How do you find the vertex and intercepts for #y=x^2 – 12x + 36#?

Replaced equation (with 2 solutions for #x#) into a corresponding function with vertex and intercepts.

1 Answer
Alan P.
Nov 13, 2015

Vertex at #(6,0)#
y-intercept: #36#
x-intercept: 6#

Explanation:

#y=x^2-12x+36#

#rArr y = (x-6)^2+0#
this is the vertex form of a parabola with vertex at #(6,0)#

The y-intercept is the value of #y# when #x=0#
#color(white)("XXX")y=(0)^2-12(0)+36 = 36#

The x-intercept is the value of #x# when #y=0#
#color(white)("XXX")0=x^2-12x+36 = (x-6)(x-6)#
#color(white)("XXX")rArr x=6# (single intercept point)
graph{x^2-12x+36 [-25.27, 54.73, -1.6, 38.4]}

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