Answer
a) Vertex: $(-30,250)$
b) Narrow
c) Upward
d) $X_{min} = -36, X_{max} = -24, Y_{min} = 0, Y_{max} = 1000$
Work Step by Step
Given $$\begin{aligned}
f(x) &= 20(x+30)^2+250.
\end{aligned}$$ a) The vertex of the parabola can be easily read from the above equation to be the point $(h,k)=(-30,250)$.
b) We see that the value of the multiplying constant is $a = 20$. Since the absolute value of $|a| = |20|= 20> 1$, the parabola is narrow 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. $$\begin{aligned}
X_{min} &= -36\\
X_{max}& = -24\\
Y_{min}& = 0\\
Y_{max}& = 1000.
\end{aligned}$$
