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#
