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12.3: Horizontal and Vertical Asymptotes

Difficulty Level: At Grade Created by: CK-12
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Practice Horizontal and Vertical Asymptotes

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Horizontal and Vertical Asymptotes

We said that a horizontal asymptote is the value of y\begin{align*}y\end{align*} that the function approaches for large values of |x|\begin{align*}|x|\end{align*}. When we plug in large values of x\begin{align*}x\end{align*} in our function, higher powers of x\begin{align*}x\end{align*} get larger much quickly than lower powers of x\begin{align*}x\end{align*}. For example, consider:

y=2x2+x13x24x+3\begin{align*}y = \frac{2x^2 + x - 1}{3x^2 - 4x + 3}\end{align*}

If we plug in a large value of x\begin{align*}x\end{align*}, say x=100\begin{align*}x = 100\end{align*}, we get:

y=2(100)2+(100)13(100)24(100)+3=20000+100130000400+2\begin{align*}y = \frac{2(100)^2 + (100) - 1}{3(100)^2 - 4(100) + 3} = \frac{20000 + 100 - 1}{30000 - 400 + 2}\end{align*}

We can see that the beginning terms in the numerator and denominator are much bigger than the other terms in each expression. One way to find the horizontal asymptote of a rational function is to ignore all terms in the numerator and denominator except for the highest powers.

In this example the horizontal asymptote is y=2x23x2\begin{align*}y = \frac{2x^2}{3x^2}\end{align*}, which simplifies to y=23\begin{align*}y = \frac{2}{3}\end{align*}.

In the function above, the highest power of x\begin{align*}x\end{align*} was the same in the numerator as in the denominator. Now consider a function where the power in the numerator is less than the power in the denominator:

y=xx2+3\begin{align*}y = \frac{x}{x^2 + 3}\end{align*}

As before, we ignore all the terms except the highest power of x\begin{align*}x\end{align*} in the numerator and the denominator. That gives us y=xx2\begin{align*}y = \frac{x}{x^2}\end{align*}, which simplifies to y=1x\begin{align*}y = \frac{1}{x}\end{align*}.

For large values of x\begin{align*}x\end{align*}, the value of y\begin{align*}y\end{align*} gets closer and closer to zero. Therefore the horizontal asymptote is y=0\begin{align*}y = 0\end{align*}.

To summarize:

• Find vertical asymptotes by setting the denominator equal to zero and solving for x\begin{align*}x\end{align*}.
• For horizontal asymptotes, we must consider several cases:
• If the highest power of x\begin{align*}x\end{align*} in the numerator is less than the highest power of x\begin{align*}x\end{align*} in the denominator, then the horizontal asymptote is at y=0\begin{align*}y = 0\end{align*}.
• If the highest power of x\begin{align*}x\end{align*} in the numerator is the same as the highest power of x\begin{align*}x\end{align*} in the denominator, then the horizontal asymptote is at y=coefficient of highest power of xcoefficient of highest power of x\begin{align*}y = \frac{\text{coefficient of highest power of } x}{\text{coefficient of highest power of } x}\end{align*}.
• If the highest power of x\begin{align*}x\end{align*} in the numerator is greater than the highest power of x\begin{align*}x\end{align*} in the denominator, then we don’t have a horizontal asymptote; we could have what is called an oblique (slant) asymptote, or no asymptote at all.

Finding Asymptotes

1. Find the vertical and horizontal asymptotes for y=1x1\begin{align*}y = \frac {1}{x-1}\end{align*}.

Vertical asymptotes:

Set the denominator equal to zero. x1=0x=1\begin{align*}x-1=0\Rightarrow x=1\end{align*} is the vertical asymptote.

Horizontal asymptote:

Keep only the highest powers of x\begin{align*}x\end{align*}. y=1xy=0\begin{align*}y=\frac {1}{x}\Rightarrow y=0\end{align*} is the horizontal asymptote.

2. Find the vertical and horizontal asymptotes for y=3x4x+2\begin{align*}y= \frac {3x}{4x+2}\end{align*}.

Vertical asymptotes:

Set the denominator equal to zero. 4x+2=0x=12\begin{align*}4x+2=0\Rightarrow x=-\frac{1}{2}\end{align*} is the vertical asymptote.

Horizontal asymptote:

Keep only the highest powers of x\begin{align*}x\end{align*}. y=3x4xy=34\begin{align*}y=\frac {3x}{4x}\Rightarrow y=\frac{3}{4}\end{align*} is the horizontal asymptote.

3. Find the vertical and horizontal asymptotes for y=x3x23x+2\begin{align*}y=\frac {x^3}{x^2-3x+2}\end{align*}.

Vertical asymptotes:

Set the denominator equal to zero: x23x+2=0\begin{align*}x^2 - 3x + 2 =0\end{align*}

Factor: (x2)(x1)=0\begin{align*}(x - 2)(x - 1) = 0\end{align*}

Solve: x=2\begin{align*}x = 2\end{align*} and x=1\begin{align*}x = 1\end{align*} are the vertical asymptotes.

Horizontal asymptote. There is no horizontal asymptote because the power of the numerator is larger than the power of the denominator.

Notice the function in part d had more than one vertical asymptote. Here’s another function with two vertical asymptotes.

Graphing Functions

Graph the function y=x2x24\begin{align*}y = \frac{-x^2}{x^2 - 4}\end{align*}.

Set the denominator equal to zero:x24=0Factor:(x2)(x+2)=0Solve:  x=2,x=2\begin{align*}\text{Set the denominator equal to zero:} \qquad x^2 - 4 = 0\!\\ \\ \text{Factor:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (x - 2)(x + 2) = 0\!\\ \\ \text{Solve:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ x = 2, x = -2\end{align*}

We find that the function is undefined for x=2\begin{align*}x = 2\end{align*} and x=2\begin{align*}x = -2\end{align*}, so we know that there are vertical asymptotes at these values of x\begin{align*}x\end{align*}.

We can also find the horizontal asymptote by the method we outlined above. It’s at y=x2x2\begin{align*}y = \frac{-x^2}{x^2}\end{align*}, or y=1\begin{align*}y = -1\end{align*}.

So, we start plotting the function by drawing the vertical and horizontal asymptotes on the graph.

Now, let’s make a table of values. Because our function has a lot of detail we must make sure that we pick enough values for our table to determine the behavior of the function accurately. We must make sure especially that we pick values close to the vertical asymptotes.

x\begin{align*}x\end{align*} y=x2x24\begin{align*}y = \frac {-x^2}{x^2-4}\end{align*}
\begin{align*}-5\end{align*} \begin{align*}y = \frac {-(-5)^2}{(-5)^2-4} = \frac {-25}{21} = -1.19\end{align*}
-4 \begin{align*}y = \frac {-(-4)^2}{(-4)^2-4} = \frac {-16}{12} = -1.33\end{align*}
-3 \begin{align*}y = \frac {-(-3)^2}{(-3)^2-4} = \frac {-9}{5} = -1.8\end{align*}
-2.5 \begin{align*}y = \frac {-(-2.5)^2}{(-2.5)^2-4} = \frac {-6.25}{2.25} = -2.8\end{align*}
-1.5 \begin{align*}y = \frac {-(-1.5)^2}{(-1.5)^2-4} = \frac {-2.25}{-1.75} = 1.3\end{align*}
-1 \begin{align*}y = \frac {-(-1)^2}{(-1)^2-4} = \frac {-1}{-3} = 0.33\end{align*}
0 \begin{align*}y = \frac {-0^2}{(0)^2-4} = \frac {0}{-4} = 0\end{align*}
1 \begin{align*}y= \frac {-1^2}{(1)^2-4} = \frac {-1}{-3} = 0.33\end{align*}
1.5 \begin{align*}y = \frac {-1.5^2}{(1.5)^2-4} = \frac {-2.25}{-1.75} = 1.3\end{align*}
2.5 \begin{align*}y = \frac {-2.5^2}{(2.5)^2-4} = \frac {-6.25}{2.25} = -2.8\end{align*}
3 \begin{align*}y = \frac {-3^2}{(3)^2-4} = \frac {-9}{5} = -1.8\end{align*}
4 \begin{align*}y = \frac {-4^2}{(4)^2-4} = \frac {-16}{12} = -1.33\end{align*}
5 \begin{align*}y = \frac {-5^2}{(5)^2-4} = \frac {-25}{21} = -1.19\end{align*}

Here is the resulting graph.

Example

Example 1

Find the vertical and horizontal asymptotes for \begin{align*}y=\frac {x^2-2}{2x^2+3}\end{align*}.

Vertical asymptotes:

Set the denominator equal to zero: \begin{align*} 2x^2+3 = 0 \Rightarrow 2x^2 = -3 \Rightarrow x^2 = -\frac{3}{2}\end{align*}. Since there are no solutions to this equation, there is no vertical asymptote.

Horizontal asymptote:

Keep only the highest powers of \begin{align*}x\end{align*}. \begin{align*}y=\frac {x^2}{2x^2} \Rightarrow y= \frac {1}{2}\end{align*} is the horizontal asymptote.

Review

Find all the vertical and horizontal asymptotes of the following rational functions.

1. \begin{align*}y=\frac {4}{x+2}\end{align*}
2. \begin{align*}y=\frac {5x-1}{2x-6}\end{align*}
3. \begin{align*}y=\frac {10}{x}\end{align*}
4. \begin{align*}y=\frac {2}{x}-5\end{align*}
5. \begin{align*}y=\frac {x + 1}{x^2}\end{align*}
6. \begin{align*}y=\frac {4x^2}{4x^2+1}\end{align*}
7. \begin{align*}y=\frac {2x}{x^2-9}\end{align*}
8. \begin{align*}y=\frac {3x^2}{x^2-4}\end{align*}
9. \begin{align*}y=\frac {1}{x^2+4x+3}\end{align*}
10. \begin{align*}y=\frac {2x+5}{x^2-2x-8}\end{align*}

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Color Highlighted Text Notes

Vocabulary Language: English

Asymptotes

An asymptote is a line on the graph of a function representing a value toward which the function may approach, but does not reach (with certain exceptions).

Hole

A hole exists on the graph of a rational function at any input value that causes both the numerator and denominator of the function to be equal to zero.

Horizontal Asymptote

A horizontal asymptote is a horizontal line that indicates where a function flattens out as the independent variable gets very large or very small. A function may touch or pass through a horizontal asymptote.

Intercept

The intercepts of a curve are the locations where the curve intersects the $x$ and $y$ axes. An $x$ intercept is a point at which the curve intersects the $x$-axis. A $y$ intercept is a point at which the curve intersects the $y$-axis.

Intercepts

The intercepts of a curve are the locations where the curve intersects the $x$ and $y$ axes. An $x$ intercept is a point at which the curve intersects the $x$-axis. A $y$ intercept is a point at which the curve intersects the $y$-axis.

Oblique Asymptote

An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division.

Oblique Asymptotes

An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division.

points of discontinuity

The points of discontinuity for a function are the input values of the function where the function is discontinuous.

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Date Created:
Oct 01, 2012
Apr 11, 2016
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