+0

# Suppose a parabola has an axis of symmetry at x = 6...

-2
451
3
+2448

I don't understand how to put it in vertex form.

Feb 20, 2018

#1
+18080
+3

Vertex form is     a (x-h)^2 +k = 0    where h,k is the vertex   so let's start with that h,k = 6,-1

a (x-6)^2 +  -1 = y     (this info was given in the question)

and the point  7,-3 given as a solution to this equation ..now we can substitute this into the equation to find 'a'

a(7-6)^2 -1 = -3

a-1 = -3

a=-2

so the equation becomes

-2(x-6)^2 -1 = y

Feb 20, 2018
edited by ElectricPavlov  Feb 20, 2018

#1
+18080
+3

Vertex form is     a (x-h)^2 +k = 0    where h,k is the vertex   so let's start with that h,k = 6,-1

a (x-6)^2 +  -1 = y     (this info was given in the question)

and the point  7,-3 given as a solution to this equation ..now we can substitute this into the equation to find 'a'

a(7-6)^2 -1 = -3

a-1 = -3

a=-2

so the equation becomes

-2(x-6)^2 -1 = y

ElectricPavlov Feb 20, 2018
edited by ElectricPavlov  Feb 20, 2018
#2
+865
+3

Which is D. Great job!

ant101  Feb 20, 2018
#3
+2448
+1

ty!!

RainbowPanda  Feb 20, 2018