Answer
$$
f(x)=\frac{3}{2}x^{2}-x-4
$$
The $x$-intercepts are
$$
x_{1}=2,\:x_{2}=-\frac{4}{3}
$$
The $y$-intercept is -4.
Vertex parabola is $\left(\frac{1}{3} , \frac{-25}{6} \right)$.
The axis is $x=\frac{1}{3} $ , the vertical line through the vertex.
Plotting the vertex, the $y$-intercept,the $x$-intercepts. and the point $\left(\frac{1}{3} , \frac{-25}{6} \right)$. gives the graph in Figure
Work Step by Step
$$
f(x)=\frac{3}{2}x^{2}-x-4
$$
The $x$-intercepts can be found by letting $f(x)=0$ to get
$$
f(x)=\frac{3}{2}x^{2}-x-4=3x^2-2x-8=0
$$
This does not appear to factor, so we’ll try the quadratic formula.
$$
\begin{split}
\:x_{1,\:2}&=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:3\left(-8\right)}}{2\cdot \:3} \\
&=\frac{2\pm \sqrt{100}}{6} \\
&=\frac{1\pm 5}{3}
\end{split}
$$
from which
$$
x_{1}=2,\:x_{2}=-\frac{4}{3}
$$
are the $x$-intercepts.
To find the $y$-intercept , set $x=0 $
$$
f(x)=\frac{3}{2}(0)^{2}-(0)-4=-4.
$$
So the $y$-intercept is -4.
The $x$-coordinate of the vertex is :
$$
x=\frac{-b}{2a}=-\frac{-16}{2(.-2)}=\frac{-(-1)}{2(\frac{3}{2})}=\frac{1}{3}
$$
Substituting this into the equation gives
$$
f(\frac{1}{3})=\frac{3}{2}(\frac{1}{3})^{2}-(\frac{1}{3})-4=\frac{1}{6}-\frac{2}{6}-\frac{24}{6}=\frac{-25}{6}
$$
The vertex is $\left(\frac{1}{3} , \frac{-25}{6} \right)$.
The axis is $x=\frac{1}{3} $ , the vertical line through the vertex.
Plotting the vertex, the $y$-intercept,the $x$-intercepts. and the point $\left(\frac{1}{3} , \frac{-25}{6} \right)$. gives the graph in Figure