Question 312851: Find a rational function that satisfies the given conditions.
a. Vertical asymptotes x = -4, x = 2
b. Vertical asymptote x = -4, x = 5; horizontal asymptote y = ˝; x –intercept (-2, 0)
Answer by solver91311(24713)   (Show Source):
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If a rational function has a vertical asymptote at  , then the denominator polynomial has a zero at  .
Since your first problem has two vertical asymptotes, the denominator polynomial must have two zeros,  and  . That means that the denominator polynomial has two factors,  and  . Multiplying these factors gives:  . The numerator can be anything, so:
\ =\ \frac{A}{x^2\ +2x\ -\ 8})
Second problem:
Again, two vertical asymptotes so the denominator is a quadratic with factors of and  and is therefore  .
Since there is a non-zero horizontal asymptote,  , the numerator and denominator polynomials are of the same degree and their lead coefficients are in the ratio  . Since the lead coefficient on the denominator is  , the lead coefficient in the numerator must be
Since there is an  -intercept at ) , the numerator polynomial must have a factor of  . Given that the lead coefficient must be , the other factor of the quadratic polynomial numerator must be of the form  . That gives us \left(x\ +\ 2\right)\ =\ \frac{x^2}{2}\ +\ (2\,+\,a)x\ +\ 2a) . But if is the ONLY -intercept, then  , which means that  , and therefore the numerator becomes 
Putting it altogether:
\ =\ \frac{\frac{x^2}{2}\ +\ 6x\ +\ 8}{x^2\ -\ x\ -\ 20})
John
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