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The Slant Asymptote Calculator is a free online tool that displays the asymptote value for a given function. STUDYQUERIES’s slant asymptote calculator tool makes the calculation quicker, and it displays the asymptote value in a fraction of a second.

**How to Use the Slant Asymptote Calculator?**

You can use the slant asymptote calculator by following these steps:

**Step 1:**Enter the function into the input field**Step 2:**To calculate the slant asymptote, click “Calculate Slant Asymptote”**Step 3:**In the new window, the asymptotic value and graph will be displayed

Slant Asymptote Calculator

**What is Meant by Slant Asymptote?**

Slanted or oblique asymptotes are also possible in graphs and functions. What happens when a function’s asymptote is itself a function? In this article, we will explore a unique element of rational functions – slant asymptotes.

*Slant asymptotes represent the linear functions guiding the end behaviors of a rational function from both ends.*

Knowing about Slant asymptotes will allow us to predict how graphs will behave at extreme values of the variable \(x)\. Since this article will focus on the slant asymptotes of a rational function, here are a few important properties of rational functions:

- Here you can learn about rational functions and their graphs.
- Ensure you are familiar with horizontal and vertical concepts.

We will also need to review our understanding of graphing linear equations when we learn about graphing slant asymptotes. Do you wish to expand your knowledge of slant asymptotes? Let’s start by defining this term.

**Definition**

Oblique asymptotes are also known as slant asymptotes. That’s because of its slanted form representing a linear function graph, \(\mathbf{y=mx+b}\). Slant asymptotes can only occur for rational functions whose numerator’s degree is exactly one degree higher than its denominator’s degree.

The slant asymptotes are linear functions that we can use to predict the end behavior of rational functions, as shown in the example below.

As can be seen from the graph, \(\mathbf{f(x)}\)’s slant asymptote is represented by a dashed line guiding the graph’s behavior. We can also see that \(\mathbf{y=\frac{1}{2}x+1}\) is a linear function of the form, \(\mathbf{y=mx+b}\).

The slanted asymptote gives us an idea of how the curve of \(\mathbf{f(x)}\) behaves as it approaches \(-\infty\) and \(+\infty\). The graph of \(\mathbf{f(x)}\) also confirms what we already know: that slant asymptotes will be linear (and slanted).

Notice how \(\mathbf{f(x)}\) has no horizontal asymptotes? It’s because a rational function may only have a horizontal asymptote or a slant asymptote, but never both.

**Slant Asymptote Rules For Rational Functions**

When finding the oblique asymptote of a rational function, we always make sure to check the degrees of the numerator and denominator to confirm if a function has an oblique asymptote. See to it that the numerator’s degree is exactly one degree higher.

**Rule 1: If the numerator is a multiple of the denominator, the oblique asymptote will be the simplified form of the function.**

Let’s say we have $$\mathbf{f(x)=\frac{x^2-9}{x-3}}$$

\(\mathbf{x^2-9}\), is equivalent to \(\mathbf{(x+3)(x-3)}\) in factored form, so the denominator is a factor of the numerator.

The simplified form \mathbf{f(x)}\) is $$\mathbf{\frac{(x-3)(x+3)}{x-3}=x+3}$$

This means that the function has an oblique asymptote at \(\mathbf{y=x+3}\).

It’s helpful to keep this in mind since canceling out factors will be a much faster approach.

**Rule 2: If the numerator is not a multiple of the denominator, use long division or synthetic division to find the quotient of the function.**

Suppose we have $$\mathbf{f(x)=\frac{x^2-6x+9}{x-1}}$$

We can see that the numerator has a higher degree (by exactly one degree), so \(\mathbf{f(x)}\) must have a slant asymptote.

We can use synthetic division to find the quotient of \(\mathbf{x^2-6x+9}\) and \(\mathbf{x-1}\). (Make sure to review your knowledge on dividing polynomials.)

This shows that the quotient is \(\mathbf{x-5}\). We can also confirm this through long division as shown below.

From these two methods, we can see that \(\mathbf{f(x)=x-5+\frac{4}{x+1}}\), so focusing on the quotient, the slant asymptote of \(\mathbf{f(x)}\) is found at \(\mathbf{y=x-5}\).

**How To Find The Slant Asymptote?**

We need to refresh our memory on the following topics when determining a rational function’s Slant asymptote:

- Review on how we can perform
on polynomials.*long divisions* - We will also need to use
, so it’s best to refresh your knowledge.*synthetic division*

We will be able to determine which of the two methods is best based on the numerator and denominator’s forms.

Since \(\mathbf{f(x)=\frac{p(x)}{q(x)}}\), is a rational function with \(\mathbf{p(x)}\) having one degree higher than \(\mathbf{q(x)}\), we can find the quotient of \(\mathbf{\frac{p(x)}{q(x)}}\) to find the Slant asymptote.

$$\mathbf{f(x)=Quotient+\frac{Remainder}{q(x)}}$$

The Slant asymptote can be found by focusing only on the quotient and disregarding the remainder.

**Step By Step Guide To Find Slant Asymptote**

**Check your polynomial’s numerator and denominator. Ensure that the degree of the numerator (the highest exponent in the numerator) is greater than the degree of the denominator. If so, a slant asymptote exists and can be found.**

As an example, look at the polynomial $$\mathbf{\frac{x^2 + 5x + 2}{x + 3}}$$. The degree of its numerator is greater than the degree of its denominator because the numerator has a power of \(2\ in\ the\ term\ x^2\) while the denominator has a power of only 1. Therefore, you can find the slant asymptote. The graph of this polynomial is shown in the picture.

**Make a long division problem. Put the numerator (the dividend) inside the division box and the denominator (the divisor) outside.**

For the example above, set up a long division problem with \(\mathbf{x^2 + 5x + 2}\) as the dividend and \(\mathbf{x + 3}\) as the divisor.

**Determine the first factor. Look for a factor that, when multiplied by the highest degree term in the denominator, will result in the same term as the highest degree term in the dividend. Write it above the division box.**

Using the example above, you should look for a factor that, when multiplied by \(\mathbf{x}\), would result in the highest degree of \(\mathbf{x^2}\). Here, that is \(\mathbf{x}\). Place \(\mathbf{x}\) above the division box.

**Calculate the product of the factor and the whole divisor.**Multiply your product and write it below the dividend.

To solve this problem, multiply \(\mathbf{x}\) by \(\mathbf{x + 3}\) and get \(\mathbf{x^2 + 3x}\). Put it under the dividend as shown.

**Subtract.**Take the lower expression under the division box and subtract it from the upper expression. Draw a line and note the result of your subtraction underneath it.

In the example above, subtract \(\mathbf{x^2 + 3x}\) from \(\mathbf{x^2 + 5x + 2}\). Draw a line and note the result, \(\mathbf{2x + 2}\), underneath it, as shown.

**Continue dividing.**You can repeat these steps using the result of your subtraction problem as your new dividend.

Note that if you multiply \(\mathbf{2}\) by the highest term of the divisor \(\mathbf{(x)}\), you get the highest degree term of the dividend, which is now \(\mathbf{2x + 2}\). The \(\mathbf{2}\) is added to the first factor, making \(\mathbf{x + 2}\). This is written on top of the division box. Under the dividend, write the product of the factor and the divisor, then subtract again.

**Stop when you get an equation of a line.**It is not necessary to perform the long division all the way to the end. Only continue until you get a line equation of the form \(\mathbf{ax + b}\), where a and b can be any numbers.

You can now stop at the example above. The equation for your line is \(\mathbf{x + 2}\).

**Draw the line alongside the graph of the polynomial.**Graph your line to verify that it is actually an asymptote.

In the example above, you would need to graph \(\mathbf{x + 2}\) to see that the line moves alongside the graph of your polynomial but never touches it, as shown below. So \(\mathbf{x + 2}\) is indeed a slant asymptote of your polynomial.

**Conclusion**

We have learned a lot about oblique asymptotes already, so let’s summarize the important properties of oblique asymptotes before we proceed.

- The function has a slant asymptote if its numerator is exactly one degree higher than its denominator.
- The slant asymptote has a general form of \(\mathbf{y=mx+c}\), so we expect it to return a linear function.
- Utilize the intercepts of the slant asymptote to plot the linear function.
- You should also refresh your knowledge on the previous topics we have discussed in this article.

**FAQs**

**How do you find slant asymptotes?**

The oblique or slant asymptote is found by dividing the numerator by the denominator. A slant asymptote exists, since the degree of the numerator is 1 greater than the degree of the denominator. The equation \(\mathbf{y=mx+c}\) is a slant asymptote.

**Can a slant asymptote be a horizontal asymptote?**

A graph can have both a vertical and a slant asymptote, but it CANNOT have both a horizontal and slant asymptote. You draw a slant asymptote on the graph by putting a dashed horizontal (left and right) line going through \(\mathbf{y = mx + b}\).

**How do you know if a graph crosses a slant asymptote?**

If there is a slant asymptote, \(\mathbf{y=mx+b}\), then set the rational function equal to \(\mathbf{mx+b}\) and solve for \(\mathbf{x}\). If \(\mathbf{x}\) is a real number, then the line crosses the slant asymptote. Substitute this number into \(\mathbf{y=mx+b}\) and solve for \(\mathbf{y}\). This will give us the point where the rational function crosses the slant asymptote.

**Are oblique and slant asymptotes the same thing?**

Vertical asymptotes occur at the values where a rational function has a denominator of zero. An oblique or slant asymptote is an asymptote along a line, where. Oblique asymptotes occur when the degree of the denominator of a rational function is one less than the degree of the numerator.