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

1 Answer
Jan 15, 2016

Vertex: #(2,0)#
y-intercept: #4#
x-intercept: #2#

Explanation:

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

#y=x^2-4x+4#

#color(white)("X")=(x-color(red)(2))^2+color(blue)(0)#
is therefore a parabola with vertex at #(color(red)(2),color(blue)(0))#

#bar(color(white)("XXXXXXXXXXXXXXXXXXXXXXXXXXXX"))#

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

Given #y=x^2-4x+4#
when #x=0#
#color(white)("XXX")y=4#

#bar(color(white)("XXXXXXXXXXXXXXXXXXXXXXXXXXXX"))#

The x intercept is the value(s) of #x# when #y=0#

Given #y=x^2-4x-4#
when #y=0#
#color(white)("XXX")0=x^2-4x+4#

#color(white)("XXX")x^2-4x+4=0#

#color(white)("XXX")(x-2)^2=0#

#color(white)("XXX")(x-2)=0#

#color(white)("XXX")x=2#