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Probably the easiest way to do that is with a bit of calculus. You can take it's derivative, which gives you the slope of the tangent to that parabola. The point at which that slope is equal to zero is where the vertex lies. For instance:

f(x) = 6x2 + 7

f'(x) = 12x

f'(x) = 0 when x = 0, so the vertex is as point (0, 7)

g(x) = 2(x - 3)2 + 1

g(x) = 2x2 - 12x + 10

g'(x) = 4x - 12

g'(x) = 0 when x = 3, so the vertex is at point (3, 1)

More generically:

f(x) = a(x - b)2 + c

f(x) = ax2 - 2abx + b2 + c

f'(x) = 2ax - 2ab

f'(x) = 0 when x = b, so the vertex is at point (b, c)

which conveniently demonstrates another technique of finding the vertex of the parabola. If you convert it to the format f(x) = a(x - b)2 + c, your vertex will always be at point (b, c).

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Q: How do you solve for vertex of a parabola?
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