Explanation:
For a nonlinear graph whose slope changes at each point, the average rate of change between any two points [tex](x_{1},f(x_{1}) \ and \ (x_{2},f(x_{2})[/tex] is the slope of the line through the two points.
[tex]ARC=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}} =\frac{Change \ in \ y}{Change \ in \ x}=m_{sec}[/tex]
So here we have the function:
[tex]h(t)=(t+3)^2+5[/tex]
and want to compute the ARC over the interval:
[tex]-5\leq t\leq -1[/tex]
So:
[tex]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}[/tex]
