SOLUTION: Find the vertex of the parabala y=-x^2+2

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Find the vertex of the parabala y=-x^2+2      Log On


   

Question 1052457: Find the vertex of the parabala y=-x^2+2
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Try to understand this as if in standard form.

y-0=-1%28x%5E2-2%29

y-0=-1%28%28x-0%29%5E2-2%29

y-0=-1%28x-0%29%5E2%2B2

y-2=-1%28x-0%29%5E2


Vertex (0,2)

Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!

Find the vertex of the parabala y=-x^2+2
x-coordinate of vertex is found at x+=+-+b%2F%282a%29 =====> x+=+0%2F%282a%29+=+0

Substituting 0 for x, y-coordinate of vertex is: matrix%281%2C3%2C+y+=+-+%280%29%5E2+%2B+2%2C+or%2C+y+=+2%29

Vertex of parabola: highlight_green%28matrix%281%2C3%2C+%22%280%22%2C+%22%2C%22%2C+%222%29%22%29%29

It's that simple.....nothing complicated!