SOLUTION: How to determine the equation of a parabola with vertex (7,-2) and passing the point (6,0)? Graph the parabola. So far I have gotten y=a(x+b)^2+c y=a(x+6)^2+0

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: How to determine the equation of a parabola with vertex (7,-2) and passing the point (6,0)? Graph the parabola. So far I have gotten y=a(x+b)^2+c y=a(x+6)^2+0      Log On


   

Question 1036483: How to determine the equation of a parabola with vertex (7,-2) and passing the point (6,0)? Graph the parabola.
So far I have gotten
y=a(x+b)^2+c
y=a(x+6)^2+0
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Standard Form, y=a%28x-h%29%5E2%2Bk has vertex (h,k). You are given the vertex and you can solve for the value of a using the other given point.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

How to determine the equation of a parabola with vertex (7,-2) and passing the point (6,0)? Graph the parabola.
So far I have gotten
y=a(x+b)^2+c
y=a(x+6)^2+0
No, no, no!! The vertex form of a parabola is the equation: y+=+a%28x+-+h%29%5E2+%2B+k. In this case, (h, k) = (7, 2), and (x, y) = (6, 0). 

You substituted (x, y) for (h, k). That's incorrect! Try again!