SOLUTION: equation of the quadratic function whose graph passes through each set of points(-3,2),(-1,0),(1,6) how do I set this up and solve

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: equation of the quadratic function whose graph passes through each set of points(-3,2),(-1,0),(1,6) how do I set this up and solve       Log On


   



Question 1128483: equation of the quadratic function whose graph passes through each set of points(-3,2),(-1,0),(1,6) how do I set this up and solve

Found 2 solutions by josmiceli, MathLover1:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
(-3,2), (-1,0),(1,6)
Use the general equation:
+y+=+a%2Ax%5E2+%2B+b%2Ax+%2B+c+
-----------------------------
(-3,2)
(1) +2+=+a%2A%28-3%29%5E2+%2B+b%2A%28-3%29+%2B+c+
(-1,0)
(2) +0+=+a%2A%28-1%29%5E2+%2B+b%2A%28-1%29+%2B+c+
(1,6)
(3) +6+=+a%2A1%5E2+%2B+b%2A1+%2B+c+
---------------------------------------
This is 3 equations with 3 unknowns, so
you can solve by any means to fill in
a, b, and c

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

equation of the quadratic function is:
y=ax%5E2%2Bbx%2Bc
if graph passes through each set of points (-3,2),(-1,0),(1,6), use them to set up system of equations and calculate a,b, and c
y=ax%5E2%2Bbx%2Bc....use (-3,2)
2=a%28-3%29%5E2%2Bb%28-3%29%2Bc
2=9a-3b%2Bc
c=2-9a%2B3b..........eq.1

y=ax%5E2%2Bbx%2Bc....use (-1,0)
0=a%28-1%29%5E2%2Bb%28-1%29%2Bc
0=a-b%2Bc
c=b-a..........eq.2

y=ax%5E2%2Bbx%2Bc....use (1,6)
6=a%28-1%29%5E2%2Bb%281%29%2Bc
6=a%2Bb%2Bc
c=6-a-b.........eq.3


from eq.1 and eq.2 we have:

2-9a%2B3b=b-a
3b-b=9a-a-2
2b=8a-2
b=4a-1...........(a}

from eq.2 and eq.3 we have:
b-cross%28a%29=6-cross%28a%29-b
b=6-b
b%2Bb=6
2b=6
highlight%28b=3%29

go to b=4a-1...........(a}, substitute b
3=4a-1
3%2B1=4a
4=4a
highlight%28a=1%29

go to c=b-a..........eq.2, substitute b and a

c=3-1
highlight%28c=2%29

your equation is: y=x%5E2%2B3x%2B2

check points on a graph:
(-3,2),(-1,0),(1,6)