Given: f(x) = -x^2 + 5x +2
Vertex form: f(x) = a(x - h)^2 + k, where "vertex" = (h, k) and a is a constant.
There are two ways to find vertex form.
- Use the vertex formula "vertex" = (-B/(2A), f(-B/(2A))) where the equation is in the form Ax^2 + Bx + C = 0.
" "But a second point is required to find the constant a.
" "-B/(2A) = -5/(-2) = 5/2
" "f(5/2) = -(5/2)^2 + 5/1*(5/2) + 2 = -25/4 + 25/2 + 2
" "f(5/2) = -25/4 +50/4 + 8/4 = 33/4
" "f(x) = a(x-5/2)^2 + 33/4
A second point can be selected by letting x = 0 in the original equation => f(0) = 2
" "2 = a (0 - 5/2)^2 +33/4
" "8/4 - 33/4 = 25/4 a
" "-25/4 = 25/4 a
" " a = -1
f(x) = -(x-5/2)^2 + 33/4
The second method is by using completing of the square .
Group the x terms together and factor out a negative:
f(x) = -x^2 + 5x +2 = -(x^2 - 5x) + 2
Take half of the x-term and add the square of this halved value because it was subtracted when the square was completed and seen below:
-(x - 5/2)^2 = -(x - 5/2)(x - 5/2) = -(x^2 -5x +25/4) = -x^2 + 5x - 25/4
As you can see 25/4 was subtracted. This must be added to keep the equation the same:
f(x) = -(x - 5/2)^2 + 2 - (-1)(5/2)^2
f(x) = -(x - 5/2)^2 + 8/4 + 25/4
f(x) = -(x - 5/2)^2 +33/4
