Answer
$$
f(x)=-2x^{2}+16x-21
$$
The $x$-intercepts are
$$
x_{1}=\frac{8-\sqrt{22}}{2} \approx 1.655 ,\:x_{2}=\frac{8+\sqrt{22}}{2} \approx 6.345
$$
The $y$-intercept is-21.
Vertex parabola is $\left(4 , 11\right)$.
The axis is $x=4$ , the vertical line through the vertex.
Plotting the vertex, the $y$-intercept,the $x$-intercepts. and the point $\left(4 , 11\right)$. gives the graph in Figure
Work Step by Step
$$
f(x)=-2x^{2}+16x-21
$$
The $x$-intercepts can be found by letting $f(x)=0$ to get
$$
f(x)=-2x^{2}+16x-21=0
$$
This does not appear to factor, so we’ll try the quadratic formula.
$$
\begin{split}
\:x_{1,\:2}& =\frac{-16\pm \sqrt{16^2-4\left(-2\right)\left(-21\right)}}{2\left(-2\right)} \\
&=\frac{-16\pm 2 \sqrt{22}}{-4}
\end{split}
$$
from which
$$
x_{1}=\frac{8-\sqrt{22}}{2} \approx 1.655 ,\:x_{2}=\frac{8+\sqrt{22}}{2} \approx 6.345
$$
are the $x$-intercepts.
To find the $y$-intercept , set $x=0 $
$$
f(x)=-2(0)^{2}+16(0)-21=-21.
$$
So the $y$-intercept is -21.
The $x$-coordinate of the vertex is :
$$
x=\frac{-b}{2a}=-\frac{-16}{2(.-2)}=\frac{-16}{-4}=4
$$
Substituting this into the equation gives
$$
f(x)=-2(4)^{2}+16(4)-21 =-32+64-21=11
$$
The vertex is $\left(4 , 11\right)$.
The axis is $x=4 $ , the vertical line through the vertex.
Plotting the vertex, the $y$-intercept,the $x$-intercepts. and the point $\left(4 , 11\right)$ gives the graph in Figure
