Question

which rule describes the function whose graph is shown?
$f(x)=\

$$\begin{cases}x + 6, & x < -4 \\\\ x^2, & -4 \\leq x < 4 \\\\ 6 - x, & x \\geq 4\\end{cases}$$

$
$h(x)=\

$$\begin{cases}x + 4, & x < -2 \\\\ x^2, & -2 \\leq x < 2 \\\\ 4 - x, & x \\geq 2\\end{cases}$$

$
$g(x)=\

$$\begin{cases}x + 6, & x < -2 \\\\ x^2, & -2 \\leq x < 2 \\\\ 6 - x, & x \\geq 2\\end{cases}$$

$

Explanation:

Step1: Analyze left linear segment

The left line passes through $(-4,2)$ and has a slope of 1. Using point-slope form $y - y_1 = m(x - x_1)$:
$y - 2 = 1(x + 4) \implies y = x + 6$, valid for $x < -2$? No, check the vertex of the parabola: the parabola starts at $x=-2$ (vertex at $(0,0)$), so left line is for $x < -2$, and at $x=-2$, $y=(-2)^2=4$, which matches $x+6$ at $x=-2$: $-2+6=4$.

Step2: Analyze middle quadratic segment

The parabola is $y=x^2$, with domain $-2 \leq x < 2$ (ends at $x=2$, $y=4$).

Step3: Analyze right linear segment

The right line passes through $(2,4)$ and has a slope of -1. Using point-slope form:
$y - 4 = -1(x - 2) \implies y = 6 - x$, valid for $x \geq 2$.

Step4: Match to given functions

Compare to the piecewise function $g(x)$:
$g(x)=

$$\begin{cases} x + 6, & x < -2 \\ x^2, & -2 \leq x < 2 \\ 6 - x, & x \geq 2 \end{cases}$$

$
This matches all segments of the graph.

Answer:

$g(x)=

$$\begin{cases} x + 6, & x < -2 \\ x^2, & -2 \leq x < 2 \\ 6 - x, & x \geq 2 \end{cases}$$

$