SOLUTION: Find the values of a, b, and c such that the equation y = ax2 + bx + c has ordered pair solutions (-3, -23), (-1, -9) and (3, -5)

Algebra ->  Equations -> SOLUTION: Find the values of a, b, and c such that the equation y = ax2 + bx + c has ordered pair solutions (-3, -23), (-1, -9) and (3, -5)      Log On


   

Question 985911: Find the values of a, b, and c such that the equation y = ax2 + bx + c has ordered pair solutions (-3, -23), (-1, -9) and (3, -5)
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
y+=+ax%5E2+%2B+bx+%2B+c+
for (-3, -23)
-23=+a%28-3%29%5E2+%2B+b%28-3%29+%2B+c
-23=+9a+-3b%2B+c ........eq.1

for (-1,+-9)
-9=+a%28-1%29%5E2+%2B+b%28-1%29+%2B+c+
-9=+a+-b%2B+c ........eq.2

for (3, -5)
-5=+a%283%29%5E2+%2B+b%283%29+%2B+c
-5=+9a+%2B3b%2B+c ........eq.3

solve system:
-23=+9a+-3b%2B+c ........eq.1
-9=+a+-b%2B+c ........eq.2
-5=+9a+%2B3b%2B+c ........eq.3
----------------------------------------------
go with
-23=+9a+-3b%2B+c ........eq.1
-5=+9a+%2B3b%2B+c ........eq.3
----------------------------------------subtract
-23-%28-5%29=+9a+-3b%2B+c+-9a-3b-c
-23%2B5=-6b
-18=+-6b
b=-18%2F-6
highlight%28b=3%29

go to eq.2 and eq.3, substitute 3 for b
-9=+a+-3%2B+c ........eq.2
-5=+9a+%2B9%2B+c ........eq.3
-----------------------------------------subtract eq.2 from eq.3
-5-%28-9%29=+9a-a+%2B9-%28-3%29%2B+c-c
-5%2B9=+8a+%2B9%2B3
4=+8a+%2B12
8a=4-12
8a=-8
a=8%2F-8
highlight%28a=-1%29
go to eq. -5=+9a+%2B3b%2B+c ........eq.3 substitute -1 for a and 3 for b
-5=+9%28-1%29+%2B3%2A3%2B+c+........eq.3
-5=+-cross%289%29+%2Bcross%289%29%2B+c+
highlight%28c=-5%29
so, your equation is: y+=+-x%5E2+%2B+3x+-5+

here is a graph with given points: