Sketch the following graphs: y = -3x2, y = -2x2 - 8, and y = x2 − 9 using the table method for x ","-2, -1, 0, 1, 2. State the domain and range for each.

$$\begin{array}{|c|c|} \hline x & y = -3x^2 \\ \hline -2 & -12 \\ -1 & -3 \\ 0 & 0 \\ 1 & -3 \\ 2 & -12 \\ \hline \end{array}$$

The graph is a parabola opening downwards with its vertex at $(0,0)$.

Domain: $\boxed{x \in \mathbb{R}}$
Range: $\boxed{y \le 0}$

1(b) $y = -2x^2 - 8$

$$\begin{array}{|c|c|} \hline x & y = -2x^2 - 8 \\ \hline -2 & -16 \\ -1 & -10 \\ 0 & -8 \\ 1 & -10 \\ 2 & -16 \\ \hline \end{array}$$

The graph is a parabola opening downwards with its vertex at $(0,-8)$.

Domain: $\boxed{x \in \mathbb{R}}$
Range: $\boxed{y \le -8}$

1(c) $y = x^2 - 9$

$$\begin{array}{|c|c|} \hline x & y = x^2 - 9 \\ \hline -2 & -5 \\ -1 & -8 \\ 0 & -9 \\ 1 & -8 \\ 2 & -5 \\ \hline \end{array}$$

The graph is a parabola opening upwards with its vertex at $(0,-9)$.

Domain: $\boxed{x \in \mathbb{R}}$
Range: $\boxed{y \ge -9}$

2. Given: $y = 4x^2 - 8$. Determine the following:

2(a) Shape of the graph

The graph is a parabola opening upwards.

2(b) x-intercept(s)

$$4x^2 = 8$$ $$x^2 = \frac{8}{4}$$ $$x^2 = 2$$ $$x = \pm \sqrt{2}$$

The x-intercepts