Posted by **AJ** on Wednesday, March 23, 2011 at 7:12am.

Find the equation of the parabola with axis parallel to the x-axis and passing through (3/2,1),(5,0) and (-1,2).

- Analytic Goemetry -
**Reiny**, Wednesday, March 23, 2011 at 9:16am
In this case we don't know anything about the vertex, so we have to use the more general equation of the form

x = ay^2 + by + c

using (3/2 , 1) --- 3/2 = a + b + c

using (5,0 ) ------ 5 = 0+0+c or c=5

using (-1,2) ----- -1 = 4a + 2b + c

from the 2nd we know c=5

so in the first: a + b = -7/2 (#1)

in the 3rd : 4a + 2b = -6 or 2a + b = -3 (#3)

subtract #3 - #1 ----> a = 1/2

back in #1 :

1/2 + b = -7/2 ----- b = -4

so equation is

x = (1/2)y^2 - 4y + 5

Check to see of all 3 points satisfy this equation.

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