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

Question

Savannah Anderson · Accepted Answer

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#

Eva Abramson · Answer

undefined The vertex form of a quadratic equation is $ y = a(x - h)^2 + k $, where $ (h, k) $ is the vertex. Given the vertex is $ (3, -6) $, and the y-intercept is 2, you can plug in these values and solve for $ a $.

First, substitute $ h = 3 $ and $ k = -6 $:

$ y = a(x - 3)^2 - 6 $

Then, plug in the y-intercept, which occurs when $ x = 0 $:

$ 2 = a(0 - 3)^2 - 6 $

Solve for $ a $:

$ 2 = 9a - 6 $

$ 9a = 8 $

$ a = \frac{8}{9} $

So, the equation in vertex form is $ y = \frac{8}{9}(x - 3)^2 - 6 $.