H(t)=(t+3)^2+5 What is the average rate of change of h over the interval -5<_t<_-1

Mathematics

1 answer:

Zina [86]4 years ago

7 0

<h2>Explanation:</h2>

For a nonlinear graph whose slope changes at each point, the average rate of change between any two points $(x_{1},f(x_{1}) \ and \ (x_{2},f(x_{2})$ is the slope of the line through the two points.

$ARC=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}} =\frac{Change \ in \ y}{Change \ in \ x}=m_{sec}$

So here we have the function:

$h(t)=(t+3)^2+5$

and want to compute the ARC over the interval:

$-5\leq t\leq -1$

So:

$t_{1}=-5 \\ \\ h({t_{1}})=(-5+3)^2+5=(-2)^2+5=9 \\ \\ \\ t_{2}=-1 \\ \\ h({t_{2}})=(-1+3)^2+5=(2)^2+5=9 \\ \\ \\ So: \\ \\ \\ ARC=\frac{9-9}{-1-(-5)} \\ \\ \boxed{ARC=0}$

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