SOLUTION: Find an equation in standard form of the parabola described. Vertex at (−3, −3); passes through (0, 0). standard form: (y-k)^2 =4p(x-h)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find an equation in standard form of the parabola described. Vertex at (−3, −3); passes through (0, 0). standard form: (y-k)^2 =4p(x-h)      Log On


   

Question 1162554: Find an equation in standard form of the parabola described.
Vertex at (−3, −3); passes through (0, 0).
standard form: (y-k)^2 =4p(x-h)
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

standard form:
%28y-k%29%5E2+=4p%28x-h%29 where h and k are coordinates of vertex
since given vertex at (-3, -3), we have
%28y-%28-3%29%29%5E2+=4p%28x-%28-3%29%29
%28y%2B3%29%5E2+=4p%28x%2B3%29 .........eq.1

since passes through (0,+0), we have
%28y%2B3%29%5E2+=4p%28x%2B3%29
%280%2B3%29%5E2+=4p%280%2B3%29
9+=12p
p=9%2F12
p=3%2F4
so, your equation is:
%28y%2B3%29%5E2+=4%283%2F4%29%28x%2B3%29
%28y%2B3%29%5E2+=3%28x%2B3%29