Equation for a rational function

Question

Find an equation for the rational function which has a zero at x=6, a hole at x=5, and asymptotes at x=2 and y=3? Thank you!

Answer 1

To find the zeros, we set the numerator of the rational function equal to zero.  Since x=6 is your zero, the numerator of the function will be (x - 6). 

 

x=2 is a vertical asymptote.  This is found by setting the denominator equal to zero.  The denominator of the function is (x - 2).

 

y=3 is a horizontal asymptote.  This was obtained by examining the degree of the function.  This asymptote is a constant when the degree of the leading term of numerator and denominator are equal to each other and dividing their coefficients.

 

So far, the function looks like this:

 

f(x) = 3(x - 6) / (x - 2)

 

Next is to utilize the hole.  x=5 is a discontinuity, but not a vertical asymptote.  We can multiply the function by (x - 5) / (x - 5) , since they cancel each other out.  The final form of f(x) when expanded is

 

 

f(x) = (3(x - 6)(x - 5)) / ((x - 2)((x - 5))

 

f(x) = 3(x² - 11x + 30) / (x² - 7x + 10)

 

 

 

f(x) = (3x² - 33x + 90) / (x² - 7x + 10)