From Algebraic Features to a Sketch
A rational function can have several distinct pieces of graph information encoded in its formula. Its denominator determines which inputs are excluded; its numerator helps locate zeros; cancellation can reveal holes; and the degrees of the polynomials, or polynomial division, describe end behavior. Bringing these features together gives a useful sketch without requiring a point-by-point table.
The crucial habit is to keep two expressions in view: the original function, which determines the domain, and a reduced expression, which makes the remaining zeros and asymptotes easier to identify. Simplifying does not put excluded inputs back into the domain.
A Reliable Assembly Process
Start by factoring the numerator and denominator and recording every real zero of the original denominator as an excluded input. Then cancel common factors to obtain a reduced expression. The cancelled factors help locate possible holes, while any denominator factors that remain help locate vertical asymptotes.
Once the domain and reduced expression are clear, identify the real zeros of the reduced numerator that are in the domain. Use the degrees or polynomial division to find the horizontal or slant asymptote, when one exists. Finally, divide the real line at every zero and excluded input, and determine the sign of the function on each resulting interval. These features tell you where the graph can lie, where it is missing, and how it behaves at important boundaries.
Set the original denominator equal to zero and exclude every real solution.
Cancel common factors to expose the simplified formula while keeping the original domain restrictions.
Check the reduced numerator and denominator at the relevant real inputs.
Use degree comparison or division for an end asymptote, then test signs between the key inputs.
Plot the features and indicate whether each branch lies above or below the \(x\)-axis on its intervals.
Holes and Vertical Asymptotes Must Be Distinguished
Suppose cancellation produces a reduced expression \(N(x)/D(x)\), while an input \(a\) remains excluded because the original denominator was zero there. If \(D(a)\ne 0\), the reduced expression has a finite value at \(a\). The original graph has a hole at \((a,N(a)/D(a))\): the reduced expression supplies the height of the missing point, but the original function is still undefined there.
If instead \(D(a)=0\) and \(N(a)\ne 0\), the reduced expression has a vertical asymptote at \(x=a\). A factor may have been canceled only partially, so an excluded input does not automatically give a hole. When the reduced denominator still vanishes, cancellation has not removed the unbounded behavior.
Worked Example: A Hole, a Zero, and a Vertical Asymptote
Worked Example: A Hole, a Zero, and a Vertical Asymptote
Sketch the main features of \(f(x)=\frac{(x-2)(x+1)}{(x-2)(x-3)}\).
1. Domain and reduction. The original denominator is zero when \(x=2\) or \(x=3\), so the domain is \(\mathbb{R}\setminus\{2,3\}\). For inputs in this domain, canceling \(x-2\) gives \(f(x)=\frac{x+1}{x-3}\). The excluded inputs remain excluded.
2. Hole and vertical asymptote. At \(x=2\), the reduced denominator is \(2-3=-1\ne0\), so this is a hole. Its height is \(\frac{2+1}{2-3}=\frac{3}{-1}=-3\), giving the hole \((2,-3)\). At \(x=3\), the reduced denominator is zero and the reduced numerator is \(3+1=4\ne0\), so \(x=3\) is a vertical asymptote.
3. Zero and horizontal asymptote. The reduced numerator is zero at \(x=-1\), and the denominator there is \(-1-3=-4\ne0\). Thus the graph crosses the \(x\)-axis at \((-1,0)\). The reduced numerator and denominator have the same degree and leading coefficient \(1\), so the horizontal asymptote is \(y=1\).
4. Interval signs and behavior. The critical inputs divide the real line into \((-\infty,-1)\), \((-1,2)\), \((2,3)\), and \((3,\infty)\). Using \(\frac{x+1}{x-3}\), the sign is positive on the first interval, negative on the second and third, and positive on the fourth. For example, at \(x=0\), \(f(0)=1/(-3)=-1/3\), confirming the negative sign on \((-1,2)\). Near \(x=3\), the numerator is positive; the denominator is negative to the left and positive to the right. Therefore \(f(x)\) tends to \(-\infty\) from the left and \(+\infty\) from the right.
5. Check. At an allowed input such as \(x=0\), the original formula gives \(\frac{(-2)(1)}{(-2)(-3)}=-2/6=-1/3\), matching the reduced formula. The value \(x=2\) is still excluded from the original formula, even though the reduced formula gives \(-3\) there. This confirms the hole’s location without incorrectly restoring that input to the domain.
Why Interval Signs Give Branch Placement
To determine the sign, list the real zeros of the numerator and the real zeros of the denominator. They divide the domain into intervals. On any one such interval, neither the numerator nor the denominator is zero, so the function cannot change from positive to negative without passing through a zero or an excluded input. A test value in each interval therefore determines whether that entire branch is above or below the \(x\)-axis.
Factored expressions make the sign test especially direct: determine the sign of each factor at the chosen test value, multiply the numerator signs, multiply the denominator signs, and divide the results. A zero at an endpoint is not part of an open interval, and an excluded input is never a point on the graph. If a zero has even multiplicity, its factor does not change sign across that zero; if it has odd multiplicity, its factor does. The interval test remains a dependable way to check the result.
Worked Example: Partial Cancellation Leaves an Asymptote
Worked Example: Partial Cancellation Leaves an Asymptote
Analyze \(g(x)=\frac{x-1}{(x-1)^2(x+2)}\), paying particular attention to \(x=1\).
1. Domain and reduction. The original denominator vanishes at \(x=1\) and \(x=-2\), so both inputs are excluded. Cancel one factor \(x-1\) to obtain \(g(x)=\frac{1}{(x-1)(x+2)}\), but keep the original exclusions. The denominator still vanishes at both excluded inputs.
2. Asymptotes and zeros. The reduced numerator is \(1\), so there are no zeros. At \(x=1\), the reduced denominator is zero while the numerator is \(1\ne0\); therefore \(x=1\) is a vertical asymptote, not a hole. At \(x=-2\), the reduced denominator is also zero and the numerator is nonzero, so \(x=-2\) is another vertical asymptote. Since the numerator has degree \(0\) and the denominator degree \(2\), the horizontal asymptote is \(y=0\).
3. Signs and one-sided behavior. The intervals are \((-\infty,-2)\), \((-2,1)\), and \((1,\infty)\). For \(x=-3\), the denominator is \((-4)(-1)=4\), so \(g(-3)=1/4>0\). For \(x=0\), it is \((-1)(2)=-2\), so \(g(0)=-1/2<0\). For \(x=2\), it is \((1)(4)=4\), so \(g(2)=1/4>0\). Near \(-2\), the factor \(x-1\) is negative: the product \((x-1)(x+2)\) is positive on the left and negative on the right, giving \(+\infty\) from the left and \(-\infty\) from the right. Near \(1\), \(x+2\) is positive, so the product is negative on the left and positive on the right, giving \(-\infty\) from the left and \(+\infty\) from the right.
4. Check. At \(x=0\), the original expression is \(\frac{-1}{(-1)^2(2)}=-1/2\), matching the reduced expression. At \(x=1\), the original denominator is zero and the reduced denominator \((1-1)(1+2)\) is also zero. There is no finite reduced value to assign as a hole height, confirming the vertical asymptote.
Horizontal and Slant Asymptotes
For end behavior, compare the degrees of the reduced numerator and denominator. If the numerator degree is less than the denominator degree, the horizontal asymptote is \(y=0\). If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. When the numerator degree is exactly one more than the denominator degree, polynomial division gives a linear quotient; its graph is the slant asymptote. In each case the asymptote describes behavior as \(x\) becomes large in magnitude, not a line the graph must avoid crossing.
For a slant asymptote, write the function as a linear quotient plus a remainder divided by the denominator. The remainder term tends to zero as \(x\) tends to either end of the real line, which explains why the graph approaches the quotient line.
Worked Example: A Repeated Zero and a Slant Asymptote
Worked Example: A Repeated Zero and a Slant Asymptote
Sketch the main features of \(h(x)=\frac{(x-2)(x-3)^2}{(x-2)(x-1)}\).
1. Domain, reduction, and hole. The original denominator is zero at \(x=2\) and \(x=1\), so these inputs are excluded. Canceling \(x-2\) gives \(h(x)=\frac{(x-3)^2}{x-1}\) on the original domain. At \(x=2\), the reduced denominator is \(1\ne0\), and the reduced value is \(\frac{(2-3)^2}{2-1}=1\). Thus there is a hole at \((2,1)\). At \(x=1\), the reduced denominator is zero while the numerator is \((1-3)^2=4\ne0\), so \(x=1\) is a vertical asymptote.
2. Zero and division. The reduced numerator is zero at \(x=3\), with multiplicity two, and the denominator there is \(2\ne0\). Thus \((3,0)\) is a zero. Divide \(x^2-6x+9\) by \(x-1\): the quotient is \(x-5\), because \((x-1)(x-5)=x^2-6x+5\); the remainder is \(4\), since \(x^2-6x+9=(x^2-6x+5)+4\). Therefore \(h(x)=x-5+\frac{4}{x-1}\), and the slant asymptote is \(y=x-5\).
3. Signs and branch behavior. The numerator \((x-3)^2\) is positive except at \(x=3\), where it is zero. The denominator \(x-1\) is negative for \(x<1\) and positive for \(x>1\). Hence the function is negative on \((-\infty,1)\), positive on \((1,3)\) and \((3,\infty)\), and undefined at \(x=2\) within the latter two intervals. The sign remains positive on both sides of \(x=3\), as expected for a double zero. Near \(x=1\), the numerator approaches \(4>0\), so the function tends to \(-\infty\) from the left and \(+\infty\) from the right. The remainder term \(\frac{4}{x-1}\) is negative for \(x<1\) and positive for \(x>1\), so the graph lies below the slant line on the left and above it on the right.
4. Check. At \(x=0\), the reduced formula gives \(\frac{9}{-1}=-9\). The division form gives \(0-5+\frac{4}{-1}=-5-4=-9\), confirming the division. At \(x=2\), the reduced formula gives \(1\), but the original denominator is zero; this verifies that the point \((2,1)\) is missing. At \(x=3\), the reduced formula gives \(0\), confirming the stated zero.
A Common Mistake: Canceling Away a Domain Restriction
A tempting but incorrect conclusion is that canceling \(x-1\) from \(\frac{x-1}{(x-1)^2}\) makes the function defined at \(x=1\). The cancellation gives \(\frac{1}{x-1}\), but this reduced expression is still undefined at \(x=1\). The correct domain excludes \(1\), and the reduced denominator still vanishes there while the numerator does not, so \(x=1\) is a vertical asymptote, not a hole.
Even when cancellation does leave a finite reduced value at an excluded input, that value identifies the height of a hole; it does not make the original function defined there. Always check the original denominator first, then inspect the reduced denominator at each excluded input.
Turning the Features into a Sketch
Mark the horizontal or slant asymptote as a guide for the graph’s ends, and draw a dashed vertical line at each vertical asymptote. Plot each zero, and use an open circle for each hole at its computed height. Then label the intervals with the signs of the function. The signs indicate whether each branch is above or below the \(x\)-axis, while the one-sided behavior near an asymptote indicates which direction the branch goes there.
These features constrain the sketch, but they do not specify every bend or turning point. A sketch based only on algebraic features should show the known information clearly without claiming an unsupported exact shape between those features.
Check Your Understanding
For each question, distinguish carefully between the original domain and the reduced formula.
- If a factor cancels but the reduced denominator is still zero at that input, what feature should you check for, and what additional condition confirms it?
- For \(r(x)=\frac{(x+1)(x-2)}{(x-2)(x-4)}\), identify the original domain, the hole, and the vertical asymptote.
- Why is a rational function’s sign constant on an interval containing no numerator zero or denominator zero?
- What is the horizontal asymptote when the reduced numerator has degree \(2\) and the denominator has degree \(3\)?
- For \(s(x)=\frac{(x-1)^2}{x+2}\), does the function change sign across its zero at \(x=1\)? Explain using the factor’s multiplicity.
