SOLUTION: FIND THE EQUATION OF THE PARABOLA WITH VERTICAL AXIS THAT PASSES THROUGH THE POINT (0,2) AND THE POINTS OF INTERSECTION OF THE PARABOLAS x² + 2x + 3y + 4 = 0 AND x²- 3x + y+ 3 =

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Question 1187080: FIND THE EQUATION OF THE PARABOLA WITH VERTICAL AXIS THAT PASSES THROUGH THE POINT (0,2) AND THE POINTS OF INTERSECTION OF THE PARABOLAS x² + 2x + 3y + 4 = 0 AND x²- 3x + y+ 3 = 0.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

vertical axis passes through (0,2)
find the points of intersection of the parabolas
x%5E2%2B2x%2B3y%2B4=0...........eq.1
x%5E2-3x%2By%2B3=0........eq.2
------------------------------------sumtract eq.2 from eq.1
x%5E2%2B2x%2B3y%2B4-%28x%5E2-3x%2By%2B3%29=0+
x%5E2%2B2x%2B3y%2B4-x%5E2%2B3x-y-3=0+
5x%2B2y%2B1=0 .............solve for y
2y=-5x-1
y=%28-5x-1%29%2F2

go to
x%5E2-3x%2By%2B3=0........eq.2, substitute y
x%5E2-3x%2B%28-5x-1%29%2F2%2B3=0........ solve for x, multiply by+2
2x%5E2-6x-5x-1%2B6=0
2x%5E2-11x%2B5=0............factor
%28x+-+5%29+%282+x+-+1%29+=+0
solutions: x=5, x=1%2F2

go to
y=%28-5x-1%29%2F2, substitute x
y=%28-5%2A5-1%29%2F2=-13
y=%28-5%2A%281%2F2%29-1%29%2F2=-7%2F4

intersection points are: (5,-13) and (1%2F2,-7%2F4)
and to find parabola containing the three points (0,2), (5,-13) and (1%2F2,-7%2F4), you could setup three equations using these points starting in a format
y=ax%5E2%2Bbx%2Bc+
2=a%2A0%5E2%2Bb%2A0%2Bc+
2=c+.........1)
-13=a%2A5%5E2%2Bb%2A5%2B2+
0=25a%2B5b%2B15+..........divide by 5 and solve for+b
-5a-3=b+...............2)
-7%2F4=a%2A%281%2F2%29%5E2%2Bb%2A%281%2F2%29%2B2+
0=%281%2F4%29a%2B%281%2F2%29b%2B2%2B7%2F4+....... multiply by 4 andsolve for+b
0=a%2B2b%2B8%2B7+
0=a%2B2b%2B15+
-a-15=+2b++
-a%2F2-15%2F2=+b++....3)
equal 2) and 3)
-5a-3=-a%2F2-15%2F2......solve for+a
15%2F2-3=5a-a%2F2+
9%2F2=9a%2F2+
a=1
then
-5%2A1-3=b+...............2)
b=-8

your equation is
y=ax%5E2%2Bbx%2Bc+......substitute a=1,b=-8, c=2
y=+x%5E2-8x%2B2+