Question 1167632: Find the equation of a parabola with axis parallel to the x-axis and
passing through (3/2, 1), (5, 0) and ( - 1, 2). Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! x=ay^2+by+c
when x=3/2
a+b+c=3/2
c=5
4a+2b+c=-1. 4a+2b=-6; 2a+b=-3; b=-2a-3
substitute into first
a-2a-3+5=3/2
-a=-1/2
a=1/2
Therefore b=-4
and the function is f(y)=(1/2)y^2-4y+5
so 2x=y^2-8y+10
(2x-10)=y^2-8
2x+6=(y-4)^2, completing the square
y=4+/- sqrt (2x+6)
