What is the vertex form of #y= (x - 3) (x - 2) #?

2 Answers
Jan 31, 2016

#y = (x - 5/2)^2 - 1/4#.

Explanation:

Firstly, we expand the right hand side,

#y = x^2 - 5x + 6#

Now we complete the square and do a bit of algebraic simplification,

#y = x^2 - 5x + (5/2)^2 - (5/2)^2 + 6#

#y = (x - 5/2)^2 - 25/4 + 6#

#y = (x - 5/2)^2 - 25/4 + 24/4#

#y = (x - 5/2)^2 - 1/4#.

Jan 31, 2016

vertex form: #y=1(x-5/2)^2+(-1/4)#

Explanation:

The general vertex form is:
#color(white)("XXX")y=m(x-color(blue)(a))^2+color(cyan)(b)#
with a vertex at #(color(blue)(a),color(cyan)(b))#
(So that's our target).

Given
#color(white)("XXX")y=(x-3)(x-2)#
Expanding the right-side by multiplying:
#color(white)("XXX")y=x^2-5x+6#
Complete the square
#color(white)("XXX")y=color(green)(x^2-5x)color(red)(+(5/2)^2)+6color(red)(-25/4)#
Re-write as a squared binomial and simplified constant
#color(white)("XXX")y=(x-color(blue)(5/2))^2+color(cyan)("("-1/4")")#
which is in the general form (assuming a default value #m=1#)

The graph below for #y=(x-2)(x-3)# helps verify that this solution is reasonable.
graph{(x-2)(x-3) [-0.45, 10.647, -2.48, 3.07]}