FInd the x-coordinates where horizontal tangents for F(x)=2x^2+4x-12 occur. The rightmost one occurs?
That is an upward-concave parabola with only one horizontal tangent, which occurs at the minimum point. You can rewrite the function, by completing the square, as
y = F(x) = 2(x^2 + 2x -6)
= 2[x^2 +2x +1 -7]
= 2[(x+1)^2 -7]
Clearly, the minimum (and horizontal tangent) is at x = -1, y = -14.