# Question Video: Finding the Vertical and Horizontal Asymptotes of a Given Function Mathematics • 10th Grade

Determine the vertical and horizontal asymptotes of the function 𝑓(𝑥) = −1 + ( 3/𝑥) − (4/𝑥²).

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### Video Transcript

Determine the vertical and horizontal asymptotes of the function 𝑓 of 𝑥 equals negative one plus three over 𝑥 minus four over 𝑥 squared.

Let’s start with the vertical asymptotes. Vertical asymptotes are vertical lines where the denominator in a rational function equals zero. In this function, we see that we have 𝑥 and 𝑥 squared in the denominator. And that means 𝑥 cannot be equal to zero. So we say that the vertical asymptote of this function is at 𝑥 equals zero. The function doesn’t exist at 𝑥 equals zero.

Now horizontal asymptotes are a little bit different. Horizontal asymptotes indicate general behaviour usually far off to sides of the graph. And when we look for horizontal asymptotes, we need to consider a few cases. We have the case that the degree of the denominator is larger than the degree of the numerator. And the case where the degree of the numerator is equal to the degree of the denominator. For now, we won’t consider the cases where the degree of the numerator is larger than the degree of the denominator.

At first glance, it might seem like the degree of the numerator here is zero because there’re no 𝑥-values in the numerator. But we need to see the case when we add all three of these terms together. We need a common denominator. If we use 𝑥 squared as our common denominator, we can rewrite negative one as negative 𝑥 squared over 𝑥 squared.

And then, to convert three over 𝑥 into a fraction with the denominator of 𝑥 squared or multiply three over 𝑥 by 𝑥 over 𝑥. And that becomes three 𝑥 over 𝑥 squared. And the four over 𝑥 squared doesn’t change. And we’ve rewritten our function to look like this, negative 𝑥 squared plus three 𝑥 minus four all over 𝑥 squared. The degree, the highest exponent of our numerator, is two. And the degree of our denominator, the highest exponent in the denominator, is also two. So we have the second case. The degree in our numerator equals the degree in our denominator.

When this is the case, there is an horizontal asymptote at the place 𝑦 equals the numerator’s leading coefficient over the denominator’s leading coefficient. The leading coefficient in our numerator is negative one. And the leading coefficient in our denominator is positive one. Therefore, we did have a horizontal asymptote at negative one over one or, more simply, at 𝑦 equals negative one.

Our vertical asymptote is at 𝑥 equals zero. And our horizontal asymptote is at 𝑦 equals negative one.