SOLUTION: Write the equation of a parabola that as a vertical axis and passes through the origin, (3,-2), and (-2,2).

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Question 636143: Write the equation of a parabola that as a vertical axis and passes through the origin, (3,-2), and (-2,2).
Found 2 solutions by lwsshak3, solver91311:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Write the equation of a parabola that as a vertical axis and passes through the origin, (3,-2), and (-2,2).
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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

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


A parabola is represented by a general form quadratic function: .

If it passes through a point , then the following will hold:



For the three points you were given:







From which the following 3X3 system can be derived:







Substituting the first into the other two gives us a 2X2 linear system:





Solve the 2X2 by whatever means you find convenient and then you will have the coefficients for your function.

John

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