How do you write the vertex form equation of the parabola #y = x^2 - 4x + 3#?

1 Answer
Oct 17, 2017

#y=(x-color(red)2)^2+color(blue)((-1))#
with vertex at #(color(red)2,color(blue)(-1))#

Explanation:

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

+----------------------------------------------------------------------------------+
| The process used (below) is often referred to as#color(white)("XXXxx")# |
| #color(white)("XXX")#completing the square#color(white)("XXXXXXXXXXXXXX..x")#|
+----------------------------------------------------------------------------------+

In order to convert the given equation: #y=x^2-4x+3#
into vertex form, we need to a term of the form #(x-color(red)a)^2#

Since #(x-color(red)a)^2=(x^2+2color(red)ax+color(red)a^2)#
and
since the coefficient of #x# in the given equation is #(-4)#
then
#color(white)("XXX")2color(red)a=-4color(white)("xxx")rarrcolor(white)("xxx")color(red)a=-2#
and
#color(white)("XXXXXXXXXXXXXXX")color(red)a^2=color(green)4#

That is the first term of the vertex form must be
#color(white)("XXX")(x-color(red)2)^2#

We need to add #color(green)4# to #x^2-4x# from the original equation to make #(x-color(red)2)^2#
...but if we are going to add #color(green)4# then we will also need to subtract #color(green)4# so the equation will not really change.

#y=x^2-4x+3#
#color(white)("XXX")#will therefore become
#y=underbrace(x^2-4xcolor(green)(+4))+underbrace(3color(green)(-4))#
#color(white)("XXX")#re-writing as a squared binomial and simplifying the constants
#y=underbrace((x-color(red)2)^2)+underbrace(color(blue)((-1)))#
#color(white)("XXX")#which is the vertex form