SOLUTION: What is the vertex and axis of symmerty for y=-2x^2+16x-16

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Question 962767: What is the vertex and axis of symmerty for y=-2x^2+16x-16
Found 2 solutions by lwsshak3, MathLover1:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
What is the vertex and axis of symmetry for
y=-2x^2+16x-16
vertex form of equation for a parabola: y=A(x-h)^2+k, (h,k)=coordinates of the vertex
complete the square:
y=-2(x^2-8x+16)+32-16
y=-2(x-4)^2+16
vertex: (4, 16)
axis of symmetry: x=4

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
+y=-2x%5E2%2B16x-16 to find the vertex, first write your equation in vertex form y=%28x-h%29%5E2%2Bk where h and k are x and y coordinates of the vertex
+y=%28-2x%5E2%2B16x%29-16....complete the square
+y=-2%28x%5E2-8x%29-16
+y=-2%28x%5E2-8x%2B_%29-%28-2%29%2A_-16
+y=-2%28x%5E2-8x%2B4%5E2%29-%28-2%29%2A4%5E2-16
+y=-2%28x-4%29%5E2%2B2%2A16-16
+y=-2%28x-4%29%5E2%2B32-16
+y=-2%28x-4%29%5E2%2B16
so, h=4 and k=16 and vertex is at (4,16)
the axis of symmetry is the line that runs down its 'center' and divides the graph into two perfect halves; in your case the axis of symmetry is a line x=4