SOLUTION: name the vertex for the parabola with equation y=3x²-12x+16

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Question 38295: name the vertex for the parabola with equation y=3x²-12x+16
Found 2 solutions by Nate, Paul:
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
y=3x%5E2-12x%2B16 original
y-16=3x%5E2-12x add constant
%28y%2F3%29-%2816%2F3%29=x%5E2-4x divide by three
%28y%2F3%29-%2816%2F3%29%2B%2812%2F3%29=%28x-2%29%5E2 complete the square
%28y%2F3%29-%284%2F3%29=%28x-2%29%5E2 combine
%28y%2F3%29=%28x-2%29%5E2%2B%284%2F3%29 add
y=3%28x-2%29%5E2%2B4 multiply by three
Use stantard form y=a%28x-h%29%5E2%2Bk where the vertex is (h,k).
Vertex is (2,4).

Answer by Paul(988) About Me  (Show Source):
You can put this solution on YOUR website!
To find the vertex, comlpete the square.
Using intercepts; y=0
0=3x%5E2-12x%2B16
3%28x%5E2-4x%29%2B16=0
3%28x%5E2-4x%2B4-4%29%2B16=0
3%28x%5E2-4x%2B4%29-12%2B16=0
3%28x-2%29%5E2%2B4=0
Vertex:
(2,4)
Paul.