Why a Hole Needs Special Attention
A rational function can have a formula with a factor that appears in both its numerator and denominator. Canceling that factor gives a simpler expression, but it does not put the excluded input back into the function’s domain. The graph of the simplified expression may therefore have one point missing: a hole.
That missing point is easy to overlook. It can even lie where the simplified curve crosses an axis or changes direction. A complete sketch must keep track of the original domain as well as the reduced formula. It must also distinguish a hole from a vertical asymptote, which represents unbounded behavior rather than one missing finite point.
Definition: A Removable Discontinuity and Its Hole
Suppose a rational function \(f\) has a numerator and denominator with a common factor \(x-c\). After canceling the common factor, write the reduced expression as \(g(x)\). The original function is still undefined at \(x=c\), because its original denominator was zero there. If \(g\) is defined at \(c\), the limit of \(f(x)\) as \(x\) approaches \(c\) is \(g(c)\). The graph has a hole at \((c,g(c))\).
By contrast, if a real zero of the denominator remains after all common factors are canceled, the reduced rational expression is unbounded near that zero. That input gives a vertical asymptote, not a hole. The hole is marked as an open circle at its limiting coordinate; a vertical asymptote is a line the graph approaches while its values grow without bound in magnitude.
A Justification for the Hole’s Coordinates
The hole’s height comes from the reduced expression, not from substituting into the original expression. To see why, suppose the original function can be written as
where \(Q(c)\ne 0\). For every \(x\ne c\) where the expression is defined, cancellation gives
Since \(Q(c)\ne 0\), the reduced rational expression \(g\) is defined and continuous at \(c\). Thus its limit there equals its value there:
The equality of \(f\) and \(g\) holds near \(c\) except at \(c\) itself, which is enough to determine the limit. But \(f(c)\) remains undefined. That is precisely why the graph has a hole rather than an ordinary point at \((c,g(c))\).
A Reliable Sketching Sequence
For a rational function with common factors, keep the original expression visible while simplifying. The following sequence prevents a canceled factor from being mistaken for a restored input.
Find the real inputs that make the original denominator zero. Each remains excluded, even if its factor later cancels.
Cancel common factors, then evaluate the reduced expression at each canceled input to find the hole’s height.
Use the reduced denominator to identify vertical asymptotes. Find axis crossings from the reduced expression, but discard any candidate input excluded from the original domain.
Determine where the function increases or decreases, splitting intervals at holes, vertical asymptotes, and valid points where the derivative is zero or undefined.
Draw each branch with its intercepts and asymptotic behavior, and mark every hole as an open circle rather than a filled point.
Here, a hole is an interval boundary for organizing a sign chart because the function is not defined there. It is not a point on the graph, and a derivative value at that input cannot make it a critical point of the original function.
Worked Example: A Hole and a Vertical Asymptote
Worked Example: A Hole and a Vertical Asymptote
Sketch the important features of
1. Domain and hole. The original denominator is zero at \(x=2\) and \(x=3\), so both inputs are excluded. Cancel \(x-2\) to obtain the reduced expression
The reduced expression at \(x=2\) is
Therefore the hole is \((2,-3)\). Checking the limit gives the same result: for \(x\ne2,3\), \(f(x)=g(x)\), so \(\lim_{x\to2}f(x)=\lim_{x\to2}(x+1)/(x-3)=-3\).
2. Vertical asymptote and end behavior. The reduced denominator is zero at \(x=3\), while the reduced numerator there is \(4\), not zero. For \(x\) close to \(3\), the numerator is positive. The denominator is negative on the left of \(3\) and positive on the right. Hence
Since numerator and denominator have the same degree and leading coefficient, the horizontal asymptote is \(y=1\).
3. Intercepts. The reduced numerator is zero at \(x=-1\), which is in the original domain, so the \(x\)-intercept is \((-1,0)\). At \(x=0\), \(f(0)=1/(-3)=-1/3\), so the \(y\)-intercept is \((0,-1/3)\).
4. Derivative signs. Differentiate the reduced expression on its domain:
The denominator is positive for every \(x\ne3\), so \(g'(x)<0\) throughout the domain. The graph decreases on each interval \((-\infty,2)\), \((2,3)\), and \((3,\infty)\). The hole separates the first two intervals, even though there is no vertical asymptote at \(x=2\).
Check. Substitution into the reduced expression confirms the hole height: \(g(2)=-3\). The derivative calculation simplifies to \(-4/(x-3)^2\), which is negative wherever it is defined. The sketch must show the open circle at \((2,-3)\), not a filled point.
Intercepts and the Original Domain
To find an \(x\)-intercept, set the reduced numerator equal to zero, then check that the candidate input belongs to the original domain. A numerator zero at an excluded input does not give an intercept: there is no function value there. The same domain check applies when evaluating the function at \(x=0\) to seek a \(y\)-intercept.
This check matters when a canceled factor also makes the reduced expression zero. In that case the reduced curve passes through the hole’s limiting coordinate, but the original graph does not contain that point. An open circle is still required, and it is not an axis crossing of the original function.
Worked Example: The Hole Is Not an Intercept
Worked Example: The Hole Is Not an Intercept
Analyze
1. Domain and hole. The original denominator vanishes at \(x=1\) and \(x=3\), so those inputs are excluded. Cancel one factor \(x-1\):
At the canceled input, the reduced expression has value
The hole is therefore \((1,0)\). Although this coordinate lies on the \(x\)-axis, it is not an \(x\)-intercept of \(f\), because \(f(1)\) is undefined.
2. Vertical asymptote and slant asymptote. The remaining denominator zero is \(x=3\). The reduced numerator there is \((3-1)^2=4>0\), so the function tends to \(-\infty\) from the left and \(+\infty\) from the right. Division gives
As \(x\) tends to either infinity, \(4/(x-3)\) tends to zero. Thus the slant asymptote is \(y=x+1\).
3. Intercepts. The only zero of the reduced numerator is \(x=1\), which is excluded. There is no \(x\)-intercept. Since \(x=0\) is allowed, the \(y\)-intercept is
4. Derivative signs. Differentiate the reduced expression using its quotient rule, then factor:
The denominator is positive away from \(x=3\). The numerator is positive for \(x<1\), negative for \(1<x<5\), and positive for \(x>5\). Accounting for the exclusions, the function increases on \((-\infty,1)\), decreases on \((1,3)\) and \((3,5)\), then increases on \((5,\infty)\). At \(x=5\), a valid derivative-zero point, the function has value \(g(5)=16/2=8\), and the derivative changes from negative to positive, giving a local minimum at \((5,8)\). The derivative is zero at \(x=1\) in the reduced formula, but \(x=1\) is excluded, so it is not a point of the original graph.
Check. The hole limit is \(g(1)=0\), while the original function is undefined at \(1\), confirming that \((1,0)\) must be open. Also \(g(5)=8\), and the factored derivative changes sign from negative to positive at \(5\), confirming the local minimum.
Worked Example: A Hole Away from the Intercepts
Worked Example: A Hole Away from the Intercepts
Sketch the main features of
1. Domain and hole. The original denominator is zero at \(x=1\) and \(x=-1\). After canceling \(x-1\), the reduced expression is
The canceled input gives the limiting value
Thus there is a hole at \((1,3/2)\), which is not on either axis.
2. Asymptotes and intercepts. The remaining denominator zero is \(x=-1\). The reduced numerator there is \(1\); the denominator changes from negative to positive, so \(f(x)\) tends to \(-\infty\) from the left and \(+\infty\) from the right. The degrees and leading coefficients match, giving horizontal asymptote \(y=1\). The reduced numerator is zero at \(x=-2\), an allowed input, so the \(x\)-intercept is \((-2,0)\). At \(x=0\), \(f(0)=2\), giving the \(y\)-intercept \((0,2)\).
3. Derivative signs. The reduced expression can be written \(g(x)=1+1/(x+1)\). Therefore
wherever it is defined. The graph decreases on \((-\infty,-1)\), \((-1,1)\), and \((1,\infty)\), with a vertical asymptote at \(-1\) and a hole at \(1\).
Check. The reduced value at the excluded input is \(g(1)=3/2\), confirming the open circle. The intercepts also verify directly: \(g(-2)=0\) and \(g(0)=2\).
Common Mistake: Treating Cancellation as a Domain Change
A frequent error is to cancel a factor and then claim that the original function is defined at the canceled input. For example, from
one might write \(f(x)=(x+1)/(x-3)\) and then calculate \(f(2)=-3\). The calculation gives the value of the reduced expression at \(2\), not the value of the original function. The original denominator is zero at \(2\), so \(f(2)\) does not exist. The correction is to say that the limit is \(-3\) and mark an open circle at \((2,-3)\).
A related mistake is to call every original denominator zero a vertical asymptote. The cancellation check must come first. A canceled factor with a finite reduced value gives a hole; a denominator zero that remains in the reduced expression gives the vertical asymptote. Keeping these features distinct makes the sketch agree with both the formula and the function’s domain.
Putting the Sketch Together
A rational-function sketch with holes should display the graph’s actual domain, not merely the shape of its reduced expression. Use the reduced formula to calculate limiting values, intercepts, derivatives, and asymptotic behavior, while retaining every exclusion from the original denominator. At a canceled input, draw an open circle at the reduced expression’s value. At an uncanceled denominator zero, draw the vertical asymptote and show the appropriate one-sided behavior.
Finally, use the derivative sign separately on each interval in the domain. An open circle can interrupt an otherwise smooth-looking branch, and the derivative of the reduced expression at that input does not fill it in. With these checks, a single sketch can represent the formula’s algebra, its domain, and its local and global behavior consistently.
Check Your Understanding
For each question, keep the original domain in view as you reason about the reduced expression.
- If a factor \(x-c\) cancels and the reduced expression has value \(L\) at \(c\), what point should be marked on the graph, and why is it not a filled point?
- What distinguishes a hole from a vertical asymptote after common factors have been canceled?
- Why must a candidate zero of the reduced numerator be checked against the original domain before calling it an \(x\)-intercept?
- If the reduced derivative is zero at a hole’s input, does that input become a critical point of the original function? Explain.
- For \(f(x)=\frac{(x-1)(x+2)}{(x-1)(x+1)}\), state the hole, the vertical asymptote, and the sign of the derivative on each interval of the domain.
