Question

graph the absolute value function\\( h(x) = 3|x + 2| - 5 \\)\\( f(x) = \

$$\begin{cases} \\square, & x \\leq \\square \\\\ \\square, & x > \\square \\end{cases}$$

\\)

Explanation:

Step1: Identify vertex of the function

The vertex form of an absolute value function is $h(x)=a|x-h|+k$, where the vertex is $(h,k)$. For $h(x)=3|x+2|-5$, rewrite $x+2$ as $x-(-2)$, so vertex is $(-2, -5)$. The value of $x$ where the absolute value changes behavior is $x=-2$.

Step2: Split function for $x\leq-2$

When $x\leq-2$, $x+2\leq0$, so $|x+2|=-(x+2)$. Substitute into the function:
$h(x)=3(-(x+2))-5 = -3x-6-5=-3x-11$

Step3: Split function for $x>-2$

When $x>-2$, $x+2>0$, so $|x+2|=x+2$. Substitute into the function:
$h(x)=3(x+2)-5 = 3x+6-5=3x+1$

Step4: Verify vertex on both pieces

For $x=-2$, $-3(-2)-11=6-11=-5$ and $3(-2)+1=-6+1=-5$, which matches the vertex $y$-value.

Answer:

$$ f(x) =
LATEXBLOCK0
$$

To graph the function:

1. Plot the vertex point $(-2, -5)$.
2. For $x\leq-2$, use the line $y=-3x-11$: pick $x=-3$, $y=-3(-3)-11=9-11=-2$, so plot $(-3,-2)$ and draw a line from this point through $(-2,-5)$ extending left.
3. For $x>-2$, use the line $y=3x+1$: pick $x=0$, $y=3(0)+1=1$, so plot $(0,1)$ and draw a line from this point through $(-2,-5)$ extending right.