SOLUTION: 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 = ½; x –interce
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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 = ½; x –intercept (-2, 0)
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
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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