SOLUTION: The graph of a polynomial has x-intercepts at 1, -2, and 10 and passes through the point (2,-32). Determine the equation of the function in factored form.
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Question 265636: The graph of a polynomial has x-intercepts at 1, -2, and 10 and passes through the point (2,-32). Determine the equation of the function in factored form.
Found 2 solutions by roseo, solver91311:
Answer by roseo(33) (Show Source): You can put this solution on YOUR website!
Because 1,-2, and 10 are x-intercepts or zeros of the polynomial then (x-1), (x+2), and (x-10) are factors of the polynomial.
So we know that
y = a(x-1)(x+2)(x-10) is our polynomial and all we have to do is find a and the point (2,-32) gives us an x and y to put in the function to determine a.
-32 = a (2-1)(2+2)(2-10)
-32 = a (1)(4)(-8)
-32 = -32a
therefore a = 1
So the polynomial y = 1(x-1)(x+2)(x-10) will pass through (2,-32)
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The factored form of all polynomial functions that have 
-intercepts at 1, -2, and 10 is:
\ =\ a(x\,-\,1)(x\,+\,2)(x\,-\,10))
However, we want the particular function such that \ =\ -32)
when 
, so substitute:
\,-\,1\right)\left((2)\,+\,2\right)\left((2)\,-\,10\right)\ =\ -32)
and solve for 
\left(4\right)\left(-8\right)\ =\ -32)
Hence:
\ =\ (x\,-\,1)(x\,+\,2)(x\,-\,10))
John
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