Answer
a) Vertex: $(-20,50)$
b) Wide
c) Upward
d) $X_{min} = -25, X_{max} = -15, Y_{min} = 0, Y_{max} = 500$
Work Step by Step
Given $$\begin{aligned}
f(x) &= 0.002(x+20)^2+50.
\end{aligned}$$ a) The vertex of the parabola can be easily read from the above equation to be the point, $(-20,50)$.
b) We see that the value of the multiplying constant is $a = 20$. Since the absolute value of, $|a| = |0.002|= 0.002< 1$, the parabola is wide compared to $f(x) = x^2$.
c) Since the constant $a$ is a positive number, the parabola will open downward.
d) Use your calculator to determine a suitable graphing window for the parabola. Here is a suggestion. $$\begin{aligned}
X_{min} &= -25\\
X_{max}& = -15\\
Y_{min}& = 0\\
Y_{max}& = 500.
\end{aligned}$$
