Horizontal Asymptotes by Degree

Far from the origin

A rational function, one polynomial divided by another, can have a vertical asymptote where its denominator is zero. At the far ends of its graph, where x is very large, either positive or negative, the graph may instead get closer and closer to a horizontal line y = c. That line is called a horizontal asymptote.

Take $y=\frac{\left(2x+1\right)}{\left(x+3\right)}$y = (2x + 1)/(x + 3). At x = 1000 it is $\frac{2001}{1003}$2001/1003, which is about 1.995. To see why the value settles near 2, divide the numerator and the denominator by x. Dividing the top and the bottom of a fraction by the same nonzero number does not change its value, so for every x other than 0, $y=\frac{\left(2+\frac{1}{x}\right)}{\left(1+\frac{3}{x}\right)}$y = (2 + 1/x)/(1 + 3/x).

The fractions in $\frac{1}{x}$1/x shrink to nothing

Now let x grow. $\frac{1}{x}$1/x and $\frac{3}{x}$3/x get smaller and smaller: at x = 1000 they are 0.001 and 0.003. So the numerator $2+\frac{1}{x}$2 + 1/x gets closer and closer to 2, the denominator $1+\frac{3}{x}$1 + 3/x gets closer and closer to 1, and y gets closer and closer to 2 ÷ 1 = 2. The same happens when x is a large negative number, since $\frac{1}{x}$1/x and $\frac{3}{x}$3/x shrink toward 0 from below.

The 2 and the 1 are the leading coefficients: the numbers in front of the highest power of x on the top and on the bottom. Far from the origin every other term is small beside the leading terms, 2x and x, and their ratio is $2x/x=2$2x / x = 2.

Values of $y=\frac{\left(2x+1\right)}{\left(x+3\right)}$y = (2x + 1)/(x + 3) far from the origin, rounded. On the right they climb toward 2 from below, and on the left they fall toward 2 from above.

Below on one side, above on the other

The table shows the curve approaching y = 2 from two different sides. To see why, work out how far the curve is from the line: $y-2=\frac{\left(2x+1\right)}{\left(x+3\right)}-2=\frac{\left(2x+1-2x-6\right)}{\left(x+3\right)}=-\frac{5}{\left(x+3\right)}$y − 2 = (2x + 1)/(x + 3) − 2 = (2x + 1 − 2x − 6)/(x + 3) = −5/(x + 3).

For large positive x the denominator x + 3 is positive, so y − 2 is negative and the curve is just below the line. For large negative x the denominator is negative, so y − 2 is positive and the curve is just above the line. On both sides $\frac{5}{\left(x+3\right)}$5/(x + 3) shrinks in size as x moves farther from the origin, so the gap closes. The curve also has a vertical asymptote at x = −3, where its denominator is zero.

The graph of $y=\frac{\left(2x+1\right)}{\left(x+3\right)}$y = (2x + 1)/(x + 3), with its asymptotes dashed. Far to the right the curve runs just below y = 2, and far to the left just above it. The vertical asymptote is x = −3.

When the bottom has the higher degree

Take $y=\frac{\left(x+1\right)}{\left(x^2+1\right)}$y = (x + 1)/(x² + 1). The highest power of x anywhere in the fraction is $x^2$x², so divide the numerator and the denominator by $x^2$x²: $y=\frac{\left(\frac{1}{x}+\frac{1}{x^2}\right)}{\left(1+\frac{1}{x^2}\right)}$y = (1/x + 1/x²)/(1 + 1/x²). As x grows, $\frac{1}{x}$1/x and $\frac{1}{x^2}$1/x² shrink to 0, so the numerator goes to 0 and the denominator goes to 1. So y goes to 0 ÷ 1 = 0, and the horizontal asymptote is y = 0, the x-axis.

The denominator grows like $x^2$x² and the numerator only like x, so the denominator grows much faster: at x = 10, $y=\frac{11}{101}$y = 11/101, which is about 0.11, and at x = 100, $y=\frac{101}{10001}$y = 101/10001, which is about 0.01.

Dividing by the highest power of x is what makes the method work in every case. After the division, every term has a power of x in its denominator and shrinks to 0, except the terms that held that highest power, which become plain numbers. Those numbers decide where the curve settles.

A curve may cross its horizontal asymptote

This curve crosses its own asymptote. The numerator x + 1 is zero at x = −1, so the graph passes through (−1, 0), on the line y = 0. To the left of −1 the numerator is negative and the denominator is positive, so the curve is below the x-axis there and rises back toward it from below.

That is allowed. A horizontal asymptote describes only what the curve does far from the origin; near the origin the curve can meet it, or cross it, any number of times. A vertical asymptote is different: at x = −3 in the first example the function has no value at all, so the curve can never be on that line.

The graph of $y=\frac{\left(x+1\right)}{\left(x^2+1\right)}$y = (x + 1)/(x² + 1). It crosses the x-axis at (−1, 0), and far out on both sides it flattens onto its horizontal asymptote, y = 0.

When the top has the higher degree

Take $y=\frac{\left(5x^2+4\right)}{\left(x+3\right)}$y = (5x² + 4)/(x + 3). Divide the numerator and the denominator by $x^2$x²: $y=\frac{\left(5+\frac{4}{x^2}\right)}{\left(\frac{1}{x}+\frac{3}{x^2}\right)}$y = (5 + 4/x²)/(1/x + 3/x²). As x grows, the numerator goes to 5, but the denominator goes to 0, and dividing 5 by a smaller and smaller number gives a larger and larger answer.

So the values grow without bound: at x = 100, $y=\frac{50004}{103}$y = 50004/103, which is about 485, and at x = 1000 it is about 4985. No horizontal line fits, so there is no horizontal asymptote. When the top is exactly one degree higher than the bottom, as here, the curve runs close to a slanted line instead, called a slant asymptote, which long division finds.

Compare the degrees

The degree of a polynomial is its highest power of x. Comparing the degree of the numerator with the degree of the denominator picks one of three cases.

If the numerator has the lower degree, the horizontal asymptote is y = 0. If the degrees are equal, it is $y=a/b$y = a / b, where a is the leading coefficient of the numerator and b is the leading coefficient of the denominator. If the numerator has the higher degree, there is no horizontal asymptote.

The equal case works for any degree. Take $f\left(x\right)=\frac{\left(2x^2+5x-3\right)}{\left(x^2-4\right)}$f(x) = (2x² + 5x − 3)/(x² − 4). Both degrees are 2, so divide by $x^2$x²: $f\left(x\right)=\frac{\left(2+\frac{5}{x}-\frac{3}{x^2}\right)}{\left(1-\frac{4}{x^2}\right)}$f(x) = (2 + 5/x − 3/x²)/(1 − 4/x²). Far out, the numerator goes to 2 and the denominator to 1, so the horizontal asymptote is $y=2/1=2$y = 2 / 1 = 2.

The curve is $f\left(x\right)=\frac{\left(2x^2+5x-3\right)}{\left(x^2-4\right)}$f(x) = (2x² + 5x − 3)/(x² − 4), and the dashed line is y = 2. Near the origin the branches beside its vertical asymptotes, x = −2 and x = 2, fill the picture. Zoom out: at x = 1000 the lower terms 5x − 3 make only about a quarter of one percent of the numerator, and the curve lies along y = 2.

Crossing, with equal degrees

Take $y=\frac{\left(2x^2+4x-2\right)}{\left(x^2+1\right)}$y = (2x² + 4x − 2)/(x² + 1). The degrees are equal and the leading coefficients are 2 and 1, so the horizontal asymptote is y = 2. The gap is $y-2=\frac{\left(2x^2+4x-2-2x^2-2\right)}{\left(x^2+1\right)}=\frac{\left(4x-4\right)}{\left(x^2+1\right)}$y − 2 = (2x² + 4x − 2 − 2x² − 2)/(x² + 1) = (4x − 4)/(x² + 1).

The gap is zero at x = 1, so the curve crosses y = 2 at (1, 2). For x greater than 1 the gap is positive and the curve is above the line; for x less than 1 it is below. Far out the gap shrinks to 0, because its denominator has the higher degree.

The graph of $y=\frac{\left(2x^2+4x-2\right)}{\left(x^2+1\right)}$y = (2x² + 4x − 2)/(x² + 1). It crosses its horizontal asymptote y = 2 at (1, 2), rises above it, and then comes back down toward it from above.

The usual mistakes

Reading the asymptote from the constant terms. For $y=\frac{\left(2x+1\right)}{\left(x+3\right)}$y = (2x + 1)/(x + 3), the 1 and the 3 are the terms that stop mattering far out, so the asymptote is not $y=\frac{1}{3}$y = 1/3. It comes from the leading coefficients: y = 2.

Answering y = 0 when the degrees are equal. y = 0 needs the denominator to have the higher degree. When both have the same degree they grow at the same rate, and their ratio settles on the ratio of the leading coefficients.

Answering y = 1, or the ratio of the leading coefficients, when the degrees differ. For $\frac{\left(x+1\right)}{\left(x^2+1\right)}$(x + 1)/(x² + 1) the denominator outgrows the numerator, so the asymptote is y = 0. For $\frac{\left(5x^2+4\right)}{\left(x+3\right)}$(5x² + 4)/(x + 3) the numerator outgrows the denominator, so there is none at all, not y = 5.

Thinking the curve can never touch its horizontal asymptote. It can, and it can cross it; the asymptote describes only the far ends.