SOLUTION: Use a cubic function whose graph passes through the points (-3,0),(0,-6),(-2,0),(1,0)

Algebra ->  Rational-functions -> SOLUTION: Use a cubic function whose graph passes through the points (-3,0),(0,-6),(-2,0),(1,0)      Log On


   

Question 691019: Use a cubic function whose graph passes through the points (-3,0),(0,-6),(-2,0),(1,0)
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Since the graph passes through the points (-3,0),(-2,0), and (1,0),
your cubic function has zeros at x=-3, x=-2, and x=1.
If a polynomial has a zero at x=a, it has (x-a) as a factor.
So your cubic function has %28x%2B3%29%28x%2B2%29%28x-1%29=x%5E3%2Bax%5E2%2Bx-6 as a factor.
Since that is already a cubic, the only other factor we need is a real number b
Your cubic function is
f%28x%29=b%28x%2B3%29%28x%2B2%29%28x-1%29=b%28x%5E3%2Bax%5E2%2Bx-6%29
For x=0, we know that f(0)=-6, because the graph passes through (0,-6).
Then, f%280%29=b%28-6%29=-6 --> b=1 and
highlight%28f%28x%29=%28x%2B3%29%28x%2B2%29%28x-1%29=x%5E3%2Bax%5E2%2Bx-6%29