Question

on which interval does the function $h(x) = -5x^2 + 8$ have an average rate of change equal to -30?
$x = -4$ to $x = 4$
$x = -2$ to $x = 4$
$x = 0$ to $x = 2$
$x = 2$ to $x = 4$

Explanation:

Step1: Recall the average rate of change formula

The average rate of change of a function \( h(x) \) over the interval \([a, b]\) is given by \(\frac{h(b) - h(a)}{b - a}\).

Step2: Test the interval \( x=-2 \) to \( x = 4 \)

First, find \( h(-2) \) and \( h(4) \).
For \( h(x)=-5x^{2}+8 \),
\( h(-2)=-5(-2)^{2}+8=-5(4)+8=-20 + 8=-12 \)
\( h(4)=-5(4)^{2}+8=-5(16)+8=-80 + 8=-72 \)
Now, calculate the average rate of change: \(\frac{h(4)-h(-2)}{4-(-2)}=\frac{-72-(-12)}{6}=\frac{-72 + 12}{6}=\frac{-60}{6}=-10\) (not - 30, so this is not the interval)

Wait, let's test \( x = 2 \) to \( x=4 \)
\( h(2)=-5(2)^{2}+8=-5(4)+8=-20 + 8=-12 \)
\( h(4)=-72 \) (as above)
Average rate of change: \(\frac{h(4)-h(2)}{4 - 2}=\frac{-72-(-12)}{2}=\frac{-72 + 12}{2}=\frac{-60}{2}=-30\)

Wait, let's check other intervals too.

For \( x=-4 \) to \( x = 4 \)
\( h(-4)=-5(-4)^{2}+8=-5(16)+8=-80 + 8=-72 \)
\( h(4)=-72 \)
Average rate of change: \(\frac{h(4)-h(-4)}{4-(-4)}=\frac{-72-(-72)}{8}=\frac{0}{8}=0\)

For \( x = 0 \) to \( x=2 \)
\( h(0)=-5(0)^{2}+8=8 \)
\( h(2)=-12 \)
Average rate of change: \(\frac{h(2)-h(0)}{2-0}=\frac{-12 - 8}{2}=\frac{-20}{2}=-10\)

Wait, when we test \( x = 2 \) to \( x = 4 \):
\( h(2)=-5*(2)^2 + 8=-20 + 8=-12 \)
\( h(4)=-5*(4)^2+8=-80 + 8=-72 \)
Average rate of change: \(\frac{h(4)-h(2)}{4 - 2}=\frac{-72-(-12)}{2}=\frac{-60}{2}=-30\)

Wait, also let's check \( x=-2 \) to \( x = 4 \) again:
\( h(-2)=-5*(-2)^2+8=-20 + 8=-12 \)
\( h(4)=-72 \)
Average rate of change: \(\frac{-72-(-12)}{4-(-2)}=\frac{-60}{6}=-10\) (wrong)

Wait, maybe I made a mistake. Wait the interval \( x = 2 \) to \( x = 4 \):

\( h(2)=-5*(2)^2 + 8=-20 + 8=-12 \)

\( h(4)=-5*(4)^2+8=-80 + 8=-72 \)

Difference in \( y \): \(-72-(-12)=-60\)

Difference in \( x \): \(4 - 2 = 2\)

So average rate of change: \(\frac{-60}{2}=-30\). Yes, that's correct.

Wait, also check the interval \( x=-2 \) to \( x = 4 \):

\( h(-2)=-12 \), \( h(4)=-72 \), difference in \( x \) is \(6\), difference in \( y \) is \(-60\), so \(\frac{-60}{6}=-10\). Not -30.

For \( x = 2 \) to \( x = 4 \), we get -30. Let's check another interval. Wait, maybe the interval \( x=-4 \) to \( x = 4 \): \( h(-4)=-72 \), \( h(4)=-72 \), so rate is 0.

\( x = 0 \) to \( x=2 \): \( h(0)=8 \), \( h(2)=-12 \), difference in \( y \) is \(-20\), difference in \( x \) is 2, so rate is -10.

So the correct interval is \( x = 2 \) to \( x = 4 \)

Wait, but let's check the options again. The options are:

1. \( x=-4 \) to \( x = 4 \)

1. \( x=-2 \) to \( x = 4 \)

1. \( x = 0 \) to \( x=2 \)

1. \( x = 2 \) to \( x = 4 \)

So the answer is \( x = 2 \) to \( x = 4 \)

Answer:

\( x = 2 \) to \( x = 4 \)