the height of a ball (in meters) is modeled by $h(t)=5t^{2}+20t + 2$ find the average rate of change of…

Explanation:

Step1: Calculate ( h(1) )

Substitute ( t = 1 ) into ( h(t)=5t^{2}+20t + 2 ) ( h(1)=5\times(1)^{2}+20\times1 + 2=5 + 20+2=27 )

Step2: Calculate ( h(3) )

Substitute ( t = 3 ) into ( h(t)=5t^{2}+20t + 2 ) ( h(3)=5\times(3)^{2}+20\times3 + 2=5\times9+60 + 2=45+60 + 2=107 )

Step3: Use the average rate of change formula

The average rate of change formula is ( \frac{h(b)-h(a)}{b - a} ), where ( a = 1 ), ( b = 3 ) ( \frac{h(3)-h(1)}{3 - 1}=\frac{107 - 27}{2}=\frac{80}{2}=40 ) Wait, there is a mistake. The function should be ( h(t)=- 5t^{2}+20t + 2 ) (because the parabola opens down - ward, coefficient of ( t^{2}) is negative).

Step1: Recalculate ( h(1) ) with ( h(t)=-5t^{2}+20t + 2 )

Substitute ( t = 1 ) ( h(1)=-5\times(1)^{2}+20\times1 + 2=-5 + 20+2=17 )

Step2: Recalculate ( h(3) ) with ( h(t)=-5t^{2}+20t + 2 )

Substitute ( t = 3 ) ( h(3)=-5\times(3)^{2}+20\times3 + 2=-5\times9+60 + 2=-45+60 + 2=17 )

Step3: Use the average rate of change formula again

( \frac{h(3)-h(1)}{3 - 1}=\frac{17-17}{2}=0 )

Answer:

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