How do you write the quadratic in vertex form given vertex is (3,-6). and y intercept is 2?

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Answer 1 · 2015-05-08T17:11:28

The general vertex form of a quadratic is
#y=m(x-a)^2+b#
where the vertex is at #(a,b)#

We are told that the vertex is at #(3,-6)#
so this becomes
#y=m(x-3)^2-6#
#= mx^2-6mx+9m -6#

We are also told that when #x=0# then #y=2# (that's what it means to say the y-intercept is #2#).

So #m(0)^2-6m(0) +9m -6 =2#

#9m = 8#

#m=8/9#

And the equation of the quadratic in vertex form is
#y=8/9(x-3)^2-6#