SOLUTION: Parabola questions are the hardest can someone help me figure this out show steps please. The parabola given by the equation y=-x^2+8 x-8 has its vertex at (h,k) for: h= an

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Parabola questions are the hardest can someone help me figure this out show steps please. The parabola given by the equation y=-x^2+8 x-8 has its vertex at (h,k) for: h= an      Log On


   

Question 622707: Parabola questions are the hardest can someone help me figure this out show steps please.

The parabola given by the equation y=-x^2+8 x-8 has its vertex at (h,k) for:
h=
and
k=
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

y = -x^2+ 8x- 8  |putting into Vertex form a%28x+-+h%29%5E2+%2B+k by completing the Square

y = -(x-4)^2 + 16 - 8

y = -(x-4)^2 + 8   V(4,8)   (h is 4 and k is 8)



Note: In general: y = ax^2 + bx + c

If you need a shortcut on this: y = -x^2+ 8x- 8, (a = -1, b = 8 and c = -8)

In general: y = ax^2 + bx + c .... going to y = a(x-h)^2 + k ... following is true

 h+=+-b%2F2a+=+-8%2F-2+=+4

 k+=+c+-+b%5E2%2F4a+=++-8+-+64%2F-4+=+8