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.
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.