How do you write the Vertex form equation of the parabola #y = –x^2 +12x – 4 #?

1 Answer
May 8, 2017

#y=-(x-6)^2+32#

Explanation:

The general vertex form of a parabola is:
#color(white)("XXX")y=color(green)m(x-color(red)a)^2+color(blue)bcolor(white)("XXX")#with vertex at #(color(red)a,color(blue)b)#

Given:
#color(white)("XXX")y=-x^2+12x-4#

Extract the #color(green)m# factor from the first 2 terms:
#color(white)("XXX")y=(color(green)(-1))(x^2-12x)-4#

Complete the square
#color(white)("XXX")y=(color(green)(-1))(x^2-12xcolor(magenta)(+(12/2)^2))-4color(magenta)(-(color(green)(-1))*((12/2)^2)#

Simplify
#color(white)("XXX")y=(color(green)(-1))(x^2-12xcolor(magenta)(+6^2))-4color(magenta)(+36)#

Write as a squared binomial and a simplified constant
#color(white)("XXX")y=(-1)(x-color(red)6)^2+color(blue)32#
which is a specific vertex form with vertex at #(color(red)6,color(blue)(32))#

Here is a graph of the original equation for verification purposes:
enter image source here