Answer
Vertex: $(1,-8)$
$x$-intercepts: $-1$; $3$
$y$-intercept: $-6$
Work Step by Step
Bring the function to the vertex form $f(x)=a(x-h)^2+k$:
$$\begin{align*}
f(x)&=2x^2-4x-6\\
&=2(x^2-2x+1)-8\\
&=2(x-1)^2-8.
\end{align*}$$
Identify the constants $a$, $h$, $k$:
$$\begin{align*}
a&=2\\
h&=1\\
k&=-8.
\end{align*}$$
Determine the vertex of the function:
$$(h,k)=(1,-8).$$
The axis of symmetry is $x=1$. It opens upward because $a>0$.
Determine the $x$-intercepts by solving the equation $f(x)=0$:
$$\begin{align*}
2(x-1)^2-8&=0\\
(x-1)^2&=4\\
x-1&=\pm 2\\
x-1=-2&\text{ or }x-1=2\\
x=-1&\text{ or }x=3.
\end{align*}$$
The $x$-intercepts are $-1$ and $3$.
Determine the $y$-intercept:
$$f(0)=2(0-1)^2-8=-6.$$
The $y$-intercept is $-6$.
We plot the vertex $(1,-8)$ and the intercept points $(-1,0)$,
$(3,0)$ and $(0,-6)$, then join them to sketch the graph of the function:
