Question

graph the function.
$r(x)=\

$$\begin{cases}x^2 - 4 & \\text{for } x \\leq 2 \\\\ 3x - 6 & \\text{for } x > 2\\end{cases}$$

$

Explanation:

Step1: Analyze first piece ($x \leq 2$)

The function is $r(x) = x^2 - 4$, a parabola.

• Vertex: At $x=0$, $r(0)=0^2-4=-4$, so $(0, -4)$.
• Intercepts:
• $x$-intercepts: $x^2-4=0 \implies x=\pm2$, so $(-2,0)$ and $(2,0)$.
• $y$-intercept: $(0,-4)$.
• Plot points for $x \leq 2$: e.g., $(-1, -3)$, $(1, -3)$, $(2, 0)$ (closed dot, since $x\leq2$ includes 2).

Step2: Analyze second piece ($x > 2$)

The function is $r(x) = 3x - 6$, a line.

• Slope $m=3$, $y$-intercept $(0,-6)$.
• At $x=2$, $r(2)=3(2)-6=0$ (open dot, since $x>2$ excludes 2).
• Plot points for $x > 2$: e.g., $(3, 3)$, $(4, 6)$.

Step3: Combine the two graphs

Draw the parabola $x^2-4$ for all $x \leq 2$ (with closed dot at $(2,0)$), and the line $3x-6$ for all $x > 2$ (with open dot at $(2,0)$; note the open dot overlaps the closed dot here, so the graph appears continuous at $x=2$).

Answer:

The graph consists of:

1. The left half (and vertex) of the parabola $y=x^2-4$ for $x \leq 2$, including the point $(2,0)$.
2. The ray of the line $y=3x-6$ for $x > 2$, starting from the open dot at $(2,0)$ (visually coinciding with the closed dot) and extending upward to the right.

Key plotted points:

• Parabola: $(-2,0)$, $(0,-4)$, $(1,-3)$, $(2,0)$ (closed)
• Line: $(3,3)$, $(4,6)$, $(2,0)$ (open)