Answer
Vertex: $(8,-20)$
Vertical intercept: $(0,236)$
Horizontal intercepts: $(5.76,0),(10.24,0)$
Work Step by Step
Given $$
h(x)=4(x-8)^2-20.
$$ The parabola is in its standard vertex form, $f(x) = a(x-h)^2+k$. The vertex can be directly read from the equation. The vertex is $(8,-20)$. The horizontal intercept is found by setting the $y$ component of the parabola to zero and solving for $x$.
$$
\begin{aligned}
4(x-8)^2-20&=0 \\
4(x-8)^2&=20 \\
\frac{4(x-8)^2}{4}&=\frac{20}{4} \\
(x-8)^2& = 5\\
x-8&= \pm\sqrt{5}\\
x &=8 \pm\sqrt{5}\\
x&= 8 -\sqrt{5}\\
&= 5.76\\
x&= 8 +\sqrt{5}\\
&= 10.24
\end{aligned}
$$ The horizontal intercepts are $(5.76,0)$ and $(10.24,0)$.
The graph has a vertical intercept for $x= 0$ Find the value of $y$ at $x= 0$. $$
\begin{aligned}
f(0)&=4(0-8)^2-20 \\
y&= 236.
\end{aligned}
$$ The vertical intercept is $(0,236)$.
