Video: Finding the Vertical and Horizontal Asymptotes of a Rational Function

Find the vertical and horizontal asymptotes of the function 𝑓(π‘₯) = (3π‘₯Β² βˆ’ 1)/(5π‘₯Β² + 3).

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Video Transcript

Find the vertical and horizontal asymptotes of the function 𝑓 of π‘₯ equals three π‘₯ squared minus one over five π‘₯ squared plus three.

What is a vertical asymptote of a function? It’s a vertical line that the graph of the function gets closer and closer to. And there are various ways that the function can do this. The function could approach infinity from the left and negative infinity from the right or vice versa. Or both left-hand and right-hand limits could be infinity. In which case, the limit itself would be infinity or negative Infinity. Or one of the one-sided limits could be finite where the other is infinite.

Again, there are various ways this could happen for a general function. But we have a special kind of function, a rational function. And the vertical asymptotes of rational functions come from zeros of the denominator. This isn’t to say that every zero of a denominator of a rational function gives a vertical asymptote. For example, the rational function π‘₯ minus one over π‘₯ minus one has a zero at π‘₯ equals one. But the graph of this function is just the straight line 𝑦 equals one with one point, the point with π‘₯-coordinate one removed. But the line π‘₯ equals one isn’t a vertical asymptote of this function. So a zero of the denominator doesn’t guarantee an asymptote. But if there is an asymptote, it will definitely come from a zero of the denominator.

What are the zeros of our denominator, five π‘₯ squared plus three? Subtracting three from both sides and dividing by five we see that π‘₯ squared is a negative number, negative three-fifths. And there’s no real number whose square is negative. And so, this equation has no real solutions. And so there are no vertical asymptotes.

We move on to finding any horizontal asymptotes there might be. A horizontal asymptote is a horizontal line that the graph of the function approaches. There are fewer cases to consider for horizontal asymptotes. The line 𝑦 equals 𝐿 is a horizontal asymptote of the graph of 𝑓 of π‘₯, if either the limit of 𝑓 of π‘₯, as π‘₯ approaches negative infinity, is 𝐿 or the limit of 𝑓 of π‘₯, as π‘₯ approaches infinity, is 𝐿. Well, there’s no limit to the number of vertical asymptotes a function’s graph could have. It can only have two horizontal asymptotes, one as π‘₯ approaches infinity and the other as π‘₯ approaches negative infinity.

To find any horizontal asymptotes therefore, we find the limit of 𝑓 of π‘₯ as π‘₯ approaches negative infinity and the limit of 𝑓 of π‘₯ as π‘₯ approaches infinity. We use the definition of 𝑓 of π‘₯. The trick to evaluating this limit is to divide both numerator and denominator by the highest power of π‘₯ you can see. In this case, π‘₯ squared. After simplifying, we get the limit of three minus one over π‘₯ squared over five plus three over π‘₯ squared as π‘₯ approaches negative infinity. And using the fact that the limit of a quotient is the quotient of the limit, as long as the value of the limit in the denominator is nonzero, and also that the limit of a sum or difference is the sum or difference of the limits, we end up with this.

The limit of the constant function three, and as π‘₯ approaches negative infinity, is just three. And similarly, the limit of the constant function five, as π‘₯ approaches negative infinity, is just five. So we can get rid of these limit signs. And the other limits, the limit of one over π‘₯ squared as π‘₯ approaches negative infinity and the limit of three over π‘₯ squared as π‘₯ approaches negative infinity, are both zero. We’re left with just three over five. And so, using our fact about horizontal asymptotes, the line 𝑦 equals three over five is a horizontal asymptote of the graph of 𝑓 of π‘₯. That was the limit of 𝑓of π‘₯, as π‘₯ approaches negative infinity. How about the limit of 𝑓 of π‘₯, as π‘₯ approaches just infinity? We just replace all the references to negative infinity by positive infinity. And we see that we get the same answer, three-fifths. So we don’t get a second horizontal asymptote. There’s only one with equation 𝑦 equals three-fifths.

This is therefore our final answer. The function has no vertical asymptote and a horizontal asymptote at 𝑦 equals three-fifths. We managed to do this all without graphing the function 𝑓 of π‘₯ equals three π‘₯ squared minus one over five π‘₯ squared plus three. You may now like to graph the function to see that the answer we got is indeed right.

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