Steve S. answered • 02/12/14

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Tutoring in Precalculus, Trig, and Differential Calculus

How do you use the vertex formula to find the coordinates of the vertex of the graph of a parabola?

A simple parabola: y = a x^2 with vertex at (0,0);

translated by <h,k>:

y - k = a(x - h)^2 with vertex at (h,k), or

y = a(x - h)^2 + k with vertex at (h,k).

If you have a vertex form equation just pick out the h and k for the vertex coordinates.

E.g.,

y = -3(x+7)^2 - 11, or

y = -3(x-(-7))^2 + (-11),

has a vertex at (-7,-11).

BTW, where does h = -b/(2a) come from?

Multiply out the general vertex form:

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

y = a(x^2 - 2hx + h^2) + k

y = a x^2 + (-2ah)x + (ah^2 + k)

Now compare to standard form:

y = a x^2 + b x + c

and it's easily seen that:

b = -2ah => h = -b/(2a), and

c = ah^2 + k => k = c - ah^2

You can also solve the general vertex form for its zeros to get an alternative quadratic formula:

y = a(x - h)^2 + k = 0

(x - h)^2 = -k/a

|x - h| = √(-k/a) <= imaginary if k and a have same signs

x = h ± √(-k/a)

If -k/a > 0 then 2 real roots,

if -k/a = 0 then 1 real root, or

if -k/a < 0 then 2 complex conjugate roots.