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

Answer 1

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.