# cubic polynomials

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• May 3rd 2008, 05:16 PM
trouble55
cubic polynomials
ok so I feel really stupid right now cause I can't figure this out:

The graph of a cubic polynomial function has x-intercepts -3,-1,and 2. The graph also passes through the point A(1,-24)

Determine the function in the factored form a(x-r1)(x-r2)(x-r3) where r1 ect represents the roots algebraically

Umm How do you solve for the a value? I know its vertical strech but can't figure it out.
Oh and I know a=3 but I can't do it algebraically
• May 3rd 2008, 05:55 PM
Soroban
Hello, trouble55!

Quote:

The graph of a cubic polynomial function has x-intercepts: -3,-1,and 2.
The graph also passes through the point A(1,-24)

Determine the function in the factored form: .$\displaystyle a(x-r_1)(x-r_2)(x-r_3)$

We know the three roots, so we have: .$\displaystyle y \;=\;a(x+3)(x+1)(x-2)\;\;{\color{blue}[1]}$

Why do you suppose they gave us another point? .$\displaystyle A(1,\,\text{-}24)$
. . This says: when $\displaystyle x = 1,\:y = \text{-}24$

Substitute into [1]: .$\displaystyle \text{-}24 \:=\:a(1+3)(1+1)(1-2) \quad\Rightarrow\quad \text{-}8a \:=\:\text{-}24 \quad\Rightarrow\quad a \:=\:3$

Therefore: . $\displaystyle y \;=\,3(x+3)(x+1)(x-2)$

• May 3rd 2008, 06:01 PM
trouble55
Thanks, (Happy)
I can't belive it was that simple!