Why Rational Functions Need a Complete Analysis
A rational function is a quotient of polynomials. Its graph can have features that polynomial graphs do not: excluded inputs, holes, vertical asymptotes, and end behavior that approaches a line rather than growing without bound. Derivatives add information about turning points and concavity, but they must be interpreted on the function’s actual domain.
A useful analysis keeps these features connected. The domain determines where branches can exist; limits describe how branches behave at excluded inputs and at the ends of the graph; derivatives describe how each branch rises, falls, and bends.
A Reliable Order of Analysis
Start with the original expression, not just a simplified one. Canceling factors can make the formula easier to analyze, but it does not put excluded inputs back into the domain. Then find intercepts, asymptotes and limits, critical points, and concavity intervals. Finally, use that information to describe each branch.
Set the original denominator unequal to zero. Mark every excluded input, including inputs whose factors later cancel.
The \(x\)-intercepts are domain inputs where the numerator is zero after accounting for common factors. The \(y\)-intercept is \(f(0)\), if \(0\) is in the domain.
Use limits at excluded inputs to distinguish holes from vertical asymptotes. Use degree comparison or polynomial division for end behavior.
Find critical numbers in the domain from \(f'(x)=0\) or from points where \(f'\) is undefined but \(f\) exists. Use \(f'\) and \(f''\) sign charts to describe monotonicity, extrema, and concavity.
Use the domain boundaries, limits, intercepts, and derivative signs together. A sign chart interval cannot cross an excluded input.
Holes and Vertical Asymptotes
Suppose a factor \(x-a\) appears in both the numerator and denominator. Canceling it simplifies the formula for inputs other than \(a\), but the original function is still undefined at \(a\). To decide what happens there, examine the limit of the simplified expression as \(x\) approaches \(a\).
More generally, after reducing the fraction, any denominator factor that still vanishes at an excluded input can cause unbounded behavior. Checking the limit is the decisive test. A hole is not part of the graph, so it is not a critical point or an inflection point.
End Behavior and Branches
For a rational function, compare the degrees of the numerator and denominator to guide the end-behavior analysis. If the denominator’s degree is larger, the function approaches \(0\). If the degrees are equal, the function approaches the ratio of leading coefficients. If the numerator’s degree is exactly one greater, polynomial division gives a slant asymptote. More generally, division gives a polynomial quotient and a remainder term whose limit can clarify the end behavior.
A vertical asymptote divides the domain into separate intervals. Determine the one-sided limits on each side; the same asymptote can have different infinite behavior from its left and right. A complete sketch must also account for the function’s behavior as \(x\) tends to positive or negative infinity.
Worked Example: A Slant Asymptote and Two Critical Points
Worked Example: Analyze a Rational Function with a Slant Asymptote
Analyze \(f(x)=\frac{x^2+x-1}{x-1}\), combining its domain, intercepts, asymptotes, extrema, and concavity.
1. Domain and division. The denominator is zero at \(x=1\), so the domain is \(x\ne1\). Polynomial division gives \(f(x)=x+2+\frac{1}{x-1}\) for \(x\ne1\). The nonzero remainder shows that the discontinuity at \(1\) is not removable: as \(x\) approaches \(1\), the last term is unbounded. Thus \(x=1\) is a vertical asymptote. The quotient gives the slant asymptote \(y=x+2\).
2. Intercepts. The \(y\)-intercept is \(f(0)=\frac{-1}{-1}=1\), so it is \((0,1)\). To find the \(x\)-intercepts, solve \(x^2+x-1=0\): \(x=\frac{-1\pm\sqrt{1+4}}{2}=\frac{-1\pm\sqrt5}{2}\). Neither root is \(1\), so both are in the domain and give \(x\)-intercepts.
3. Limits at the vertical asymptote and at infinity. From the divided form, \(x+2\) remains finite near \(1\), while \(\frac{1}{x-1}\) determines the infinite behavior: \(\lim_{x\to1^-}f(x)=-\infty\) and \(\lim_{x\to1^+}f(x)=+\infty\). As \(x\) tends to either infinity, the remainder \(\frac{1}{x-1}\) tends to \(0\), confirming the slant asymptote \(y=x+2\).
4. First derivative and extrema. Differentiate the divided form: \(f'(x)=1-\frac{1}{(x-1)^2}=\frac{(x-1)^2-1}{(x-1)^2}=\frac{x(x-2)}{(x-1)^2}\). The critical numbers in the domain are \(x=0\) and \(x=2\). Since the denominator of \(f'\) is positive wherever defined, the sign is determined by \(x(x-2)\): it is positive for \(x<0\), negative for \(0<x<1\) and \(1<x<2\), and positive for \(x>2\). Thus \(f\) increases up to \(x=0\), decreases on both intervals from \(0\) to \(2\) separated by the asymptote, and increases after \(2\). The sign changes give a local maximum at \(x=0\), with \(f(0)=1\), and a local minimum at \(x=2\), with \(f(2)=4+2-1=5\) divided by \(1\), or \(5\).
5. Concavity. Differentiate again: \(f''(x)=\frac{2}{(x-1)^3}\). This is negative for \(x<1\) and positive for \(x>1\), so the graph is concave down on \((-\infty,1)\) and concave up on \((1,\infty)\). Although concavity changes across \(x=1\), that input is not in the domain, so it is not an inflection point.
Check. Differentiating \(x+2+\frac{1}{x-1}\) gives \(1-\frac{1}{(x-1)^2}\), matching \(f'\). Substituting \(x=0\) and \(x=2\) into the original function gives \(1\) and \(5\), respectively. The remainder term tends to zero at both ends, as required for the slant asymptote.
Why the Limit Test Matters: A Hole and an Asymptote in One Function
Worked Example: Keep a Canceled Input Out of the Domain
Analyze \(g(x)=\frac{(x-1)(x+1)}{(x-1)(x+2)}\) while keeping the original domain restrictions.
1. Domain and reduction. The original denominator vanishes at \(x=1\) and \(x=-2\). Thus the domain excludes both values. For other inputs, \(g(x)=\frac{x+1}{x+2}=1-\frac{1}{x+2}\). At \(x=1\), the reduced expression has finite limit \(\lim_{x\to1}g(x)=\frac{1+1}{1+2}=\frac23\). Therefore, the graph has a hole at \((1,\frac23)\). At \(x=-2\), the reduced denominator still vanishes, and \(g\) is unbounded, so \(x=-2\) is a vertical asymptote.
2. Intercepts and end behavior. The reduced numerator is zero at \(x=-1\), which is in the domain, giving the \(x\)-intercept \((-1,0)\). Also, \(g(0)=\frac12\), giving the \(y\)-intercept \((0,\frac12)\). Since \(g(x)=1-\frac{1}{x+2}\) and the fraction tends to zero at either end, the horizontal asymptote is \(y=1\).
3. Limits and derivatives. The form \(1-\frac{1}{x+2}\) shows that \(\lim_{x\to-2^-}g(x)=+\infty\) and \(\lim_{x\to-2^+}g(x)=-\infty\). For inputs in the domain, \(g'(x)=\frac{1}{(x+2)^2}>0\). Thus the function increases on each of its domain intervals, \((-\infty,-2)\), \((-2,1)\), and \((1,\infty)\); there are no critical points. The excluded hole at \(1\) does not join those intervals.
4. Concavity and check. The second derivative is \(g''(x)=-\frac{2}{(x+2)^3}\). It is positive for \(x<-2\) and negative for \(x>-2\). The graph is concave up on \((-\infty,-2)\) and concave down on \((-2,1)\) and \((1,\infty)\). There is no inflection point at \(x=1\), because \(g\) is undefined there. Differentiating \(1-\frac{1}{x+2}\) confirms both derivatives; substituting \(x=-1\) and \(x=0\) into the reduced expression confirms the intercepts.
Worked Example: Two Vertical Asymptotes and an Inflection Point
Worked Example: Describe All Four Branches
Analyze \(h(x)=\frac{x}{x^2-1}\). This example shows why derivative signs must be checked separately on intervals divided by vertical asymptotes.
1. Domain, intercepts, and end behavior. Since \(x^2-1=(x-1)(x+1)\), the domain excludes \(x=-1\) and \(x=1\). The numerator is zero at \(x=0\), where the denominator is \(-1\), so the graph has both \(x\)- and \(y\)-intercept \((0,0)\). The denominator’s degree exceeds the numerator’s by one, so the horizontal asymptote is \(y=0\). In fact, \(h(x)\) approaches \(0\) from below as \(x\to-\infty\) and from above as \(x\to+\infty\).
2. Vertical asymptotes and one-sided behavior. Near \(x=-1\), the numerator is negative. The denominator is positive just to the left of \(-1\) and negative just to the right, giving \(\lim_{x\to-1^-}h(x)=-\infty\) and \(\lim_{x\to-1^+}h(x)=+\infty\). Near \(x=1\), the numerator is positive. The denominator is negative just to the left and positive just to the right, giving \(\lim_{x\to1^-}h(x)=-\infty\) and \(\lim_{x\to1^+}h(x)=+\infty\).
3. Monotonicity and critical points. The quotient rule gives \(h'(x)=\frac{(x^2-1)-2x^2}{(x^2-1)^2}=-\frac{x^2+1}{(x^2-1)^2}\). The numerator \(x^2+1\) is always positive, and the squared denominator is positive on the domain. Therefore \(h'(x)<0\) throughout its domain. The function decreases on each branch interval \((-\infty,-1)\), \((-1,1)\), and \((1,\infty)\), and has no critical points.
4. Concavity and inflection. Differentiating \(h'\) gives \(h''(x)=\frac{2x(x^2+3)}{(x^2-1)^3}\). Because \(x^2+3>0\), the sign changes at \(x=-1,0,1\). Testing the intervals shows \(h''<0\) on \((-\infty,-1)\), \(h''>0\) on \((-1,0)\), \(h''<0\) on \((0,1)\), and \(h''>0\) on \((1,\infty)\). At \(x=0\), the graph is continuous and the concavity changes, so \((0,0)\) is an inflection point. The asymptotes \(x=-1\) and \(x=1\) cannot be inflection points because the function is not defined there.
Check. Applying the quotient rule to \(x/(x^2-1)\) gives the displayed first derivative. Differentiating that expression gives the displayed second derivative. The three branch intervals and the limits at their boundaries agree with the signs of \(h'\): each branch decreases from its left-end behavior toward its right-end behavior.
A Common Mistake: Canceling Without Checking the Limit
Consider \(r(x)=\frac{x-1}{(x-1)^2}\), whose original domain excludes \(x=1\). Canceling one factor gives \(r(x)=\frac{1}{x-1}\) for \(x\ne1\). A mistaken conclusion would be: “A factor canceled, so there is a hole at \(x=1\).” But \(\lim_{x\to1^-}r(x)=-\infty\) and \(\lim_{x\to1^+}r(x)=+\infty\). The graph has a vertical asymptote, not a hole. The correction is to reduce the expression and then check the limit: a finite limit gives a hole; unbounded behavior gives a vertical asymptote.
Putting the Sketch Together
A complete analysis is more than a list of derivative results. Use the information together: plot intercepts and any hole; draw vertical and end-behavior asymptotes; then trace each branch according to its one-sided limits, increasing or decreasing behavior, and concavity. Check that an extremum or inflection point lies in the domain. Do not connect branches across an asymptote or a hole, and do not label a missing input as an inflection point simply because concavity differs on either side.
Check Your Understanding
For each question, use the domain and interval structure of the function as part of your reasoning.
- For \(f(x)=\frac{x^2-4}{x-2}\), what is the domain, and what point is missing from the graph?
- For \(g(x)=\frac{x-1}{(x-1)^2}\), is \(x=1\) a hole or a vertical asymptote? Justify the classification using a limit.
- What end-behavior asymptote is suggested when the numerator and denominator have equal degree? What information determines its value?
- Why can a vertical asymptote not be an inflection point?
- For a rational function with vertical asymptotes at \(x=-2\) and \(x=3\), which intervals must be analyzed separately for monotonicity and concavity?
