Question

for items 1–2, identify the following information for the function of each graph: 1. domain: $(-infty, infty)$ range: $(-20, infty)$ x-intercepts: $(-4, 0)$, $(4, 0)$ y-intercepts: $(0, -20)$ interval positive: $(-infty, -4) cup (4, infty)$ interval negative: $(-4, 4)$ interval increasing: $(0, infty)$ interval decreasing: $(-infty, 0)$ average rate of change over $-5, 0$:

Explanation:

Step1: Determine the function's equation

The graph is a parabola opening upwards with vertex at \((0, -20)\) and \(x\)-intercepts at \((-4, 0)\) and \((4, 0)\). The general form of a parabola is \(y = ax^2 + bx + c\). Since the vertex is \((0, -20)\), \(b = 0\) and \(c=-20\). Using the \(x\)-intercept \((4, 0)\):
\[
0 = a(4)^2 - 20 \implies 16a = 20 \implies a=\frac{5}{4}
\]
So the function is \(y = \frac{5}{4}x^2 - 20\).

Step2: Recall the average rate of change formula

The average rate of change of a function \(f(x)\) over the interval \([a, b]\) is \(\frac{f(b) - f(a)}{b - a}\). Here, \(a = -5\) and \(b = 0\).

Step3: Calculate \(f(-5)\) and \(f(0)\)

• For \(x = -5\):

\[
f(-5)=\frac{5}{4}(-5)^2 - 20=\frac{5}{4}(25)-20=\frac{125}{4}-20=\frac{125 - 80}{4}=\frac{45}{4} = 11.25
\]

• For \(x = 0\):

\[
f(0)=\frac{5}{4}(0)^2 - 20=-20
\]

Step4: Compute the average rate of change

\[
\frac{f(0)-f(-5)}{0 - (-5)}=\frac{-20 - 11.25}{5}=\frac{-31.25}{5}=-6.25=-\frac{25}{4}
\]

Answer:

\(-\frac{25}{4}\) (or \(-6.25\))