For the quadratic equation y = 8 − 2x2, find the x-intercepts, axis of symmetry, turning point, and range. For the linear equation y = -2x, find the gradient.

$$
2x^2 = 8
$$
$$
x^2 = 4
$$
$$
x = \pm\sqrt{4}
$$
$$
x = \pm 2
$$
The x-intercepts are $\boxed{(-2, 0) \text{ and } (2, 0)}$.

• Axis of symmetry:

In $y = 8 - 2x^2$, $a = -2$ and $b = 0$. $x = -\frac{0}{2(-2)}$ $x = 0$ The axis of symmetry is $\boxed{x = 0}$.

• Turning point:

$$
y = 8 - 2(0)^2
$$
$$
y = 8 - 0
$$
$$
y = 8
$$
The turning point is $\boxed{(0, 8)}$.

• Range of y:

In $y = -2x$, $m = -2$.  
The gradient is $\boxed{-2}$.

• 1.2.2 What is the mathematical name of the graph? The graph of $y = -2x$ is a $\boxed{\text{linear function (or straight line)}}$.

$x = -2, -1, 0, 1, 2$.

• For $x = -2$, $y = 3^{-2} = \frac{1}{3^2} = \frac{1}{9}$.
• For $x = -1$, $y = 3^{-1} = \frac{1}{3}$.
• For $x = 0$, $y = 3^0 = 1$.
• For $x = 1$, $y = 3^1 = 3$.
• For $x = 2$, $y = 3^2 = 9$.

The tabulated values are: $\begin{array}{|c|c|} \hline x & y \\ \hline -2 & \frac{1}{9} \\ \hline -1 & \frac{1}{3} \\ \hline 0 & 1 \\ \hline 1 & 3 \\ \hline 2 & 9 \\ \hline \end{array}$

• 1.3.2 Use the values in question 1.3.1 to sketch the graph of $y = 3^x$ on the ADDENDUM provided. To sketch the graph, plot the points $(-2, \frac{1}{9})$, $(-1, \frac{1}{3})$, $(0, 1)$, $(1, 3)$, and $(2, 9)$ on a coordinate plane and connect them with a smooth curve.

• 1.3.3 Is the graph drawn in question 1.3.2 continuous or discontinuous? The graph of $y = 3^x$ is $\boxed{\text{continuous}}$.

• 1.3.4 Write down the new equation of y if the graph moves two units down.

Moving it two units down gives $y = 3^x - 2$.
The new equation is $\boxed{y = 3^x - 2}$.