Answer
Vertex: $(1.83,-9.92)$
Vertical intercept: $(0,-20)$
Horizontal intercepts: None
Work Step by Step
Given $$
h(x)=-3 x^2+11 x-20.
$$ The parabola is in its standard form, $f(x) = ax^2+bx+c$. The values of the constants are:
$$
\begin{aligned}
a&=-3 \\
b&=11 \\
c&=-20.
\end{aligned}
$$ Find the $x$ and $y$ values of the vertex.
$$
\begin{aligned}
x & =\frac{-b}{2 a} \\
& =\frac{-(11)}{2(-3)} \\
& =\frac{11}{6} \\
& \approx 1.83.
\end{aligned}
$$ $$\begin{aligned} y & =-3\left(\frac{11}{6}\right)^2+11\left(\frac{11}{6}\right)-20 \\ & \approx-9.92\end{aligned}$$ The vertex is: $$
\text { Vertex: }(1.83,-9.92).
$$
Find the vertical intercept.
$$
\begin{aligned}
&y=c=-20\\
&\text {Vertical intercept: }(0,-20)
\end{aligned}
$$ Find the horizontal intercept by setting $y= 0$ to find the $x$ values.
$$
\begin{aligned}
\frac{-3 x^2+11 x-20}{-3} & =\frac{0}{-3} \\
x^2-\frac{11}{3} x+\frac{20}{3} & =0 \\
x^2-\frac{11}{3} x+\left(\frac{11}{6}\right)^2 & =-\frac{20}{3}+\left(\frac{11}{6}\right)^2\\
\left(x-\frac{11}{6}\right)^2 & =\frac{121}{36}-\frac{20}{3} \cdot \frac{12}{12} \\
\left(x-\frac{11}{6}\right)^2 & =\frac{121-240}{36} \\
\left(x-\frac{11}{6}\right)^2 & =-\frac{119}{36}
\end{aligned}
$$ There is no horizontal intercept since we cannot take the square root of a negative number.
