At what points do vertical asymptotes occur? Is this always the case?
See solution.
Practice makes perfect
Let's start by reviewing what a rational function is and its main characteristics.
A rational function is a function that you can write in the form f(x) = P(x)Q(x), where P(x) and Q(x) are polynomial functions. The domain of f(x) is all real numbers for which Q(x) ≠ 0. If Q(a) = 0 we said that x=a is a point of discontinuity.
In a rational function, vertical asymptotes occur at the points of discontinuity. Recall the inverse parent function for instance, y = 1x. The zero for the denominator occurs when x=0, and we can find a vertical asymptote there.
We can see then that if Q(x) has no real zeros, we can guarantee that f(x) will not have a vertical asymptote. However, if Q(x) has a real zero and it represents removable discontinuity it will not have a vertical asymptote either. Let's review what these are and when they happen.
A rational function f(x) = P(x)Q(x) for which Q(a)=0, presents a removable discontinuity if we can redefine the value of f(x) at x=a and make the function continuous. This happens whenever (x-a)^m is a factor of P(x) and (x-a)^n a factor of Q(x) with m ≥ n.
This happens because we could divide out the common factors to find an equivalent expression that is not a quotient. However, the function would still have the gap in the value for which Q(x) =0. Let's see some examples.
Example 1
Consider the following rational function.
y_1 = x-4x-4
In this case P(4)=0 and Q(4)=0. Since they have a common zero, this function has a removable discontinuity at x=4. We can confirm this by looking at its graph.
Note that we can write the equivalent function y_1 = 1, with x≠ 4. As we can see, we could make the function continuous by defining y_1(4) = 1. And, since this is the case, the function has no vertical asymptotes.
Example 2
Now we will analyze a function where m>n.
y_2 = (x-4)^4/(x-4)^2
In this case both P(4)=0 and Q(4)=0. However, the factor causing this zero has a bigger exponent in the denominator. As we can see from its graph, it presents a removable discontinuity at x=4 and the graph has no vertical asymptotes.
Again, notice that we can write the equivalent function, y_2 = (x-4)^2, with x≠ 4. As we can see, we could make the function continuous by defining y_2(4) = 0.
Conclusion
Following the ideas discussed above, we can conclude that a rational function will have no vertical asymptotes when the denominator has no real zeros. Or, if it has them, whenever the factors causing them can be divided out with a similar factor from the numerator polynomial.
