Answer
Vertex: $(1.25,26.25)$
Vertical intercept: $(0,20)$
Horizontal intercepts: $(-1.31,0),(3.81,0)$
Work Step by Step
Given $$
f(x)=-4 x^2+10 x+20.
$$ The parabola is in its standard form, $f(x) = ax^2+bx+c$. The values of the constants are: $$
\begin{aligned}
a&=-4 \\
b&=10 \\
c&=20.
\end{aligned}
$$ Find the $x$ and $y$ values of the vertex.
$$
\begin{aligned}
x&=\frac{-b}{2a} \\
x&=\frac{-(10)}{2(-4)} \\
x&=1.25\\
y&= -4\cdot (1.25)^2+10\cdot (1.25)+20\\
&= 26.25.
\end{aligned}
$$ The vertex is $(1.25,26.25)$.
The horizontal intercept is found by setting the $y$ component of the parabola to zero and solving for $x$. $$
\begin{aligned}
3 x^2-18 x+15&=0 \\
\frac{3x^2-18 x+15}{3}&= 0\\
x^2-6 x+5&= 0 \\
x^2-5 x-x+5&= 0 \\
x(x-5)-1(x-5)&= 0\\
(x-5)(x-1)&= 0\\
\end{aligned}
$$ We have: $$
\begin{aligned}
& x=\frac{-(10) \pm \sqrt{(10)^2-4(-4)(20)}}{2(-4)} \\
& x=-\frac{-10 \pm \sqrt{420}}{8}.
\end{aligned}
$$ This gives: $$
\begin{aligned}
x&=-\frac{-10 + \sqrt{420}}{8} \\
x&\approx-1.3117\\
x&=-\frac{-10 - \sqrt{420}}{8} \\
x&\approx 3.8117.
\end{aligned} $$ The vertical intercepts are $(-1.31,0)$ and $(3.81,0)$.
The graph has a vertical intercept for $x= 0$ and $y = c = 20$. The vertical intercept is $(0,20)$.
