Answer
a) Vertex: $(1000,1000)$
b) Wide.
c) Upward
d) $X_{min} = -500, X_{max} = 2500, Y_{min} = 1000, Y_{max} = 2000$
Work Step by Step
Given $$\begin{aligned}
f(x) &= 0.0005(x-1000)^2+1000.
\end{aligned}$$ a) The vertex of the parabola can be easily read from the above equation to be the point $(1000,1000)$.
b) We see that the value of the multiplying constant is $a = 0.0005$. Since the absolute value of $|a| = |0.0005|= 0.0005< 1$, the parabola is wide compared to $f(x) = x^2$.
c) Since the constant $a$ is a positive number, the parabola will open upward.
d) Use your calculator to determine a suitable graphing window for the parabola. Here is a suggestion. $$\begin{aligned}
X_{min} &= -500\\
X_{max}& = 2500\\
Y_{min}& = 1000\\
Y_{max}& = 2000.
\end{aligned}$$
