SOLUTION: Find an equation in the form y=ax^2+bx+c for the parabola passing through the points (3,-22) (1,4) (2,-4)

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Question 1002751: Find an equation in the form y=ax^2+bx+c for the parabola passing through the points (3,-22) (1,4) (2,-4)
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

y=ax%5E2%2Bbx%2Bc for the parabola passing through the points (3,-22) (1,4) (2,-4)
use given points to set up the system and find a,b and c
(3,-22):
-22=a%2A3%5E2%2Bb%2A3%2Bc
-22=9a%2B3b%2Bc.....eq.1
(1,4)
4=a%2A1%5E2%2Bb%2A1%2Bc
4=a%2Bb%2Bc.....eq.2

(2,-4)
-4=a%2A2%5E2%2Bb%2A2%2Bc
-4=4a%2B2b%2Bc.....eq.3
so, the system is:
-22=9a%2B3b%2Bc.....eq.1
4=a%2Bb%2Bc.....eq.2
-4=4a%2B2b%2Bc.....eq.3
----------------------------
start with
-22=9a%2B3b%2Bc.....eq.1
4=a%2Bb%2Bc.....eq.2
-----------------------------subtract eq.1 from eq.2
4-%28-22%29=a%2Bb%2Bc-9a-3b-c
4%2B22=-8a-2b
26=-8a-2b.....both sides divide by 2
13=-4a-b.....solve for b
b=-4a-13...............1a
do same with eq.2 and eq.3
4=a%2Bb%2Bc.....eq.2
-4=4a%2B2b%2Bc.....eq.3
----------------------------subtract eq.2 from eq.3
-4-4=4a%2B2b%2Bc-a-b-c
-8=3a%2Bb
b=-8-3a........2a
from 1a and 2a we have:
-4a-13=-8-3a........solve for a
8-13=4a-3a
highlight%28a=-5%29
go to b=-8-3a........2a, substitute -5 for a
b=-8-3%28-5%29
b=-8%2B15
highlight%28b=7%29
go to 4=a%2Bb%2Bc.....eq.2, substitute -5 for a and 7 for b
4=-5%2B7%2Bc
4=2%2Bc
c=4-2
highlight%28c=2%29
so, your equation is: y=-5x%5E2%2B7x%2B2
see it and given points on the graph:
(3,-22) (1,4) (2,-4)