Answer
$(a)$ The average speed is $80ft/s$
The park is $800ft$ away.
They were in the park for $20$ minutes.
$(b)$ See the image below.
The original function is shrunk by the factor of $2$.
The new speed is: $40ft/s$
The new park is $400ft$ away.
$(c)$ See the image below.
The original function is shifted to the right-hand side by $10$ units.
In the new function, the trip begins $10$ minutes later.
Work Step by Step
$(a)$ In general the average speed is $\frac{f(x_2)-f(x_1)}{x_2-x_1}$
In our case we can input the values and write:
$\frac{800-0}{10-0}=80ft/s$
The park is $800ft$ away.
They were in the park for $20$ minutes. That is part of the graph where the distance between the school and park doesn't change.
$(b)$ See the image above.
The original function is shrunk by the factor of $2$.
The new speed is:
$\frac{400}{10}=40ft/s$
According to the new graph, the new park is $400ft$ away.
$(c)$ See the image below.
The original function is shifted to the right-hand side by $10$ units.
In the new function, the trip begins $10$ minutes later.
