y & = x^2 - 10x + 8 \\ \\ \text{Coefficient of } x^2 \text{ is positive} & \implies \text{Minimum curve } (\cup) \\ \\ \text{When } & x = 0, \\ y & = 0^2 - 10(0) + 8 \\ & = 8 \\ \\ \implies & y\text{-intercept is } 8 \\ \\ \text{When } & y = 0, \\ 0 & = x^2 - 10x + 8 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-10) \pm \sqrt{(-10)^2 - 4(1)(8)} \over 2(1)} \\ & = {10 \pm \sqrt{68} \over 2} \\ & = {10 \pm \sqrt{4} \sqrt{17} \over 2} \\ & = {10 \pm 2\sqrt{17} \over 2} \\ & = {10 \over 2} \pm {2\sqrt{17} \over 2} \\ & = 5 \pm \sqrt{17} \\ & \approx 9.12, 0.877 \\ \\ \implies & x\text{-intercepts are } 9.12, 0.877 \\ \\ x\text{-coordinate of turning point} & = {(5 + \sqrt{17}) + (5 - \sqrt{17}) \over 2} \\ & = {10 \over 2} \\ & = 5 \\ \\ \text{When } & x = 5, \\ y & = (5)^2 - 10(5) + 8 \\ & = -17 \\ \\ \implies & \text{Turning point is } (5, -17) \end{align}

Reflect the portion of the graph below the $x$-axis to obtain the graph of $y = |x^2 - 10x + 8|$.

(iii)

$$ x \text{-coordinates of points of intersection between } y = |x^2 - 10x + 8| \text{ and the line } y = 8$$

(a)

\begin{align} y & = 2\sqrt{x} \\ & = 2x^{1 \over 2} \end{align}

(The shape resembles the graph in the first row of the table on page 103.)

\begin{align} y & = {6 \over \sqrt{x}} \\ & = 6 \left(1 \over \sqrt{x}\right) \\ & = 6x^{-{1 \over 2}} \end{align}

(The shape resembles the graph in the third row of the table on page 103.)

\begin{align} y & = 2\sqrt{x} \phantom{000} \text{ --- (1)} \\ \\ y & = {6 \over \sqrt{x}} \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2\sqrt{x} & = {6 \over \sqrt{x}} \\ 2(\sqrt{x})^2 & = 6 \\ 2x & = 6 \\ x & = {6 \over 2} \\ & = 3 \\ \\ \text{Substitute } & x = 3 \text{ into (1),} \\ y & = 2\sqrt{3} \\ \\ \\ \therefore \text{Point} & \text{ of intersection is } (3, 2\sqrt{3}) \end{align}

(b)

\begin{align} y & = 2\sqrt{x} + 1 \phantom{000} \text{ --- (1)} \\ \\ y & = {6 \over \sqrt{x}} \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2\sqrt{x} + 1 & = {6 \over \sqrt{x}} \\ \sqrt{x}(2\sqrt{x} + 1) & = 6 \\ 2(\sqrt{x})^2 + \sqrt{x} & = 6 \\ 2(\sqrt{x})^2 + \sqrt{x} - 6 & = 0 \\ (2\sqrt{x} - 3) (\sqrt{x} + 2) & = 0 \\ \\ 2\sqrt{x} - 3 = 0 \phantom{0000}&\text{or}\phantom{000} \sqrt{x} + 2 = 0 \\ 2\sqrt{x} = 3 \phantom{0000} & \phantom{or000+2} \sqrt{x} = -2 \\ \sqrt{x} = {3 \over 2} \phantom{00()} & \phantom{or000+20.} x = (-2)^2 \\ x = \left(3 \over 2\right)^2 \phantom{} & \phantom{or000+20.} x = 4 \\ x = {9 \over 4} \phantom{00()} & \end{align}
\begin{align} \text{Substitute } & x = {9 \over 4} \text{ into (1),} \\ y & = 2\sqrt{9 \over 4} + 1 \\ & = 4 \\ \\ \text{Substitute } & x = {9 \over 4} \text{ into (2),} \\ y & = {6 \over \sqrt{9 \over 4}} \\ & = 4 \\ \\ \text{Point } & \text{of intersection is } \left({9 \over 4}, 4\right) \\ \\ \\ \text{Substitute } & x = 4 \text{ into (1),} \\ y & = 2\sqrt{4} + 1 \\ & = 5 \\ \\ \text{Substitute } & x = 4 \text{ into (2),} \\ y & = {6 \over \sqrt{4}} \\ & = 3 \\ \\ \therefore \text{Reject } & x = 4 \\ \\ \therefore \text{Only } & \text{point of intersection is } \left({9 \over 4}, 4\right) \end{align}