SOLUTION: find the vertex of the following: y=(x-5)^2 y=(x+3)^2 y=(x-1)^2 y=x^2+8 y=x^2-5 y=x^2+2

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: find the vertex of the following: y=(x-5)^2 y=(x+3)^2 y=(x-1)^2 y=x^2+8 y=x^2-5 y=x^2+2      Log On


   

Question 608521: find the vertex of the following:
y=(x-5)^2
y=(x+3)^2
y=(x-1)^2
y=x^2+8
y=x^2-5
y=x^2+2
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi

Note: the vertex form of a parabola opening up or down, y=a%28x-h%29%5E2+%2Bk where(h,k) is the vertex

find the vertex of the following:

 y=(x-5)^2  highlight%28V%285%2C0%29%29

 y=(x+3)^2  highlight%28V%28-3%2C0%29%29

 y=(x-1)^2  highlight%28V%281%2C0%29%29

 y=x^2+8    highlight%28V%280%2C8%29%29

 y=x^2-5    highlight%28V%280%2C-5%29%29

 y=x^2+2    highlight%28V%280%2C2%29%29