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1) The solution to the inequality \(\frac{x + c}{x^2 + ax + b} \le 0\) is \(x \in (-\infty,-1) \cup [1,2).\) Find a + b + c.

2) 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.

 Jul 27, 2019
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2)     4x - x^3                 x  ( 4 - x^2)           x  ( 2 - x) ( 2 + x)

       _______  =           __________   =   _______________

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

 

If we have a vertical  asymptote at  x = 1....then  (x - 1)  must be one of the factors of q(x)

And if we have a hole at x = - 2, then  (x + 2)  must also be a factor of q(x)

Since we have no horizontal asymptote, the q(x)  must just be a second power polynomial....and we will have a "slant asymptote"

And if q(3)  = 30  .....then q(x)  =   a ( 3 - 1) (3 + 2)  = -30  .....so 

a(2)(5)  =  -30

10a  = - 30

a = -3

 

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

 

Here's a graph : https://www.desmos.com/calculator/vsc0kthufu

 

If you drag your cursor along the curve....you will find that  at x = -2, y  = undefined  = "a hole" at this point

 

 

cool cool cool

 Jul 27, 2019

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