What is a rational function that satisfies the following properties: a horizontal asymptote at y=3 and a vertical asymptote of x=-5?

graph{(3x)/(x+5) [-23.33, 16.67, -5.12, 14.88]}

There are certainly many ways to write a rational function that satisfy the conditions above but this was the easiest one I can think of.

In order to determine a function for a specific horizontal line we must keep the following in mind.

1. If the degree of the denominator is bigger than the degree of the numerator, the horizontal asymptote is the line #y = 0#.
   ex: #f(x)=x/(x^2+2)#

2. If the degree of the numerator is bigger than the denominator, there is no horizontal asymptote.
   ex: #f(x) = (x^3+5)/(x^2)#

3. If the degrees of the numerator and denominator are the same, the horizontal asymptote equals the leading coefficient of the numerator divided by the leading coefficient of the denominator
   ex: #f(x)=(6x^2)/(2x^2)#

The third statement is what we have to keep in mind for this example so our rational function must have the same degree in both the numerator and denominator but also, the quotient of leading coefficients had to equal #3#.

As for the function I gave, #f(x)=(3x)/(x+5)#
Both the numerator and denominator have a degree of #1#, so the horizontal asymptote is the quotient of the leading coefficients of the numerator over the denominator: #3/1 = 3# so the horizontal asymtopte is the line #y=3#

For the Vertical asymptote we keep in mind that all it really means is where on the graph is our function undefined. Since we are talking about a rational expression, our function is undefined when the denominator is equal to #0#.

As for the function I gave, #f(x)=(3x)/(x+5)#

We set the denominator equal to #0# and solve for #x#

#x+5=0 -> x=-5#

So our vertical asymptote is the line #x=-5#

In essence, the horizontal asymptote depends on the degree of both the numerator and the denominator. The vertical asymptote is determined by setting the denominator equal to #0# and solving for #x#