equation of rational function

Question

Find an equation of a rational function f that satisfies the given conditions.
vertical asymptotes: x = −3, x = 5
horizontal asymptote: y = 2
x-intercept: −5, 3; hole at x = 0

 

 

Thanks!

Answer 1

The vertical asymptotes are found by setting the denominator of a rational function equal to zero.

 

Since vertical asymptotes are x=-3 and x=5 , your denominator is

 

(x + 3)(x - 5)

 

 

The x-intercepts are found by setting the numerator of a rational function equal to zero.

 

Since the x-intercepts are x=-5 and x=3 , you numerator is

 

(x + 5)(x - 3)

 

 

So far, your function is going to be

 

f(x) = [(x + 5)(x - 3)] / [(x + 3)(x - 5)]

 

 

If we include the horizontal asymptote y=2 as well as the hole of x=0, we get a final function of

 

f(x) = [2x(x + 5)(x - 3)] / [x(x + 3)(x - 5)]

 

 

Now we just multiply the numerator part and the denominator part.

 

f(x) = [2x(x² + 2x - 15)] / [x(x² - 2x - 15)]

 

f(x) = (2x³ + 4x² - 30x) / (x³ - 4x² - 15x)