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

 Dec 19, 2019
 #1
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You can take q(x) = -15/2*(x - 1)^2.

 Dec 19, 2019
 #2
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(-x^3  + 4x)  =  x ( 4 -x^2)   =   x  ( 2 - x) ( 2 + x)

 

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

 

Also....since we have a vertical asymptote at  x =1, then  (x - 1)   must also be a factor of q(x)

 

Since we have no horizontal asymptote, then we must have  a slant asymptote, which means that we have a higher/ lower situation with respect to the degree of the polynomial in the numerator to the degree of the polynomial in the  denominator

 

And  since   q(3)   =  -30

Then we must have  that   

-30   = a (3 - 1) ( 3 + 2)

-30  = a (2)(5)

-30  = 10a

a=-30/10  =  -3

 

So

 

q(x)  must  be   -3 ( x -1 ) ( x + 2)  =   (-3) (x^2 + x - 2)  =  -3x^2 -3x  + 6

 

 

cool cool cool

 Dec 19, 2019

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