SOLUTION: i need the vertex, the axis of symmetry of this problem....y=-x^2+6x-5...i know i have to put in form y=a(x-h)^2+k but i am lost

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: i need the vertex, the axis of symmetry of this problem....y=-x^2+6x-5...i know i have to put in form y=a(x-h)^2+k but i am lost      Log On


   

Question 175102: i need the vertex, the axis of symmetry of this problem....y=-x^2+6x-5...i know i have to put in form y=a(x-h)^2+k but i am lost
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Actually, you don't need to put it in a different form.

For any parabola described by f%28x%29=y=ax%5E2%2Bbx%2Bc, the vertex is located at (-b%2F2a,f%28-b%2F2a%29) and the equation of the axis of symmetry is x=-b%2F2a. Just plug in your values and do the arithmetic.