\begin{align} \text{When } & x = 0, \\ y & = 3(0)^5 \\ & = 0 \\ \\ x \text{ and } y & \text{-intercepts are } (0, 0) \end{align}

\begin{align} \text{When } & x = 0, \\ y & = -3(0)^4 \\ & = 0 \\ \\ x \text{ and } y & \text{-intercepts are } (0, 0) \end{align}

\begin{align} y & = {2 \over x^3} \\ & = 2 \left(1 \over x^3\right) \\ & = 2x^{-3} \end{align}

\begin{align} y & = -{2 \over x^4} \\ & = -2 \left(1 \over x^4\right) \\ & = -2x^{-4} \end{align}

\begin{align} \text{When } & x = 0, \\ y & = (0)^{1 \over 4} \\ & = 0 \\ \\ x \text{ and } y & \text{-intercepts are } (0, 0) \end{align}

\begin{align} \text{When } & x = 0, \\ y & = (0)^{2 \over 3} \\ & = 0 \\ \\ x \text{ and } y & \text{-intercepts are } (0, 0) \end{align}

\begin{align} \text{When } & x = 0, \\ y & = (0)^{3 \over 2} \\ & = 0 \\ \\ x \text{ and } y & \text{-intercepts are } (0, 0) \end{align}

\begin{align} \text{When } & x = 0, \\ y & = 3(0)^{3 \over 2} \\ & = 0 \\ \\ x \text{ and } y & \text{-intercepts are } (0, 0) \end{align}

\begin{align} \text{When } & x = 0, \\ y & = -3(0)^{3 \over 2} \\ & = 0 \\ \\ x \text{ and } y & \text{-intercepts are } (0, 0) \end{align}

\begin{align} \text{Area of circle} & = \pi \times \text{Radius} \times \text{Radius} \\ x & = \pi \times y \times y \\ x & = \pi y^2 \\ {x \over \pi} & = y^2 \\ \pm \sqrt{x \over \pi} & = y \\ \pm {\sqrt{x} \over \sqrt{\pi}} & = y \\ \pm {1 \over \sqrt{\pi}} \sqrt{x} & = y \\ \pm {1 \over \sqrt{\pi}} x^{1 \over 2} & = y \\ \\ \text{Since radius of } & \text{cannot be negative,} \\ \\ \therefore y & = {1 \over \sqrt{\pi}} x^{1 \over 2} \end{align}

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

\begin{align} y & = {2\pi \over \sqrt{10}} x^{1 \over 2} \end{align}

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

$$ \text{The graphs are reflections of each other about the } x \text{-axis}$$

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

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

\begin{align} y & = 2x - 3 \phantom{000000} [\text{Straight line } y = mx + c] \\ \\ \text{When } & x = 0, \\ y & = 2(0) - 3 \\ & = -3 \\ \\ \implies y & \text{-intercept is } - 3 \\ \\ \text{When } & y = 0, \\ 0 & = 2x - 3 \\ 3 & = 2x \\ {3 \over 2} & = x \\ 1.5 & = x \\ \\ \implies x & \text{-intercept is } 1.5 \end{align}

$$ \therefore 1 \text{ solution} $$

\begin{align} y & = 4 - x^2 \\ & = -x^2 + 4 \\ \\ \text{Since coefficient of } x^2 & \text{ is negative,} \text{ this is a maximum curve } (\cap) \\ \\ \text{When } & y = 0, \\ 0 & = -x^2 + 4 \\ x^2 & = 4 \\ x & = \pm \sqrt{4} \\ & = \pm 2 \\ \\ \implies x& \text{-intercepts} \text{ are } 2, -2 \\ \\ \text{When } x & = 0, \\ y & = -(0)^2 + 4 \\ & = 4 \\ \\ \implies y & \text{-intercept} \text{ is 4} \\ \\ x-\text{coordinate of turning point} & = {2 + (-2) \over 4} \\ & = 0 \\ \\ \text{When } & x = 0, \\ y & = -(0)^2 + 4 \\ & = 4 \\ \\ \implies & \text{Turning point}\text{ is (0, 4)} \end{align}

(Since $x \ge 0$, only the right side of $y = 4 - x^2$ is required.)

\begin{align} 4x^{1 \over 3} + x^2 & = 4 \\ \underbrace{4x^{1 \over 3}}_\text{Power function} & = \underbrace{4 - x^2}_\text{Quadratic function} \\ \\ \text{From the graph, } & \text{there is 1 point of intersection.} \\ \\ \therefore & \text{1 solution} \end{align}

\begin{align} y & = \sqrt{6x} \\ & = \sqrt{6} \sqrt{x} \\ & = \sqrt{6} x^{1 \over 2} \end{align}

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

\begin{align} y & = -\sqrt[3]{8x} \\ & = - \sqrt[3]{8} \sqrt[3]{x} \\ & = - 2 x^{1 \over 3} \end{align}

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

\begin{align} y & = {4 \over \sqrt[5]{32x^2}} \\ & = {4 \over \sqrt[5]{32} \sqrt[5]{x^2}} \\ & = {4 \over 2 (x^2)^{1 \over 5}} \\ & = {4 \over 2 x^{2 \over 5}} \\ & = {2 \over x^{2 \over 5}} \\ & = 2 \left(1 \over x^{2 \over 5}\right) \\ & = 2x^{-{2 \over 5}} \end{align}

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

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

(The shape resembles the graph in the bottom left section of the table on page 101.)

\begin{align} y & = 2x + 1 \phantom{000000} [\text{Straight line } y = mx + c] \\ \\ \text{When } & x = 0, \\ y & = 2(0) + 1 \\ & = 1 \\ \\ \implies & y\text{-intercept is } 1 \\ \\ \text{When } & y = 0, \\ 0 & = 2x + 1 \\ -2x & = 1 \\ x & = {1 \over -2} \\ & = -0.5 \\ \\ \implies & x\text{-intercept is } -0.5 \end{align}

From part (i), the points of intersection betwen the two graphs are $(-1, -1)$ and $(0.5, 2)$.

\begin{align} 5x^{1 \over 2} & = 2x^{1 \over 6} \\ (5x^{1 \over 2})^6 & = (2x^{1 \over 6})^6 \\ (5)^6 (x^{1 \over 2})^6 & = (2)^6 (x^{1 \over 6})^6 \\ 15625 x^3 & = 64x \\ 15626 x^3 - 64x & = 0 \\ x(15626 x^2 - 64) & = 0 \\ \\ x = 0 \phantom{000}&\text{or}\phantom{000} 15626x^2 - 64 = 0 \\ & \phantom{or000-64} 15626x^2 = 64 \\ & \phantom{or000-6415626} x^2 = {64 \over 15626} \\ & \phantom{or000-64156260} x = \pm \sqrt{64 \over 15626} \\ & \phantom{or000-64156260x} = {8 \over 125} \text{ or } -{8 \over 125} \text{ (Reject)} \end{align}

Note $x = -{8 \over 125}$ is rejected as $\left(-{8 \over 125}\right)^{1 \over 2} $ has no real solutions.

From part (i), the $x$-coordinates of the points of intersection are $0$ and ${8 \over 125}$

\begin{align} \text{When } & x = 0, \\ y & = 5(0)^{1 \over 2} \\ & = 0 \\ \\ \text{When } & x = {8 \over 125}, \\ y & = 5 \left(8 \over 125\right)^{1 \over 2} \\ & = 5 \sqrt{8 \over 125} \\ & = 5 \left( {\sqrt{8} \over \sqrt{125}} \right) \\ & = 5 \left( {\sqrt{4} \sqrt{2} \over \sqrt{25} \sqrt{5} } \right) \\ & = 5 \left( 2 \sqrt{2} \over 5 \sqrt{5} \right) \\ & = {2 \sqrt{2} \over \sqrt{5}} \end{align}

\begin{align} I & = \sqrt{2P \over L} \\ \\ \text{When } & L = 8, \\ I & = \sqrt{2P \over 8} \\ & = \sqrt{P \over 4} \\ & = {\sqrt{P} \over \sqrt{4}} \\ & = {P^{1 \over 2} \over 2} \\ & = {1 \over 2} P^{1 \over 2} \end{align}

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

\begin{align} I & = \sqrt{2P \over L} \\ \\ \text{When } & P = 18, \\ I & = \sqrt{2(18) \over L} \\ & = \sqrt{36 \over L} \\ & = {\sqrt{36} \over \sqrt{L}} \\ & = {6 \over L^{1 \over 2}} \\ & = 6 \left(1 \over L^{1 \over 2}\right) \\ & = 6 L^{-{1 \over 2}} \end{align}

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

\begin{align} I & = \sqrt{2P \over L} \\ \\ \text{When } & P = 18, \\ I & = \sqrt{2(18) \over L} \\ I & = \sqrt{36 \over L} \\ I & = {\sqrt{36} \over \sqrt{L}} \\ I & = {6 \over \sqrt{L}} \\ I \sqrt{L} & = 6 \\ \sqrt{L} & = {6 \over I} \\ (\sqrt{L})^2 & = \left(6 \over I\right)^2 \\ L & = {36 \over I^2} \\ L & = 36 \left(1 \over I^2\right) \\ L & = 36 I^{-2} \end{align}

(The shape resembles the graph in the bottom right section of the table on page 101.)

\begin{align} y & = {3 \over x} \\ & = 3 \left(1 \over x\right) \\ & = 3 x^{-1} \end{align}

(The shape resembles the graph in the bottom left section of the table on page 101.)

$$ y = 4x^{1 \over 5} $$

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

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

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

\begin{align} y & = 2x \phantom{00000} [\text{Straight line}] \\ \\ \text{When } & x = 0, \\ y & = 0 \\ \\ \implies & x \text{ and } y\text{-intercept are } (0, 0) \end{align}

\begin{align} y & = 3\sqrt{x} \\ & = 3x^{1 \over 2} \end{align}

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

\begin{align} y & = 2x - 2 \phantom{000000} [\text{Straight line } y = mx + c] \\ \\ \text{When } & x = 0, \\ y & = 2(0) - 2 \\ & = - 2 \\ \\ \implies & y\text{-intercept is } -2 \\ \\ \text{When } & y = 0, \\ 0 & = 2x - 2 \\ -2x & = -2 \\ x & = {-2 \over -2} \\ & = 1 \\ \\ \implies & x\text{-intercept is } 1 \end{align}

\begin{align} y & = 2x - 2 \phantom{000} \text{ --- (1)} \\ \\ y & = 3\sqrt{x} \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2x - 2 & = 3\sqrt{x} \\ 2x - 3\sqrt{x} - 2 & = 0 \\ \\ \text{From part } & \text{(i),} \\ x & = 4 \\ \\ \text{Substitute } & x = 4 \text{ into (1),} \\ y & = 2(4) - 2 \\ & = 6 \\ \\ \implies & \text{Point of intersection is } (4, 6) \end{align}

\begin{align} y & = {1 \over 2} \sqrt[3]{x} \\ y & = {1 \over 2}x^{1 \over 3} \\ \\ \\ y & = {16 \over 2x} \\ y & = {8 \over x} \\ y & = 8x^{-1} \end{align}

$$ y = 8x^{-1} \text{ is symmetrical about } y = x $$