Question 1198846
<pre>
The vertex form of a parabola is:

{{{"f(x)"= a(x-h)^2+k}}} where (h,k) is the vertex.

{{{"f(x)" = 2x^2+12x-4}}}

Factor 2 out of the first two terms only:

{{{"f(x)" = 2(x^2+6x)-4}}}

We must add and subtract the proper number inside the parentheses to
cause {{{x^2+6x}}} to become a perfect square.  To do that we take
1/2 of the coefficient of x, which is 6.  1/2 of 6 is 3.  Then we
square 3 and get 9.  So 9 is the number we must add, then subtract,
i.e., +9-9 inside the parentheses:

{{{"f(x)" = 2(x^2+6x+9-9)-4}}}

Next we factor only the first three terms inside the parentheses:

{{{"f(x)" = 2((x+3)(x+3)^""-9)-4}}}

And, as you see it factors as the square of a binomial:

{{{"f(x)" = 2((x+3)^2-9)-4}}}

Next we distribute the 2 into the parentheses leaving the (x+3)<sup>2</sup> intact.

{{{"f(x)" = 2(x+3)^2-18-4}}}

{{{"f(x)" = 2(x+3)^2-22}}}

Comparing to 

{{{"f(x)"= a(x-h)^2+k}}} where (h,k) is the vertex.

The vertex is (h,k) = (-3,-22)

Edwin</pre>