Question

graph $h(x) = -2|x + 5| - 3$.

Explanation:

Step1: Identify vertex of absolute value function

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

Step2: Determine slope of the two branches

The coefficient $a=-2$ means:

• For $x > -5$, the slope is $-2$, so the line is $y=-2(x+5)-3=-2x-13$.
• For $x < -5$, the slope is $2$, so the line is $y=2(x+5)-3=2x+7$.

Step3: Plot additional points

For $x > -5$: Let $x=-4$, $h(-4)=-2|-4+5|-3=-2(1)-3=-5$, so point $(-4,-5)$.
For $x < -5$: Let $x=-6$, $h(-6)=-2|-6+5|-3=-2(1)-3=-5$, so point $(-6,-5)$.

Step4: Sketch the graph

Draw a V-shape with vertex at $(-5,-3)$, passing through $(-4,-5)$ and $(-6,-5)$, opening downward (due to negative $a$).

Answer:

The graph is a downward-opening V with vertex at $(-5, -3)$, passing through points such as $(-4, -5)$ and $(-6, -5)$.