SOLUTION: Find the values of a, b, and c such that the equation y = ax2 + bx + c has ordered pair solutions (-1, -4), (2, -7) and (3, 0) this is just a pretest

Question 986161: Find the values of a, b, and c such that the equation y = ax2 + bx + c has ordered pair solutions (-1, -4), (2, -7) and (3, 0)
this is just a pretest
Found 3 solutions by stanbon, solver91311, MathLover1:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
Find the values of a, b, and c such that the equation y = ax2 + bx + c has ordered pair solutions (-1, -4), (2, -7) and (3, 0)
this is just a pretest
------
Form:: ax^2 + bx + c = y
Solve for a,b, and c using the 3 pairs::
a - b + c = -4
4a +2b + c = -7
9a +3b + c = 0
----------------------
Using any method you know solve for a, b, and c::
Ans::
a = 2 ; b = -3 ; c = -9
-----
Equation:
y = 2x^2 -3x - 9
======================
Cheers,
Stan H.
---------------
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!

If is an element of the solution set, then

Likewise:

and

Simplifying:

Solve the 3X3 system of equations for the values of the coefficients of the desired quadratic.

John

My calculator said it, I believe it, that settles it

Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!
Find the values of , , and such that the equation has ordered pair solutions (,), (,) and (,):
for (,)=(,)

.............eq.1

for (,)= (,)

.............eq.2

for (,)=(,)

.............eq.3

solve this system:
.............eq.1
.............eq.2
.............eq.3
------------------------------------------------

start with
.............eq.1
.............eq.2...subtract 1 from 2
-----------------------------------

.............eq.1a
go with
.............eq.2
.............eq.3..........subtract 2 from 3
--------------------------------------------

..........eq.2a
from eq.1a and 2a we have
........solve for

go to
.............eq.1a.....substitute for

go to .............eq.3 ...substitute for , for

so, your equation is:

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