SOLUTION: The curve y=ax˛+bx+c passes (2,5), (3,12) and (-1,-4). Find the equation of the curve.

Algebra ->  Linear-equations -> SOLUTION: The curve y=ax˛+bx+c passes (2,5), (3,12) and (-1,-4). Find the equation of the curve.      Log On


   

Question 959876: The curve y=ax˛+bx+c passes (2,5), (3,12) and (-1,-4). Find the equation of the curve.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

y=ax%5E2%2Bbx%2Bc passes (2,5), (3,12) and (-1,-4)
use given points, set the system of three equations, solve it for a,b, and c
y=ax%5E2%2Bbx%2Bc.....for (2,5)
5=a%2A2%5E2%2Bb%2A2%2Bc
5=4a%2B2b%2Bc
4a%2B2b%2Bc-5=0..............eq.1

y=ax%5E2%2Bbx%2Bc.....for (3,12)
12=a%2A3%5E2%2Bb%2A3%2Bc
12=9a%2B3b%2Bc
9a%2B3b%2Bc-12=0..............eq.2

y=ax%5E2%2Bbx%2Bc.....for (-1,-4)
-4=a%2A%28-1%29%5E2%2Bb%2A%28-1%29%2Bc
-4=a-b%2Bc
a-b%2Bc%2B4=0..............eq.3
so, your system is:
4a%2B2b%2Bc-5=0..............eq.1
9a%2B3b%2Bc-12=0..............eq.2
a-b%2Bc%2B4=0..............eq.3
---------------------------------
subtract eq.1 from eq.2
9a%2B3b%2Bc-12-%284a%2B2b%2Bc-5%29=0
9a%2B3b%2Bc-12-4a-2b-c%2B5=0
5a%2Bb-7=0......solve for b
b=7-5a.........substitute in eq.3

a-%287-5a%29%2Bc%2B4=0..............eq.3
a-7%2B5a%2Bc%2B4=0
6a-3%2Bc=0.........solve for c
c=3-6a
go to 4a%2B2b%2Bc-5=0..............eq.1 substitute b and c
4a%2B2%287-5a%29%2B%283-6a%29-5=0...........solve for a
4a%2B14-10a%2B3-6a-5=0
12-12a=0
12=12a
highlight%28a=1%29
now find b and c
b=7-5a
b=7-5%2A1
highlight%28b=2%29

c=3-6a
c=3-6%2A1
highlight%28c=-3%29
so, your equation is: y=x%5E2%2B2x-3