Answer
a) Vertex: $(1000,0)$
b) Wide
c) Upward
d) $X_{min} = -1000, X_{max} = 2000, Y_{min} = 0, Y_{max} = 2000$
Work Step by Step
Rewrite the given equation of the parabola so that it looks exactly like the general standard vertex form, $ f(x) = a(x-h)^2+k$.
$$\begin{aligned}
f(x) &= 0.0005(x-1000)^2\\
&= 0.0005(x-1000)^2+0.
\end{aligned}$$ a) The vertex of the parabola can be easily read from the above equation to be the point $(1000,0)$.
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 value of $a$ 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. $$\begin{aligned}
X_{min} &= -1000\\
X_{max}& = 2000\\
Y_{min}& = 0\\
Y_{max}& = 2000\\
\end{aligned}$$
