Find an equation of a rational function f that satisfies the given conditions

Question

vertical asymptotes: x = −9, x = 3
horizontal asymptote: y = 0
x-intercept: −2; f(0) = −2
hole at x = 6

Answer 1

Rational functions are functions that have a numerator part and a denominator part.

 

When we have vertical asymptotes, it means the function is undefined at the given points.  Since function is undefined when the denominator zero, the denominator part of the function becomes

 

(x + 9)(x - 3) =

 

x² + 6x - 27

 

 

x-intercepts means that if we set the numerator part equal to zero, the function becomes zero.  The numerator part of the function becomes

 

x + 2

 

 

So far, the function is

 

f(x) = (x + 2) / (x² + 6x - 27)

 

 

When we have a hole, it indicates a discontinuity of f(x), but it is not an asymptote.  So we multiply f(x) by (x - 6) / (x - 6).

 

 

f(x) = [(x + 2) / (x² + 6x - 27)] * [(x - 6) / (x - 6)]

 

f(x) = (x² - 4x - 12) / (x³ - 63x + 162)

 

 

 

Now we set the function as

 

f(x) = C(x² - 4x - 12) / (x³ - 63x + 162)

 

 

We will use the y-intercept (0, -2) to find the value of C.

 

 

-2 = C(-12 / 162)

 

-2 = (-2 / 27)C

 

C = 27

 

 

Therefore, the final function is

 

f(x) = (27x² - 108x - 324) / (x³ - 63x + 162)