Why Analyze a Rational Function in Cases?
A rational function can have several different features at once: an input excluded from its domain, a hole where a factor cancels, a vertical asymptote where the function grows without bound, and a horizontal asymptote describing its end behavior. These features are related, but they are not interchangeable. In particular, canceling a factor can simplify the formula without restoring an excluded input.
A dependable analysis begins with the original formula. First find its domain exclusions; then examine the behavior at each excluded input; finally, study what happens as \(x\) tends to positive and negative infinity. This order prevents a common error: simplifying first and forgetting where the original function was undefined.
A Case-Analysis Checklist
For each rational function, retain the original numerator and denominator while factoring. The denominator’s real zeros determine the domain exclusions. After factoring, cancel only common factors, and keep track of every excluded input even if its factor disappears. Evaluate limits at those inputs using the simplified expression when that expression is defined nearby. For vertical asymptotes, determine the one-sided signs as well as whether the function is unbounded.
Set the original denominator equal to zero. Each real solution is excluded.
Identify common factors, cancel them for analysis near an excluded input, but do not add that input back to the domain.
A finite limit indicates a hole; an infinite one-sided limit indicates a vertical asymptote and an infinite discontinuity.
Compare polynomial degrees and, when necessary, leading coefficients to determine horizontal asymptotes and end behavior.
Worked Example: A Canceled Factor Creates a Hole
Worked Example: A Canceled Factor Creates a Hole
Analyze \(f(x)=\frac{x^2-1}{x-1}\): find its domain, classify its excluded input, and describe its end behavior.
Step 1: Find the domain exclusion. The original denominator is zero when \(x-1=0\), so \(x=1\) is excluded. The domain is all real numbers except \(1\).
Step 2: Factor and simplify.
The restriction \(x\ne1\) remains, even though the formula simplifies.
Step 3: Classify the exclusion. As \(x\to1\), the simplified expression \(x+1\) tends to \(2\). The original function is undefined at \(1\), so its graph has a hole at \((1,2)\). This is a removable discontinuity, not a vertical asymptote.
Step 4: Analyze end behavior. For every allowed \(x\), \(f(x)=x+1\). Thus \(f(x)\to\infty\) as \(x\to\infty\), and \(f(x)\to-\infty\) as \(x\to-\infty\). There is no horizontal asymptote.
Check: The limit at the excluded input is \(\lim_{x\to1}f(x)=\lim_{x\to1}(x+1)=2\), matching the hole’s height. The simplified formula also confirms the stated end behavior.
Why Cancellation Distinguishes a Hole from an Asymptote
The key is the number of denominator factors that remain after common factors are canceled. Here is a local justification. Suppose \(c\) is an excluded input and the numerator and denominator each have a factor of \(x-c\). Near \(c\), write them as \(P(x)=(x-c)^m p(x)\) and \(Q(x)=(x-c)^n q(x)\), where \(p(c)\ne0\) and \(q(c)\ne0\). The integers \(m\) and \(n\) count the respective powers of that factor.
Because \(p\) and \(q\) are polynomials and \(q(c)\ne0\), the factor \(\frac{p(x)}{q(x)}\) tends to the finite, nonzero value \(\frac{p(c)}{q(c)}\) as \(x\to c\). Therefore:
- If \(m\ge n\), the expression has a finite limit. Since \(c\) was excluded from the original domain, this gives a removable discontinuity.
- If \(m<n\), a negative power of \(x-c\) remains. Its magnitude grows without bound as \(x\to c\), giving an infinite discontinuity and a vertical asymptote.
This argument explains why simply seeing a zero in the original denominator is not enough to declare a vertical asymptote. A common factor may cancel completely, or it may cancel only part of the denominator’s power.
Worked Example: A Hole and a Vertical Asymptote Together
Worked Example: A Hole and a Vertical Asymptote Together
Analyze \(g(x)=\frac{x+2}{x^2-4}\), including the domain, all discontinuities, one-sided behavior, and end behavior.
Step 1: Find the domain. Factor the original denominator:
The denominator is zero at \(x=-2\) and \(x=2\). Both inputs are excluded.
Step 2: Simplify without losing the exclusions.
Step 3: Classify \(x=-2\). The simplified expression is defined near \(-2\), and
So \(x=-2\) is a removable discontinuity, with a hole at \(\left(-2,-\frac14\right)\).
Step 4: Classify \(x=2\). The numerator of the simplified expression is positive. For \(x\to2^-\), \(x-2\) is negative and approaches zero, so \(g(x)\to-\infty\). For \(x\to2^+\), \(x-2\) is positive and approaches zero, so \(g(x)\to+\infty\). Thus \(x=2\) is an infinite discontinuity and a vertical asymptote.
Step 5: Describe end behavior. Since \(g(x)=\frac1{x-2}\) on its domain, \(g(x)\to0\) as \(x\to\infty\) and as \(x\to-\infty\). The horizontal asymptote is \(y=0\).
Check: At \(-2\), the simplified formula gives \(\frac1{-4}=-\frac14\), confirming the hole’s limiting height. Near \(2\), the denominator changes from negative to positive, confirming the two opposite infinite limits. At both ends, the denominator’s magnitude grows without bound, so the fraction tends to zero.
Horizontal Asymptotes and End Behavior
For \(f(x)=\frac{P(x)}{Q(x)}\), let \(m\) be the degree of \(P\) and \(n\) the degree of \(Q\). The degree comparison gives a quick test for finite horizontal asymptotes.
| Degree comparison | End behavior |
|---|---|
| \(m<n\) | \(f(x)\to0\) at both ends; horizontal asymptote \(y=0\). |
| \(m=n\) | \(f(x)\) tends at both ends to the ratio of the leading coefficients; that constant is the horizontal asymptote. |
| \(m>n\) | There is no horizontal asymptote. The function does not tend to a finite constant at either end. |
To see why the first two cases work, divide numerator and denominator by the highest power in the denominator. If \(m<n\), every numerator term divided by \(x^n\) tends to zero, while the denominator divided by \(x^n\) tends to its nonzero leading coefficient. If \(m=n\), the leading terms give the ratio of leading coefficients and every lower-degree term contributes a quantity tending to zero. The same reasoning applies as \(x\to-\infty\); degree parity can affect signs along the way, but not these finite limits.
A horizontal asymptote describes a limit, not a line the graph must stay on one side of. The graph can cross it. Also, an asymptote at one end is enough to call the corresponding line a horizontal asymptote, even though rational functions have the same finite limit at both ends when such an asymptote exists.
Worked Example: A Hole, a Vertical Asymptote, and a Horizontal Asymptote
Worked Example: A Hole, a Vertical Asymptote, and a Horizontal Asymptote
Analyze \(h(x)=\frac{3x^2+2x-1}{x^2-1}\), including its domain exclusions, discontinuities, and behavior at both ends.
Step 1: Find the domain. Factor numerator and denominator:
The original denominator is zero at \(x=-1\) and \(x=1\), so both are excluded.
Step 2: Simplify while retaining restrictions.
Step 3: Classify \(x=-1\).
Thus \(x=-1\) is a removable discontinuity, with a hole at \((-1,2)\).
Step 4: Classify \(x=1\). At \(x=1\), the simplified numerator is \(3(1)-1=2\), which is positive. As \(x\to1^-\), the denominator \(x-1\) is negative and approaches zero, so \(h(x)\to-\infty\). As \(x\to1^+\), the denominator is positive and approaches zero, so \(h(x)\to+\infty\). Therefore \(x=1\) is a vertical asymptote.
Step 5: Find the horizontal asymptote and end behavior. The numerator and denominator of the original function have the same degree, so the horizontal asymptote is the ratio of leading coefficients, \(y=3\). The simplified expression also shows the direction of approach:
As \(x\to\infty\), \(\frac2{x-1}\to0\) through positive values, so \(h(x)\to3\) from above. As \(x\to-\infty\), \(\frac2{x-1}\to0\) through negative values, so \(h(x)\to3\) from below.
Check: Substituting \(-1\) into the simplified expression gives \(\frac{-4}{-2}=2\), as required for the hole. Near \(1\), the numerator remains positive while the denominator changes sign, verifying the one-sided infinite limits. The remainder term \(\frac2{x-1}\) tends to zero at both ends, verifying the horizontal limit \(3\).
Common Mistakes to Avoid
A frequent error is to cancel a factor and then treat the canceled input as part of the domain. For example, the calculation \(\frac{x^2-1}{x-1}=x+1\) does not mean the original function is defined at \(x=1\). The original denominator is zero there. The correct conclusion is that the graph has a hole at \((1,2)\).
Another error is to declare every denominator zero a vertical asymptote. In \(\frac{x+2}{x^2-4}\), the input \(-2\) makes the original denominator zero, but cancellation leaves a finite limit of \(-\frac14\); that point is a hole, not a vertical asymptote. Finally, do not infer the horizontal asymptote from a few plotted values. Use the degree comparison, and report the limits at both ends.
Check Your Understanding
For each question, show enough work to justify the classification and limits.
- For \(r(x)=\frac{x^2-9}{x-3}\), state the domain and classify the excluded input.
- For \(s(x)=\frac{x-1}{x^2-1}\), identify every excluded input and determine which gives a hole and which gives a vertical asymptote.
- For \(t(x)=\frac{5x^2+1}{2x^2-3}\), find the horizontal asymptote and the limits as \(x\to\infty\) and \(x\to-\infty\).
- For \(u(x)=\frac{x+1}{x^2+1}\), state the domain, horizontal asymptote, and end behavior.
- Explain why canceling a common factor does not change the domain of the original rational function.
