Question 1113237
The vertex of a parabola is (-2, -20), and its y-intercept is (0, -12).
<pre>
We draw the approximate graph, using what is given:

{{{drawing(320,800,-6,4,-22,3,
graph(320,800,-6,4,-22,3,2x^2+8x-12),
circle(-2,-20,.1), circle(0,-12,.1), locate(-4,-20,"(-2,-20)"),
locate(0,-12,"(0,-12)")  )}}}
<pre>

The standard form of a quadratic parabola is

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

Since it is given that the vertex (h,k) = (-2, -20),

h = -2, and k = -20.

And since it passes through the point (0, -12),

x = 0 and y = -12

Substituting in

{{{y=a(x-h^"")^2+k}}},

{{{-12=a(0-(-2)^"")^2+(-20)}}}

{{{-12=a(0+2^"")^2-20}}}

{{{-12=a(2^"")^2-20}}}

{{{-12=a(4)-20}}}

{{{-12=4a-20}}}

{{{8=4a}}}

{{{2=a}}}

Now go back to the standard form:

{{{y=a(x-h^"")^2+k}}}

Substitute a = 2, h = -2, k = -20

However this time, we do not substitute anything
for x and y, for we want to leave them as variables:

{{{y=a(x-h^"")^2+k}}}

{{{y=2(x-(-2)^"")^2+(-20)}}}

Then we simplify:

{{{y=2(x+2^"")^2-20}}}

{{{y=2(x+2^"")(x+2^"")-20}}}

{{{y=2(x^2+4x+4)-20}}}

{{{y=2x^2+8x+8-20}}}

{{{y=2x^2+8x-12}}}

Edwin</pre>