Create an original rational function that has at least one asymptote (vertical, horizontal, and/or slant) and possibly a removable discontinuity.
Consider the function f(x) = (x^2 - 4)/(x + 2).
This function has a vertical asymptote at x = -2, as the denominator becomes zero at x = -2 and the function approaches infinity as x approaches -2 from both sides.
To find if there's a horizontal asymptote, we can check the limit as x approaches infinity and negative infinity.
As x approaches infinity, let's simplify the function by dividing the numerator and denominator by x:
f(x) = (x^2 - 4)/(x + 2) = (1 - 4/x^2)/(1/x + 2/x) = (1 - 4/x^2)/(1/x + 2/x) = (1 - 4/x^2)/(1 + 2/x)
As x approaches infinity, both (1 - 4/x^2) and (1 + 2/x) approach 1. Therefore, the function approaches 1/1 = 1.
As x approaches negative infinity, we can use the same simplification steps:
f(x) = (x^2 - 4)/(x + 2) = (1 - 4/x^2)/(1/x + 2/x) = (1 - 4/x^2)/(1/x + 2/x) = (1 - 4/x^2)/(3/x)
As x approaches negative infinity, both (1 - 4/x^2) and (3/x) approach 1. Therefore, the function approaches 1/1 = 1.
Since the function approaches the same value (1) as x approaches infinity and negative infinity, there is a horizontal asymptote at y = 1.
Now let's consider if there's a removable discontinuity. We can simplify the function further by factoring the numerator:
f(x) = (x^2 - 4)/(x + 2) = ((x + 2)(x - 2))/(x + 2) = x - 2
By canceling out the (x + 2) terms, we get f(x) = x - 2. This function doesn't have any removable discontinuities since we canceled out any potential denominator factors.
Therefore, the function f(x) = (x^2 - 4)/(x + 2) has a vertical asymptote at x = -2 and a horizontal asymptote at y = 1.