SOLUTION: A parabola whose axis is parallel to the y-axis passes through the points (1,1), (2,2) and (-1,5). Find its equation.

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Question 948208: A parabola whose axis is parallel to the y-axis passes through the points (1,1), (2,2) and (-1,5). Find its equation.
Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A parabola whose axis is parallel to the y-axis passes through the points (1,1), (2,2) and (-1,5). Find its equation.
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Form::
ax^2 + bx + c = y
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Using (1,1) u get:: a + b + c = 1
Using (2,2) u get: 4a + 2b + c = 2
Using (-1,5) u get: a - b + c = 5
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Using a Matrix Program I get::
a = 1
b = -2
c = 2
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Equation:
y = x^2 - 2x + 2
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Cheers,
Stan H.
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Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
y+=+ax%5E2%2B+bx+%2B+c
use given points to set up the system of equations to find a,b,and c
(1, 1) ----> a+%2B+b+%2B+c+=+1
(2,+2) ----> 4a+%2B+2b+%2B+c+=+2
(-1,5) ---> a+-b+%2B+c+=+5
Solve simultaneous equations:subtract first from second one
4a+%2B+2b+%2B+c+-%28a+%2B+b+%2B+c%29=+2-1
4a+%2B+2b+%2B+c+-a+-+b+-+c=+1
3a+%2B+b++=+1...solve for a
=>3a+=+1-b
=>a+=+1%2F3-b%2F3.......1a
add first and third one:
a+%2B+b+%2B+c+%2Ba+-b+%2B+c+=+1%2B5
2a+%2B+2c+=+6 ....solve for a
=>2a++=+6+-2c
=>a++=+3+-c.........2a
go to first eq., substitute a from 1a
1%2F3-b%2F3+%2B+b+%2B+c+=+1
2b%2F3+%2B+c+=+1-1%2F3
2b%2F3+%2B+c+=+2%2F3
2b+%2B+3c+=+2
2b+=+2-3c ....solve for b
b+=+1-3c%2F2...............1b
go to eq.2, substitute a from 2a and b from 1b
4%28+3+-c%29%2B+2%28+1-3c+%2F2%29%2B+c+=+2
12+-4c%2B+2-3c+%2B+c+=+2
14-6c+=+2
14-2+=+6c
12+=+6c
highlight%28c=2%29
now find a
=>a++=+3+-c
=>a++=+3+-2
=>highlight%28a++=+1%29
and b
b+=+1-%283%2A2%29%2F2
=>b+=+1-3
=>highlight%28b+=+-2%29
so, your equation is:

highlight%28y+=+x%5E2+-2x+%2B+2%29