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# help

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Find $$q(x)$$ if the graph of $$\frac{4x-x^3}{q(x)}$$ has a hole at $$x=-2$$, a vertical asymptote at $$x=1$$, no horizontal asymptote, and $$q(3) = -30$$.

May 6, 2019

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4x - x^3            x ( 4 - x^2)            x ( 2 - x) ( 2 +x)

______  =       ___________ =    _____________

q(x)                    q(x)                       q(x)

If we have a  "hole" at x = -2, then q(x) must have a  factor  of ( x + 2)

If we have a vertical asymptote at x = 1, then (x - 1)  must also be a factor

If we have no horizontal asymptote, the q(x) must be a second degree polynomial - the rational fuction has a s;ant asymptote

And  since   q(3) = -30....then we have that

-30 = a(3 + 2) (3 - 1)

-30  =a (5)(2)

-30 = 10a

a = -3

So....q(x)  = -3(x + 2)(x - 1) =  -3(x^2 + x - 2)  =  -3x^2 - 3x + 6

Here's a graph :  https://www.desmos.com/calculator/dz5hnv1yoa

May 6, 2019