SOLUTION: I need help with this problem of algebra 2: The general form of an equation for a parabola is y=ax^2+bx+c, where (x,y) is a point on the parabola. If three points on a parabola

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Question 669596: I need help with this problem of algebra 2:
The general form of an equation for a parabola is y=ax^2+bx+c, where (x,y) is a point on the parabola. If three points on a parabola were (2,-10)(-5,-101)and (6,-90) determine the values of a, b and c write the general form of the equation.

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The general form of an equation for a parabola is y=ax^2+bx+c, where (x,y) is a point on the parabola.
If three points on a parabola were (2,-10)(-5,-101)and (6,-90) determine the
values of a, b and c write the general form of the equation.
;
Use elimination or substitution to find a, b, c, write an equation for each pair
:
2,-10
(2^2)a + 2b + c = -10
4a + 2b + c = -10
:
-5,-101
25a - 5b + c = -101
:
6,-90
36a + 6b + c = -90
:
Eliminate c,
subtract the 1st equation from the 2nd equation
25a - 5b + c = -101
4a + 2b + c = -10
-------------------Subtracting eliminates c
21a - 7b = -91
:
subtract the 1st eq from the 3rd eq
36a + 6b + c = -90
4a + 2b + c = -10
---------------------
32a + 4b = -80
4b = -32a - 80
simplify divide by 4
b = -8a - 20
:
Replace b with (-8a-20) in eq: 21a - 7b = -91
21a - 7(-8a-20) = -91
21a + 56a + 140 = -91
77a = -91 - 140
77a = -231
a = -3
:
Find b, replace a
b = -8(-3) - 20
b = +24 - 20
b = 4
:
Find c using the 1st equation
4(-3) + 2(4) + c = -10
-12 + 8 + c = -10
-4 + c = -10
c = -10 + 4
c = -6
:
The equation:
y = -3x^2 + 4x - 6
:
I checked this on my Ti83, to ensure the given pairs are on the graph
you should do the same