Answer
See graph, axis of symmetry $x=1$,
domain $(-\infty,\infty)$, range $[-8, \infty)$
Work Step by Step
Step 1. Given $f(x)=2x^2-4x-6=2(x^2-2x+1)-8=2(x-1)^2-8$, we can identify its vertex at $(1,-8)$
Step 2. The x-intercept can be found by letting $y=0$, which gives $(x-1)^2=4$ and $x=-1,3$
Step 3. The y-intercept can be found by letting $x=0$, which gives $y=-6$
Step 4. Sketch the function as shown in the figure.
Step 5. The axis of symmetry can be identified as $x=1$
Step 6. The domain can be found to be $(-\infty,\infty)$ and the range to be $[-8, \infty)$
