SOLUTION: How do you solve for a vertex when B is not given. {{{y=X^2+5}}} i know how to do it when it is ax^2+bx+c, but i dont know what to do when its just ax^2+c

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: How do you solve for a vertex when B is not given. {{{y=X^2+5}}} i know how to do it when it is ax^2+bx+c, but i dont know what to do when its just ax^2+c      Log On


   

Question 606647: How do you solve for a vertex when B is not given. y=X%5E2%2B5 i know how to do it when it is ax^2+bx+c, but i dont know what to do when its just ax^2+c
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.

y=x%5E2%2B5 tells us "h" = 0

Vertex is Pt(0,5)