Question

1. $h(x)=\

$$\begin{cases}(x + 4)^2, & x\\leq - 2 \\\\ -2x, & -2 < x \\leq 0 \\\\ 2x, & 0 < x \\leq 2 \\\\ (x - 4)^2, & x > 2\\end{cases}$$

$
a. find $h(-1)$.
b. find $h(7)$.
c. find $h(3)$.
d. find $h(x)=0$.
e. sketch a graph of this function.

Explanation:

Step1: Identify interval for $h(-1)$

$-2 < -1 \leq 0$, use $h(x)=-2x$

Step2: Calculate $h(-1)$

$h(-1) = -2(-1) = 2$

Step3: Identify interval for $h(7)$

$7 > 2$, use $h(x)=(x-4)^2$

Step4: Calculate $h(7)$

$h(7) = (7-4)^2 = 3^2 = 9$

Step5: Identify interval for $h(3)$

$3 > 2$, use $h(x)=(x-4)^2$

Step6: Calculate $h(3)$

$h(3) = (3-4)^2 = (-1)^2 = 1$

Step7: Solve $h(x)=0$ for $x\leq-2$

Use $h(x)=(x+4)^2$:
$$(x+4)^2=0 \implies x=-4$$

Step8: Solve $h(x)=0$ for $-2

Use $h(x)=-2x$:
$$-2x=0 \implies x=0$$

Step9: Solve $h(x)=0$ for $0

Use $h(x)=2x$:
$$2x=0 \implies x=0 \text{ (not in interval, discard)}$$

Step10: Solve $h(x)=0$ for $x>2$

Use $h(x)=(x-4)^2$:
$$(x-4)^2=0 \implies x=4$$

Step11: Graph sketch guidelines

1. For $x\leq-2$: Plot parabola $y=(x+4)^2$, vertex at $(-4,0)$, right endpoint at $(-2,4)$ (closed dot).
2. For $-2
3. For $0
4. For $x>2$: Plot parabola $y=(x-4)^2$, vertex at $(4,0)$, left endpoint $(2,4)$ (open dot).

Answer:

a. $2$
b. $9$
c. $1$
d. $x=-4$, $x=0$, $x=4$
e. (Graph constructed per the guidelines in Step11: piecewise graph with a left-opening parabola, two line segments, and a right-opening parabola, with correct open/closed endpoints at $x=-2,0,2$)