Answer
$$
f(x)=\frac{1}{2}x^{2}+6x+24
$$
There are no $x$-intercepts.
The $y$-intercept is 24.
Vertex parabola is $\left(-6 , 6\right)$.
The axis is $ x=-6 $ , the vertical line through the vertex.
Plotting the vertex, the $y$-intercept,the $x$-intercepts. and the point $\left(-6 , 6\right)$. gives the graph in Figure
Work Step by Step
$$
f(x)=\frac{1}{2}x^{2}+6x+24
$$
The $x$-intercepts can be found by letting $f(x)=0$ to get
$$
f(x)=\frac{1}{2}x^{2}+6x+24=x^2+12x+48=0
$$
This does not appear to factor, so we’ll try the quadratic formula.
$$
\begin{split}
\:x_{1,\:2} &=\frac{-12\pm \sqrt{12^2-4\cdot \:1\cdot \:48}}{2\cdot \:1} \\
&=\frac{-12\pm \sqrt{-48}}{2}
\end{split}
$$
Since the radical is negative, then there are no $x$-intercepts.
To find the $y$-intercept , set $x=0 $
$$
f(x)=\frac{1}{2}(0)^{2}+6(0)+24=24.
$$
So the $y$-intercept is 24.
The $x$-coordinate of the vertex is :
$$
x=\frac{-b}{2a}=-\frac{-6}{2.\frac{1}{2}}=\frac{-6}{1}=-6
$$
Substituting this into the equation gives
$$
f(x)=\frac{1}{2}(-6)^{2}+6.(-6)+24=18-36+24=6
$$
The vertex is $\left(-6 , 6\right)$.
The axis is $x=-6 $ , the vertical line through the vertex.
Plotting the vertex, the $y$-intercept,the $x$-intercepts. and the point $\left(-6 , 6\right)$. gives the graph in Figure