What is the equation of the parabola that has a vertex at (2, -9) (2,9) and passes through point (12, -4) (12,4)?

1 Answer
Jun 10, 2018

y=1/20(x-2)^2-9y=120(x2)29 in Vertex Form of the equation

Explanation:

Given:
Vertex->(x,y)=(2-9)(x,y)=(29)
Point on curve ->(x,y)=(12,-4)(x,y)=(12,4)

Using the completed square format of a quadratic

y=a(x+b/(2a))^2+ky=a(x+b2a)2+k

y=a(xcolor(red)(-2))^2color(blue)(-9)y=a(x2)29

x_("vertex")=(-1)xx(color(red)(-2)) = +2" "xvertex=(1)×(2)=+2 Given value
y_("vertex")=color(blue)(-9)" "yvertex=9 Given value

Substituting for the given point

-4=a(12-2)^2-94=a(122)29

-4=a(100)-94=a(100)9

a=5/100=1/20a=5100=120 giving:

y=1/20(x-2)^2-9y=120(x2)29 in Vertex Form of the equation

Tony B