How do you write a quadratic equation with vertex; ( 2,3 ); point: ( 4,11 )?

1 Answer
May 1, 2017

y=2(x-2)^2+3

Explanation:

The general vertex form for a quadratic is
color(white)("XXX")color(brown)y=color(green)m(color(orange)x-color(red)a)^2+color(blue)b
with vertex at (color(red)a,color(blue)b)
and a "spread factor" of color(green)m

Given the vertex: (color(red)2,color(blue)3)
this becomes
color(white)("XXX")color(brown)y=color(green)m(color(orange)x-color(red)2)^2+color(blue)3

We are given that one solution point is (color(orange)x,color(brown)y)=(color(orange)4,color(brown)11)

So we have
color(white)("XXX")color(brown)11=color(green)m(color(orange)4-color(red)2)^2+color(blue)3

Simplifying:
color(white)("XXX")8=color(green)m(2)^2

color(white)("XXX")8 =4color(green)m

color(white)("XXX")color(green)m=2

We can substitute this back into our vertex equation, to get
color(white)("XXX")color(brown)y=color(green)2(color(orange)x-color(red)2)^2+color(blue)3

If your instructor prefers this in "standard form" we can expand the right side to get:
color(white)("XXX")y=2x^2-8x+11

For verification purposes, here is the graph of y=2(x-2)^2+3
enter image source here