SOLUTION: Determine whether a quadratic model exists for the set of values below. If so, write the model f(0)=-1,f(3)=-16,f(-1)=-8

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Determine whether a quadratic model exists for the set of values below. If so, write the model f(0)=-1,f(3)=-16,f(-1)=-8       Log On


   

Question 1003067: Determine whether a quadratic model exists for the set of values below. If so, write the model
f(0)=-1,f(3)=-16,f(-1)=-8
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=ax%5E2%2Bbx%2Bc
given:
f%280%29=-1=> it is a point (0,-1)=(x,f%28x%29)
f%283%29=-16=> it is a point (3,-16)
f%28-1%29=-8=> it is a point (-1,-8)
form the system using given points to find the coefficients a,b and c
-1=a%2A0%5E2%2Bb%2A0%2Bc
highlight%28c=+-1%29.......eq.1
-16=a%2A3%5E2%2Bb%2A3-1
-16=9a%2B3b-1
-16%2B1=9a%2B3b
-15=9a%2B3b......simplify, both sides divide by 3
-5=3a%2Bb.......solve for b
b=+-5-3a............eq.2

-8=a%28-1%29%5E2%2Bb%28-1%29-1
-8=a-b-1
b=a%2B8-1
b=a%2B7...........eq.3
since left sides in eq.2 and eq.3 are same, equal right sides
-5-3a=a%2B7...........solve for a
-5-7=a%2B3a
-12=4a
-12%2F4=a
highlight%28a=-3%29
go to b=a%2B7...........eq.3, plug in a and find b
b=+-3%2B7
highlight%28b=4%29

so, you have highlight%28a=-3%29,highlight%28b=4%29 and highlight%28c=-1%29
a quadratic model is:
f%28x%29=-3x%5E2%2B4x-1