Question

f(x) = \

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

part: 0 / 6
part 1 of 6
(a) graph the function.

Explanation:

Step1: Analyze $x \leq 2$ piece

This is a downward-opening parabola $f(x) = -x^2 + 4$.

• Vertex at $(0, 4)$ (since $f(0) = -0^2 +4=4$)
• Roots: $-x^2 +4=0 \implies x^2=4 \implies x=\pm2$
• At $x=2$, $f(2) = -(2)^2 +4=0$ (closed dot here, since $x\leq2$)

Step2: Plot parabola points

Key points for $x\leq2$:
$(-2, 0), (0, 4), (2, 0)$
Draw the left half and middle of the downward parabola, ending with a closed dot at $(2,0)$.

Step3: Analyze $x > 2$ piece

This is a linear function $f(x)=4x$.

• At $x=2$, $f(2)=4*2=8$ (open dot here, since $x>2$)
• Additional point: $x=3$, $f(3)=12$

Step4: Plot linear segment

Draw a ray starting at the open dot $(2,8)$, passing through $(3,12)$, extending rightwards.

Answer:

The graph consists of:

1. A downward-opening parabola $f(x)=-x^2+4$ for $x\leq2$, with a closed endpoint at $(2, 0)$, passing through $(-2,0)$ and $(0,4)$.
2. A ray of the line $f(x)=4x$ for $x>2$, with an open endpoint at $(2,8)$, passing through $(3,12)$ and extending to the right.