Can a rational function have both a horizontal and a slant asymptote?

No. A rational function will have a horizontal asymptote if $n \le m$ and a slant asymptote if $n = m + 1$. These conditions are mutually exclusive, so a function cannot exhibit both.

What happens if a factor cancels out from the denominator?

The $x$-value that makes the canceled factor zero corresponds to a removable discontinuity (a hole) on the graph, not a vertical asymptote, provided it does not also remain in the denominator.

Can a graph cross its own horizontal asymptote?

Yes. A horizontal asymptote describes the end behavior of the function as $x o \pm\infty$. The function may intersect this line at finite values of $x$.

How to find the asymptotes of a rational function

A rational function $f(x) = \frac{P(x)}{Q(x)}$ has asymptotes determined by the roots of its denominator and the relative degrees of its polynomials. This method applies to any function expressed as a ratio of two polynomials with real coefficients.

The setup

Define $f(x) = \frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials. Factor both the numerator and the denominator completely. Cancel any common factors to obtain the simplified form of the rational function.

The steps

Vertical Asymptotes: Set the simplified denominator equal to zero and solve for $x$ . Each root $x=c$ is a vertical asymptote. 2. Horizontal Asymptotes: Compare the degree of $P(x)$ , denoted as $n$ , and the degree of $Q(x)$ , denoted as $m$ . If $n < m$ , the horizontal asymptote is $y=0$ . If $n = m$ , the horizontal asymptote is $y = \frac{a}{b}$ , where $a$ and $b$ are the leading coefficients of $P(x)$ and $Q(x)$ respectively. 3. Slant (Oblique) Asymptotes: If $n = m + 1$ , perform polynomial long division. The quotient, ignoring the remainder, gives the equation of the slant asymptote $y = mx + c$ . If $n > m + 1$ , there are no linear asymptotes.

Checking the result

Evaluate $\lim_{x o \pm\infty} f(x)$ to verify your horizontal or slant asymptotes. Check values arbitrarily close to your calculated vertical asymptotes to confirm the function tends to positive or negative infinity.

Common errors

Failing to cancel common factors before finding vertical asymptotes is a standard error. This leads to identifying false vertical asymptotes at locations where the function actually has a removable discontinuity (a hole).

Worked example

Find all asymptotes of $f(x) = \frac{2x^2 + 5x - 3}{x - 1}$ .

Factor the numerator: $P(x) = 2x^2 + 5x - 3 = (2x - 1)(x + 3)$ . The denominator is $Q(x) = x - 1$ . There are no common factors to cancel. To find the vertical asymptote, set the denominator to zero: $x - 1 = 0$ , so the vertical asymptote is $x = 1$ . Next, compare degrees. The degree of $P(x)$ is $n=2$ and the degree of $Q(x)$ is $m=1$ . Since $n = m + 1$ , there is a slant asymptote and no horizontal asymptote. Perform polynomial long division of $(2x^2 + 5x - 3)$ by $(x - 1)$ . First term: $2x^2 / x = 2x$ . Multiply: $2x(x - 1) = 2x^2 - 2x$ . Subtract: $(2x^2 + 5x - 3) - (2x^2 - 2x) = 7x - 3$ . Second term: $7x / x = 7$ . Multiply: $7(x - 1) = 7x - 7$ . Subtract: $(7x - 3) - (7x - 7) = 4$ . The remainder is 4 and the quotient is $2x + 7$ . Therefore, the slant asymptote is $y = 2x + 7$ .

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References: OpenStax College Algebra, Chapter 5: Polynomial and Rational Functions · Stewart Calculus, Chapter 2: Limits and Derivatives