How do you find the vertex and intercepts for #y = x^2 - 4x + 3#?

1 Answer
Alan P.
Dec 21, 2015

vertex at #(2,-1)#
y-intercept: #3#
x-intercepts: #3# and #1#

Explanation:

To find the vertex rewrite the given equation #y=x^2-4x+3#
into vertex form: #y=m(x-a)^2+b# with vertex at #(a,b)#

Complete the square
#color(white)("XXX")y=x^2-4xcolor(blue)(+2^2)+3color(blue)(-4)#
Rewrite as a squared binomial and simplified constant (in vertex form)
#color(white)("XXX")y=1(x-2)^2+(-1)#
with vertex at #(2,-1)#

The y-intercept is the value of #y# when #x=0#

For #y=x^2-4x+3# when #x=0#
#color(white)("XXX")y=9^2-4xx(0)+3 = 3#

The x-intercepts are the values when #y=0#

Since #y=x^2-4x+3#
can be factored as #y=(x-3)(x-1)#
when #y=0#
#color(white)("XXX")x=3# or #x=1#

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