<PRE><B>Question 636143</B>
Write the equation of a parabola that as a vertical axis and passes through the origin, (3,-2), and (-2,2).
**
Standard form of parabola with vertical axis of symmetry: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of vertex.
For given parabola:
h=0
equation:
y=Ax^2+k
use given coordinates, (3,-2) and (-2,2), to solve for A and k
..
-2=3^2A+k
2=(-2)^2A+k
..
-2=9A+k
  2=4A+k
subtract
5A=-4
A=-4/5
k=2-4A
=2-4*-4/5
=2+16/5
=10/5+16/5
=26/5
equation: y=-4x^2/5+26/5</PRE>