Answer
$$
f(x)=\frac{1}{3}x^{2}-\frac{8}{3}x+\frac{1}{3}
$$
The $x$-intercepts are
$$
x_{1}=4+\sqrt{15} \approx 7.873 ,\:x_{2}=4-\sqrt{15}\approx 0.127.
$$
The $y$-intercept is \frac{1}{3}.
Vertex parabola is $\left(4 ,-5\right)$ .
The axis is $x=4 $ , the vertical line through the vertex.
Plotting the vertex, the $y$-intercept,the $x$-intercepts. and the point $\left(4 ,-5\right)$ gives the graph in Figure
Work Step by Step
$$
f(x)=\frac{1}{3}x^{2}-\frac{8}{3}x+\frac{1}{3}
$$
The $x$-intercepts can be found by letting $f(x)=0$ to get
$$
f(x)=\frac{1}{3}x^{2}-\frac{8}{3}x+\frac{1}{3}=x^{2}-8x+1=0
$$
This does not appear to factor, so we’ll try the quadratic formula.
$$
\begin{split}
x_{1,\:2}&=\frac{-\left(-8\right)\pm \sqrt{\left(-8\right)^2-4\cdot \:1\cdot \:1}}{2\cdot \:1} \\
&=\frac{8\pm \sqrt{60}}{2} \\
&=8 \pm \sqrt {15}
\end{split}
$$
from which
$$
x_{1}=4+\sqrt{15} \approx 7.873 ,\:x_{2}=4-\sqrt{15}\approx 0.127
$$
are the $x$-intercepts.
To find the $y$-intercept , set $x=0 $
$$
f(0)=\frac{1}{3}(0)^{2}-\frac{8}{3}(0)+\frac{1}{3}=\frac{1}{3}
$$
So the $y$-intercept is \frac{1}{3}.
The $x$-coordinate of the vertex is :
$$
x=\frac{-b}{2a}=-\frac{-16}{2(.-2)}=\frac{-\frac{-8}{3}}{2(\frac{1}{3})}=4
$$
Substituting this into the equation gives
$$
f(4)=\frac{1}{3}(4)^{2}-\frac{8}{3}(4)+\frac{1}{3} =\frac{16}{3}-\frac{32}{3}+ \frac{1}{3}=-5
$$
The vertex is $\left(4 ,-5\right)$.
The axis is $x=4 $ , the vertical line through the vertex.
Plotting the vertex, the $y$-intercept,the $x$-intercepts. and the point $\left(4 ,-5\right)$. gives the graph in Figure