Algebra 2 · Precalculus · Rational functions · Polynomials worksheets · Quadratics worksheets
Graphing Rational Functions Worksheet and Feature Finding Guide
Start with the denominator before you cancel anything. Its zeros tell you where the function is undefined, and the reduced denominator tells you which of those values are vertical asymptotes and which are holes. Four worked examples show the process, followed by 24 practice problems with answers, three interactive tools, and a one-page reference.
Part of a nine guide unit. See all the rational expressions and rational functions worksheets, listed in a recommended order.
Worksheet and cheat sheet downloads
Two files. The worksheet holds all 24 practice problems plus a three problem challenge set, with blank grids to graph on and a five page answer key that gives every feature and a completed graph for each function. The cheat sheet is one page for a binder or a classroom wall.
Graphing rational functions, free PDFs
Printed at letter size, single sided, so students can work without the key in view.
Free to copy for your own classroom or your own kitchen table. If you want to share it, please link to this page rather than rehosting the files, so students always land on the current version.
The seven features to find
A rational function is a quotient of two polynomials. Graphing one is a matter of finding a fixed list of features and then drawing the only curve that fits them. The list does not change from problem to problem, which is what makes this topic learnable in a single sitting.
| Feature | Where it comes from | What it does to the graph |
|---|---|---|
| Excluded values | Zeros of the original denominator | The graph has no point at these x-values |
| Holes | Factors that cancel completely | A single missing point, drawn as an open circle |
| Vertical asymptotes | Zeros of the reduced denominator | The curve runs to infinity beside a vertical line |
| x-intercepts | Zeros of the reduced numerator that the domain allows | Points where the curve meets the x-axis |
| y-intercept | Evaluate at x = 0, if 0 is allowed | One point on the y-axis, at most |
| End behavior | Degree of the numerator against the denominator | A horizontal line, a slant line, or a curve |
| Sign on each interval | One test value per interval | Fixes whether each piece sits above or below the x-axis |
Two of those rows are where I see errors in sessions. Vertical asymptotes come from the reduced denominator, not the original one, and excluded values come from the original denominator, not the reduced one. Reading either off the wrong form is what turns a hole into an asymptote or loses a hole entirely.
Factoring is the prerequisite for all of it. If the factoring is what slows you down, work through the factoring polynomials flowchart guide first, and simplifying rational expressions after that.
The six step routine
I teach this in a fixed order. The order matters because step 2 depends on the original form and step 3 depends on the reduced form, and doing them out of sequence is how the two get confused.
- Factor the numerator and the denominator completely. Leave both in factored form for the rest of the problem. Expanding hides everything you are about to look for.
- Write the excluded values from the original denominator. Set each denominator factor equal to zero. Write this list before cancelling anything, because cancelling erases the evidence.
- Cancel common factors, then read the reduced denominator. Each zero of the reduced denominator is a vertical asymptote. Each excluded value that is not a zero of the reduced denominator is a hole.
- Find the intercepts. Set the reduced numerator equal to zero for the x-intercepts, and discard any root that is an excluded value. Evaluate the function at x = 0 for the y-intercept, if 0 is allowed.
- Compare the degrees for end behavior. Numerator degree against denominator degree, using the reduced form. This gives a horizontal asymptote, a slant asymptote, or neither.
- Build a sign chart and test one value per interval. Cut the number line at every real zero of the reduced numerator and the reduced denominator. Test a value from each interval to place the branches above or below the axis.
The excluded values come from the denominator as it was handed to you. The vertical asymptotes come from the denominator after cancelling. A factor that cancels completely leaves an excluded value with no asymptote, which is a hole. Write the excluded list first and it stays available for the rest of the problem.
Hole or vertical asymptote
An excluded value produces one of two things on the graph, and the difference is decided by the reduced denominator.
| Function | What cancels | At that x-value |
|---|---|---|
| x2 − 4x − 2 | (x − 2) cancels completely | Hole at (2, 4). The reduced function is x + 2, so the graph is the line y = x + 2 with one point missing. |
| x − 2(x − 2)2 | (x − 2) cancels once, one copy stays below | Vertical asymptote at x = 2. The reduced function is 1 over (x − 2), and the reduced denominator is still zero at 2. |
| x + 1x − 2 | Nothing cancels | Vertical asymptote at x = 2. The numerator is 3 there and the denominator is 0. |
The middle row is the one worth slowing down for. A factor appearing in both the numerator and the denominator does not guarantee a hole. Cancel first, then read the reduced denominator, and let that decide.
Substitute the excluded value into the reduced function to get the height of the missing point. For (x2 − 4) over (x − 2), the reduced function x + 2 gives 2 + 2 = 4, so the hole is at (2, 4). An answer that names the hole without its y-coordinate is usually marked incomplete.
End behavior and the degree test
End behavior is decided by degree, comparing the reduced numerator against the reduced denominator. Write n for the numerator degree and d for the denominator degree.
| Degrees | End behavior | How to find it | Example |
|---|---|---|---|
| n < d | Horizontal asymptote y = 0 | The denominator grows faster, so the quotient shrinks toward zero | 2xx2 − 9 approaches y = 0 |
| n = d | Horizontal asymptote at the ratio of leading coefficients | Divide the leading coefficient of the numerator by the leading coefficient of the denominator | 3x2 − 2xx2 + 1 approaches y = 3 |
| n = d + 1 | Slant asymptote | Divide the numerator by the denominator. The quotient is the line, and the remainder is discarded | x2 + 2x − 1 approaches y = x + 1 |
| n > d + 1 | Neither | The quotient is a polynomial of degree 2 or more, and the graph follows that curve | x3x − 1 follows a parabola |
The slant case is polynomial long division, the same procedure as in the polynomial division worksheet. Divide, keep the quotient, discard the remainder. For (x2 + 2) divided by (x − 1) the quotient is x + 1 with remainder 3, so the slant asymptote is y = x + 1.
Use the reduced form for the degree test. In (x3 − 4x) over (x2 − 2x), the original degrees are 3 and 2, which suggests a slant asymptote. After cancelling, the function is x + 2 with holes at x = 0 and x = 2. There is no asymptote at all, and the graph is a straight line with two points removed.
Meeting the horizontal or slant asymptote
A vertical asymptote is a line the graph never touches. A horizontal or slant asymptote is different: it describes what happens far out to the left and right, and it says nothing about the middle of the graph. A curve is free to cross it, touch it, or meet it several times.
To find where a graph meets a horizontal asymptote y = k, solve f(x) = k. Written out, that means solving
N(x) − k · D(x) = 0
using the reduced numerator N and reduced denominator D, then discarding any root that is an excluded value. For a slant asymptote, replace k with the linear quotient and solve the same way.
| At a solution of N(x) − k · D(x) = 0 | What the graph does |
|---|---|
| Odd multiplicity | Crosses the asymptote and continues to the other side |
| Even multiplicity | Touches the asymptote and returns to the same side |
| No real solutions | The graph never reaches the asymptote |
| Every coefficient cancels | The graph coincides with that line wherever it is defined |
A graph can meet a horizontal asymptote more than once even when the degrees are equal. Cancelling the leading terms drops the degree, so the equation left over can have several roots: (x4 + x2) over (x4 + 1) meets y = 1 at both x = −1 and x = 1.
A repeated root does not mean the graph touches without crossing. The test is odd against even, not single against repeated. In x3 over (x4 + 1), the root at the origin has multiplicity 3, and the graph crosses y = 0 there.
Touching is not limited to the x-axis. The graph of (x4 + x2 − 2x + 2) over (x4 + 1) touches y = 1 at the point (1, 1) and stays above it on both sides.
The sign chart, and what it does not tell you
Once the features are known, one question remains: which side of the x-axis does each piece of the curve sit on? A sign chart answers it. The same tool appears in polynomial inequalities, and the procedure is identical.
Cut the number line at every real zero of the reduced numerator and every real zero of the reduced denominator. Pick one value inside each interval, substitute it into the reduced function, and record whether the result is positive or negative.
Cut at the numerator's zeros, not at the x-intercepts
Those two lists are usually the same, and when they differ the sign chart is what catches it. A zero of the numerator that sits at an excluded value is not an x-intercept, because the function has no point there. It still marks a place where the sign can change.
Here f(x) = x2 over x reduces to y = x with a hole at the origin. There are no x-intercepts, because x = 0 is excluded. The function is still negative to the left of 0 and positive to the right. Cutting only at the intercepts would give one interval and one sign, which is wrong.
No test value may be an excluded x. Substituting an excluded value gives no answer at all, not a sign. Pick a different number in the same interval.
A cut point is not always a sign change. At a factor of even multiplicity the sign stays the same on both sides. In (x + 1)2 over ((x − 2)(x + 3)), the chart cuts at −1, and the function is negative on both sides of it.
The number of intervals is not the number of branches. One vertical asymptote splits the curve into two pieces, and the sign chart for that function usually has three or four intervals. Count pieces from the asymptotes and count intervals from the cut points.
Branch direction beside a vertical asymptote
To decide whether a branch runs up or down beside a vertical asymptote, evaluate the whole reduced function at a number just to the left of the asymptote, and again just to the right. Positive means that side runs up, negative means it runs down.
Use the whole function, not the numerator alone. The other factors in the denominator carry their own signs, and leaving them out can flip the result. In (x2 − 9) over (x2 − 4) just to the left of x = 2, the numerator is negative and the denominator is also negative, so the quotient is positive and that branch runs up.
Worked example 1. Two vertical asymptotes and a horizontal asymptote
Graph f(x) = x2 − 1x2 − 4
Step 1. Factor
f(x) = (x − 1)(x + 1)(x − 2)(x + 2)
Step 2. Excluded values
The denominator factors are (x − 2) and (x + 2), so x = 2 and x = −2 are excluded. Write them down now.
Step 3. Cancel, then read the reduced denominator
Nothing cancels. The numerator factors are (x − 1) and (x + 1), and neither appears below. The reduced denominator is still (x − 2)(x + 2), so both excluded values are vertical asymptotes: x = 2 and x = −2. There are no holes.
Step 4. Intercepts
Set the numerator to zero: (x − 1)(x + 1) = 0 gives x = 1 and x = −1. Neither is excluded, so both are x-intercepts, at (1, 0) and (−1, 0). For the y-intercept, f(0) = (−1) over (−4) = 1/4, giving (0, 1/4).
Step 5. End behavior
Both degrees are 2, so the horizontal asymptote is the ratio of the leading coefficients, which is 1 over 1. The asymptote is y = 1. To check whether the graph meets it, solve (x2 − 1) − 1(x2 − 4) = 0, which simplifies to 3 = 0. There is no solution, so this graph never reaches y = 1.
Step 6. Sign chart
Cut at −2, −1, 1, and 2. Test one value from each of the five intervals, avoiding −2 and 2.
| Interval | x < −2 | −2 to −1 | −1 to 1 | 1 to 2 | x > 2 |
|---|---|---|---|---|---|
| Test value | −3 | −1.5 | 0 | 1.5 | 3 |
| f(test) | 8/5 | −5/7 | 1/4 | −5/7 | 8/5 |
| Sign | + | − | + | − | + |
Five intervals, three pieces of curve. The two vertical asymptotes cut the graph into three branches, and the middle branch changes sign twice within itself, once at each x-intercept.
Reading the branch directions
Just left of x = −2 the function is positive, so that branch runs up. Just right of x = −2 it is negative, so the next branch starts at the bottom. The middle branch rises to the x-intercept at −1, arches to (0, 1/4), and falls back through (1, 0). Just left of x = 2 the function is negative and just right of it positive, which completes the picture.
Worked example 2. A hole and an asymptote together
Graph f(x) = x2 − 2x − 3x2 − 5x + 6
Step 1. Factor
f(x) = (x − 3)(x + 1)(x − 3)(x − 2)
Step 2. Excluded values
x = 3 and x = 2, both read off the original denominator before anything cancels.
Step 3. Cancel, then read the reduced denominator
The factor (x − 3) appears once above and once below, so it cancels completely. The reduced function is (x + 1) over (x − 2). The reduced denominator is zero only at x = 2, so that is the one vertical asymptote. The excluded value x = 3 has no asymptote, which makes it a hole.
The height of the hole comes from the reduced function: (3 + 1) over (3 − 2) = 4. The hole is at (3, 4).
Step 4. Intercepts
The reduced numerator is zero at x = −1, which is allowed, so (−1, 0) is the x-intercept. At x = 0, the reduced function gives 1 over (−2), so the y-intercept is (0, −1/2).
Step 5. End behavior
The reduced degrees are 1 and 1, so the horizontal asymptote is y = 1. Solving (x + 1) − 1(x − 2) = 0 gives 3 = 0, so the graph never meets it.
Step 6. Sign chart
Cut at −1 and 2. The excluded value 3 is not a cut point, because (x − 3) is no longer in the reduced function and the sign does not change across a hole in this case. Test values must still avoid x = 3.
| Interval | x < −1 | −1 to 2 | x > 2 |
|---|---|---|---|
| Test value | −2 | 0 | 4 |
| f(test) | 1/4 | −1/2 | 5/2 |
| Sign | + | − | + |
Notice that the open circle at (3, 4) sits on the right branch, above the horizontal asymptote. The hole is a missing point on an otherwise ordinary curve, so the rest of the branch is drawn exactly as if the hole were not there.
At x = 3 the answer is not only that the function is undefined. It is a hole, and it has coordinates. A response that says x = 3 is excluded and stops there is missing the part that gets graded.
Worked example 3. A slant asymptote
Graph f(x) = x2 + 2x − 1
Steps 1 to 3
The numerator x2 + 2 does not factor over the real numbers, and it has no real zeros. The denominator is zero at x = 1, which is the only excluded value. Nothing cancels, so x = 1 is a vertical asymptote and there are no holes.
Step 4. Intercepts
The numerator is never zero, so there are no x-intercepts. At x = 0, f(0) = 2 over (−1) = −2, so the y-intercept is (0, −2).
Step 5. End behavior by long division
The numerator degree is 2 and the denominator degree is 1, so the degree gap is exactly one and there is a slant asymptote. Divide:
x2 + 2 = (x − 1)(x + 1) + 3
The quotient is x + 1 and the remainder is 3. The slant asymptote is y = x + 1. As x grows, the remainder term 3 over (x − 1) shrinks toward zero and the curve settles onto that line.
To check for a meeting point, solve (x2 + 2) − (x + 1)(x − 1) = 0. The left side simplifies to 3, so there is no solution and the graph stays off the slant asymptote everywhere.
Step 6. Sign chart
The reduced numerator has no real zeros, so the only cut point is x = 1. The function is negative for x < 1 and positive for x > 1.
The remainder is positive, so the curve sits above the line y = x + 1 on the right branch and below it on the left branch. The two branches run off in opposite directions while both approaching the same slanted line.
Worked example 4. A graph that crosses its own asymptote
Graph f(x) = x2 − 2x + 2x2 + 1
Steps 1 to 4
Neither polynomial factors over the real numbers. The denominator x2 + 1 is never zero, so there are no excluded values, no holes, and no vertical asymptotes. The numerator x2 − 2x + 2 has discriminant 4 − 8 = −4, so it has no real zeros and the graph has no x-intercepts. At x = 0, f(0) = 2 over 1 = 2, so the y-intercept is (0, 2).
Step 5. End behavior, and where the curve meets it
Equal degrees, leading coefficients 1 and 1, so the horizontal asymptote is y = 1. Now solve f(x) = 1:
(x2 − 2x + 2) − 1(x2 + 1) = 0 → −2x + 1 = 0 → x = 1/2
One solution, with multiplicity 1, so the graph crosses y = 1 at the point (1/2, 1) and continues to the other side. Substituting confirms it: f(1/2) = (1/4 − 1 + 2) divided by (1/4 + 1) = (5/4) over (5/4) = 1.
Step 6. Sign chart
The reduced numerator and denominator have no real zeros, so there are no cut points at all. The function is positive for every real x, which one test value confirms: f(0) = 2.
This graph has a single unbroken piece. It rises from below y = 1 on the right, crosses the asymptote at x = 1/2, peaks near the y-intercept, and settles back toward y = 1 going left. A horizontal asymptote describes the far left and far right of a graph, and the crossing at x = 1/2 does not conflict with that.
Solving f(x) = k takes one line, and the answer is either a set of x-values or no solution. Students who skip it draw the curve approaching the asymptote from one side across the whole graph, which is wrong whenever a crossing exists.
Tool 1. Rational function feature finder
Enter numerator and denominator polynomials with whole number coefficients and degree 8 or less, or choose one of the nine examples. The tool cancels shared factors and reports the graph’s features in the order of the six step routine, with the sign chart underneath. Some higher-degree zeros are approximate, and those are labeled. Use it to check work after you have done a problem by hand.
Accepted input: whole number coefficients, x, and the symbols + − * ^ with parentheses. Write x2 as x^2 and 2x as 2x or 2*x. Put the numerator and the denominator in their own boxes rather than typing a slash.
Tool 2. Sign chart builder
This drill splits the sign chart into its two decisions. First name the cut points, which are the real zeros of the reduced numerator and the reduced denominator. Then choose a sign for each interval those cut points create. The tool scores the two stages separately, so you can see which half needs practice.
Stage 2. Choose the sign of the function on each interval. Substitute a value from inside the interval, and do not use a value the domain excludes.
Write fractions with a slash, such as 3/2 or 3 / 2. Order does not matter. A value that is both a numerator zero and a denominator zero is listed once.
Tool 3. Build a function from its features
This tool names a set of features and asks you to write a rational function that has them. It then checks your function against each requirement one at a time and reports which ones it meets.
There is no single right answer. Any function with the listed features passes, so the tool grades the features rather than comparing your factors to a stored answer. I use this exercise with a student who can find features correctly but still cannot say why a hole differs from an asymptote.
Tool 3 works on a projector as a warm up. Put one feature list on the board, take a function from a student, and run it. When it fails a requirement the tool says which one, and the discussion that follows is usually about the difference between the original denominator and the reduced denominator. That is the distinction this whole topic rests on.
Where my students lose points
A common mistake is to cancel correctly and then forget that the original denominator still restricts the domain. The six checks below catch that error and help place the branches correctly.
1. Reading the vertical asymptotes off the original denominator
A student factors, sees (x − 3) in the denominator, and writes down a vertical asymptote at x = 3 before checking whether that factor survives the cancelling. When it cancels, there is no asymptote there, only a hole, and the graph gets an extra vertical line that does not exist.
Write the excluded values first, then cancel, then draw a line under the reduced function and take the asymptotes only from what is below that line.
2. Assuming a cancelling factor always gives a hole
This is the reverse of the first error. A factor that appears above and below cancels, but if the denominator held more than one copy of it, a copy remains and the excluded value is still an asymptote.
Read the reduced denominator to decide. Cancelling on its own does not settle whether the excluded value is a hole or an asymptote.
3. Naming a hole without its coordinates
A student correctly identifies a hole at x = 3 and writes exactly that. On most keys the hole is worth a point only when it has a y-value attached, because finding the height requires substituting into the reduced function, which is the part being tested.
4. Testing the sign with the numerator alone
To decide whether a branch runs up or down beside a vertical asymptote, some students check only the numerator, on the reasoning that the denominator is heading to zero anyway. The sign of the denominator is what decides the direction, so leaving it out turns the answer into a guess.
Evaluate the reduced function, not one part of it. A number close to the asymptote on each side settles both branches in two substitutions.
5. Treating the horizontal asymptote as a barrier
Students carry the rule for vertical asymptotes across to horizontal ones and draw the curve approaching y = k from one side across the entire graph. A horizontal asymptote describes the far left and far right of the graph only, and the curve can cross it in the middle.
Solving f(x) = k is one line of algebra and it belongs on every problem that has a horizontal or slant asymptote. The answer is either a list of x-values or no solution, and both are useful.
6. Cutting the sign chart in the wrong places
Two versions of this. The first is cutting at the x-intercepts rather than at the zeros of the reduced numerator, which loses any zero that sits at an excluded value. The second is picking a test value that the domain excludes, which produces no number at all and then gets recorded as a sign anyway.
Cut at the zeros of the reduced numerator and the reduced denominator, then choose test values that the domain allows. If an interval is short and its only convenient value is excluded, pick a decimal inside it instead.
One page cheat sheet
The whole routine, sized for one side of paper. This is the same content as the printable PDF, written out here so it is searchable and readable on a phone.
The six steps
- Factor the numerator and the denominator completely.
- Excluded values from the original denominator.
- Cancel, then read the reduced denominator: its zeros are the vertical asymptotes. Any other excluded value is a hole.
- Intercepts. Reduced numerator = 0 for the x-intercepts, discarding excluded values. Evaluate at x = 0 for the y-intercept.
- End behavior by comparing degrees in the reduced form.
- Sign chart. Cut at every real zero of the reduced numerator and the reduced denominator, then test one allowed value per interval.
Hole or asymptote
| Factor cancels completely | Hole. Height from the reduced function. |
| A copy remains in the reduced denominator | Vertical asymptote, even though something cancelled. |
| Nothing cancels | Vertical asymptote. |
Read the reduced denominator. Cancelling alone does not decide it.
Degree test, using the reduced form
| n < d | Horizontal asymptote y = 0 |
| n = d | Horizontal asymptote y = ratio of leading coefficients |
| n = d + 1 | Slant asymptote: divide, keep the quotient, drop the remainder |
| n > d + 1 | Neither. The graph follows the polynomial quotient |
Does the graph meet that asymptote
Solve N(x) − k · D(x) = 0 with the reduced N and D, then discard any excluded root. For a slant asymptote, put the linear quotient in place of k.
- Odd multiplicity: the graph crosses and continues to the other side.
- Even multiplicity: the graph touches and returns to the same side.
- No real roots: the graph never reaches it.
- Everything cancels: the graph lies on that line wherever it is defined.
Sign chart rules
- Cut at the zeros of the reduced numerator and reduced denominator, not at the x-intercepts.
- No test value may be an excluded x.
- Even multiplicity means no sign change at that cut point.
- Intervals are not branches. Count branches from the vertical asymptotes.
Branch direction
Substitute a number just left of the asymptote and again just right, into the whole reduced function. Positive means that side runs up, negative means it runs down. The other denominator factors carry signs, so the numerator alone is not enough.
Three answers that look wrong and are not
- A function with a hole and no asymptote at all, when the denominator cancels away entirely.
- A graph that crosses its own horizontal asymptote, once or several times.
- A sign chart cut point where the sign does not change.
Worked in one line
x2 − 2x − 3x2 − 5x + 6 = (x − 3)(x + 1)(x − 3)(x − 2) → excluded 3, 2 → reduced x + 1x − 2 → asymptote x = 2, hole (3, 4), x-intercept (−1, 0), y-intercept (0, −1/2), asymptote y = 1, signs + − +
24 practice problems
Work in order. Part A gives you the factored form so the reading of features becomes automatic. Part B adds the factoring. Part C is sign charts and branch direction. Part D is end behavior, slant asymptotes, and the question of whether the graph meets its asymptote. Answers for each part are behind the button under that part.
For every problem, state the excluded values, any holes with their coordinates, the vertical asymptotes, all x-intercepts and the y-intercept if they exist, and the end behavior. Parts C and D then ask for one further step, named in the heading of each part. Some of these functions have no x-intercepts, some have no asymptote at all, and one has a sign change where there is no intercept, so a blank line in your answer is not evidence that you have gone wrong.
If a feature does not exist, write "none." The answers below cover the requested features. Completed graphs are in the printable answer key.
- 1.f(x) = x + 2x − 3
- 2.f(x) = x − 4(x + 1)(x − 5)
- 3.f(x) = 3xx + 2
- 4.f(x) = (x − 1)(x + 4)(x − 1)(x + 2)
- 5.f(x) = 5(x − 2)2
- 6.f(x) = x2 − 9x + 3
Show answers for Part A
- 1. Excluded: 3. Nothing cancels, so vertical asymptote x = 3. x-intercept (−2, 0). y-intercept (0, −2/3). Degrees 1 and 1, so horizontal asymptote y = 1. Solving f(x) = 1 gives 5 = 0, so the graph never meets it. Signs: + on x < −2, − from −2 to 3, + beyond 3.
- 2. Excluded: −1 and 5. Both are vertical asymptotes. x-intercept (4, 0). y-intercept (0, 4/5), since f(0) = (−4) over (1)(−5) = 4/5. Numerator degree 1 is less than denominator degree 2, so horizontal asymptote y = 0. The graph meets y = 0 at its x-intercept (4, 0). Signs: −, +, −, + across the four intervals.
- 3. Excluded: −2, a vertical asymptote. x-intercept (0, 0), which is also the y-intercept. Degrees 1 and 1, so horizontal asymptote y = 3. Solving 3x = 3(x + 2) gives 0 = 6, so it never meets that line. Signs: + on x < −2, − from −2 to 0, + beyond 0.
- 4. Excluded: 1 and −2. The factor (x − 1) cancels completely, so x = 1 is a hole. The reduced function is (x + 4) over (x + 2), so the height is (1 + 4) over (1 + 2) = 5/3, and the hole is at (1, 5/3). Vertical asymptote x = −2. x-intercept (−4, 0). y-intercept (0, 2). Horizontal asymptote y = 1.
- 5. Excluded: 2. The factor is squared and nothing cancels, so x = 2 is a vertical asymptote. The numerator is never zero, so there are no x-intercepts. y-intercept (0, 5/4). Horizontal asymptote y = 0. The denominator is a square, so the function is positive on both sides and both branches run up.
- 6. Excluded: −3. Factor first: (x − 3)(x + 3) over (x + 3). The factor cancels completely and the denominator becomes 1, so there is no asymptote of any kind. The reduced function is x − 3, a straight line, with a hole at (−3, −6). x-intercept (3, 0). y-intercept (0, −3).
- 7.f(x) = x2 − 4x2 − x − 6
- 8.f(x) = x2 + x − 6x2 − 4
- 9.f(x) = 2x2 − 8x2 − 5x + 6
- 10.f(x) = x2 − 1x2 + 2x + 1
- 11.f(x) = x3 − 4xx2 − 2x
- 12.f(x) = x2 − 16x2 − 8x + 16
Show answers for Part B
- 7. (x − 2)(x + 2) over (x − 3)(x + 2). Excluded: 3 and −2. The factor (x + 2) cancels, so hole at (−2, 4/5). Reduced: (x − 2) over (x − 3). Vertical asymptote x = 3. x-intercept (2, 0). y-intercept (0, 2/3). Horizontal asymptote y = 1.
- 8. (x + 3)(x − 2) over (x − 2)(x + 2). Excluded: 2 and −2. The factor (x − 2) cancels, so hole at (2, 5/4). Reduced: (x + 3) over (x + 2). Vertical asymptote x = −2. x-intercept (−3, 0). y-intercept (0, 3/2). Horizontal asymptote y = 1.
- 9. 2(x − 2)(x + 2) over (x − 2)(x − 3). Excluded: 2 and 3. Hole at (2, −8), from the reduced function 2(x + 2) over (x − 3) at x = 2. Vertical asymptote x = 3. x-intercept (−2, 0). y-intercept (0, −4/3). Horizontal asymptote y = 2, from the leading coefficients 2 and 1.
- 10. (x − 1)(x + 1) over (x + 1)2. Excluded: −1 only. The factor (x + 1) cancels once, and one copy remains below, so x = −1 is a vertical asymptote and not a hole. Reduced: (x − 1) over (x + 1). x-intercept (1, 0). y-intercept (0, −1). Horizontal asymptote y = 1.
- 11. x(x − 2)(x + 2) over x(x − 2). Excluded: 0 and 2. Both factors cancel completely and the denominator becomes 1, so there is no asymptote. The reduced function is x + 2, a straight line, with holes at (0, 2) and (2, 4). x-intercept (−2, 0). There is no y-intercept, because x = 0 is excluded.
- 12. (x − 4)(x + 4) over (x − 4)2. Excluded: 4 only. The same structure as problem 10: one copy of (x − 4) remains below, so x = 4 is a vertical asymptote. Reduced: (x + 4) over (x − 4). x-intercept (−4, 0). y-intercept (0, −1). Horizontal asymptote y = 1.
- 13.f(x) = x − 1(x + 2)(x − 3)
- 14.f(x) = x2x − 1
- 15.f(x) = (x + 1)2(x − 2)(x + 3)
- 16.f(x) = x2 − 1x
- 17.f(x) = x2x
- 18.f(x) = 4 − x2x2 − 1
Show answers for Part C
- 13. Excluded: −2 and 3, both vertical asymptotes. No holes. x-intercept (1, 0). y-intercept (0, 1/6). Horizontal asymptote y = 0, met at (1, 0). Cut at −2, 1, 3. Four intervals, signs −, +, −, +, from test values −3, 0, 2, 4 giving −2/3, 1/6, −1/4, 1/2. Three branches. The left branch stays below the axis, approaching y = 0 from below on the far left and falling to negative infinity beside x = −2. The middle branch comes down from positive infinity beside x = −2, falls through the intercept at (1, 0), since the sign goes from positive to negative there, and drops to negative infinity beside x = 3. The right branch comes down from positive infinity beside x = 3 toward y = 0 from above.
- 14. Excluded: 1, a vertical asymptote. No holes. x-intercept (0, 0) with multiplicity 2, which is also the y-intercept. Slant asymptote y = x + 1, from the division x2 = (x − 1)(x + 1) + 1, and the graph never meets it since the remainder is 1. Cut at 0 and 1. Three intervals, signs −, −, +. The sign does not change at 0, because the numerator has a zero of multiplicity 2 there, so the graph touches the x-axis at the origin and turns back down.
- 15. Excluded: −3 and 2, both vertical asymptotes. No holes. x-intercept (−1, 0) with multiplicity 2. y-intercept (0, −1/6), since f(0) = 1 over (−2)(3). Horizontal asymptote y = 1, and solving f(x) = 1 gives (x + 1)2 = x2 + x − 6, so 2x + 1 = x − 6 and x = −7. The graph crosses y = 1 at (−7, 1), far to the left of the visible action. Cut at −3, −1, 2. Four intervals, signs +, −, −, +, with no sign change at −1 because of the squared factor.
- 16. Excluded: 0, a vertical asymptote. No holes. x-intercepts (−1, 0) and (1, 0). There is no y-intercept, because x = 0 is excluded. Slant asymptote y = x, since x2 − 1 divided by x is x with remainder −1, and the graph never meets it. Cut at −1, 0, 1. Four intervals, signs −, +, −, +.
- 17. Excluded: 0. The factor cancels completely, so there is a hole at (0, 0) and no asymptote. The reduced function is y = x, a straight line. There are no x-intercepts, because the only candidate is excluded. The sign chart still cuts at 0, the zero of the reduced numerator, and the signs are −, +. This is the case where a sign change happens without an intercept.
- 18. (2 − x)(2 + x) over (x − 1)(x + 1). Excluded: 1 and −1, both vertical asymptotes. Cut at −2, −1, 1, 2. Five intervals, signs −, +, −, +, −. x-intercepts (−2, 0) and (2, 0). y-intercept (0, −4). Horizontal asymptote y = −1, from leading coefficients −1 and 1. Solving f(x) = −1 gives 4 − x2 = 1 − x2, so 4 = 1 and the graph never meets it.
- 19.f(x) = x2 + 1x − 2
- 20.f(x) = 3x2 − 2xx2 + 1
- 21.f(x) = x2 − 4x + 3x − 1
- 22.f(x) = 2xx2 − 9
- 23.f(x) = x3x2 − 1
- 24.f(x) = x2 + 2x + 1x2 + 1
Show answers for Part D
- 19. Excluded: 2, a vertical asymptote. The numerator has no real zeros, so there are no x-intercepts. y-intercept (0, −1/2). Degrees 2 and 1, so a slant asymptote: x2 + 1 = (x − 2)(x + 2) + 5, giving y = x + 2. Solving f(x) = x + 2 leaves 5 = 0, so the graph never meets it. Signs: − on x < 2, + beyond.
- 20. The denominator x2 + 1 is never zero, so there are no excluded values, no holes, and no vertical asymptotes. Factor the numerator as x(3x − 2): x-intercepts (0, 0) and (2/3, 0). y-intercept (0, 0). Equal degrees, so horizontal asymptote y = 3. Solving (3x2 − 2x) − 3(x2 + 1) = 0 gives −2x − 3 = 0, so the graph crosses y = 3 at (−3/2, 3). One unbroken piece.
- 21. (x − 1)(x − 3) over (x − 1). Excluded: 1. The factor cancels completely and the denominator becomes 1, so there is no asymptote. The reduced function is x − 3, a straight line, with a hole at (1, −2). x-intercept (3, 0). y-intercept (0, −3). The degree comparison on the original form suggests a slant asymptote, which is why the cancelling has to be done first.
- 22. Excluded: 3 and −3, both vertical asymptotes. x-intercept (0, 0), which is also the y-intercept. Numerator degree 1 is less than denominator degree 2, so horizontal asymptote y = 0, and the graph meets it at the origin. Cut at −3, 0, 3. Signs −, +, −, +. Three branches.
- 23. Excluded: 1 and −1, both vertical asymptotes. x-intercept (0, 0) with multiplicity 3, also the y-intercept. Degrees 3 and 2, so a slant asymptote: x3 = (x2 − 1)(x) + x, giving y = x. Solving f(x) = x gives x3 − x(x2 − 1) = x = 0, so the graph crosses its slant asymptote at the origin. Cut at −1, 0, 1. Signs −, +, −, +.
- 24. The denominator is never zero, so no excluded values and no vertical asymptotes. The numerator is (x + 1)2, so there is one x-intercept at (−1, 0) with multiplicity 2, where the graph touches the axis without crossing. y-intercept (0, 1). Equal degrees, so horizontal asymptote y = 1. Solving (x2 + 2x + 1) − (x2 + 1) = 0 gives 2x = 0, so the graph crosses y = 1 at (0, 1). The function is positive everywhere except at x = −1, where it is zero.
- C1.Find every point where the graph of f(x) = x4 + x2 − 2x + 2x4 + 1 meets its horizontal asymptote, and say whether it crosses or touches at each one.
- C2.The graph of f(x) = x3x4 + 1 meets its horizontal asymptote at a root of multiplicity 3. Does it cross or touch there? Explain which property of the multiplicity decides it.
- C3.Write a rational function with a vertical asymptote at x = −2, a hole at x = 3, an x-intercept at x = 1, and a horizontal asymptote y = 2. Then write a second function, with different factors, that has the same four features.
Show answers for the challenge set
- C1. The denominator x4 + 1 is never zero, so nothing is excluded. Equal degrees with leading coefficients 1 and 1, so the horizontal asymptote is y = 1. Solve (x4 + x2 − 2x + 2) − 1(x4 + 1) = 0, which reduces to x2 − 2x + 1 = 0, that is (x − 1)2 = 0. The only solution is x = 1, with multiplicity 2. Even multiplicity, so the graph touches y = 1 at the point (1, 1) and stays above it on both sides. Touching an asymptote is not limited to the x-axis.
- C2. It crosses. Solving f(x) = 0 gives x3 = 0, a root of multiplicity 3, and 3 is odd. The test is odd against even multiplicity, not single against repeated. An odd multiplicity means the expression f(x) − k changes sign as x passes through the root, which is what crossing means. A repeated root of even multiplicity, such as the one in C1, does not change sign, so the graph returns to the side it came from.
- C3. Many functions work, which is the point of the question. One answer is f(x) = 2(x − 1)(x − 3) over ((x + 2)(x − 3)). Check it: (x − 3) cancels completely, so x = 3 is a hole, and the reduced function 2(x − 1) over (x + 2) gives the height 2(2) over 5 = 4/5, putting the hole at (3, 4/5). The reduced denominator is (x + 2), so the vertical asymptote is x = −2. The reduced numerator is zero at x = 1. The reduced degrees are 1 and 1 with leading coefficients 2 and 1, so the horizontal asymptote is y = 2.
A second answer built from different factors is f(x) = 2(x − 1)(x − 3)(x2 + 1) over ((x + 2)(x − 3)(x2 + 2)). The factors x2 + 1 and x2 + 2 have no real zeros, so they add no asymptote and no hole. Again (x − 3) cancels completely, giving a hole at x = 3, this time at height 8/11. The vertical asymptote is still x = −2, the x-intercept is still x = 1, and the reduced degrees are now 3 and 3 with leading coefficients 2 and 1, so the horizontal asymptote is still y = 2.
What the two answers share is not a factor list. The denominator’s copies of (x − 3) have to cancel completely, so the numerator needs at least as many of them, and extra copies in the numerator are allowed. What is fixed is the ratio of the leading coefficients after cancelling, which the horizontal asymptote pins at 2. Use Tool 1 above to check the features of either function you write.
Standards and further reading
This page is built against one Common Core standard and three AP Precalculus topics. The Common Core standard names zeros, asymptotes, and end behavior in one sentence, which is the same list the six step routine works through.
| Standard or topic | What it asks for | Where it appears above |
|---|---|---|
| CCSS HSF.IF.C.7d | Graph rational functions, identifying zeros and asymptotes when suitable factorizations are available, and showing end behavior. | The six step routine, all four worked examples, Parts A through D |
| CCSS HSF.IF.B.4 | Interpret key features of a graph, including intercepts and end behavior, for a function that models a relationship. | The seven features table and the feature finder tool |
| AP Precalculus 1.7 and 1.8 | Describe the end behavior of rational functions, and the behavior near a vertical asymptote in terms of the numerator and denominator. | End behavior and the degree test, branch direction |
| AP Precalculus 1.10 | Determine holes in the graph of a rational function, and distinguish them from vertical asymptotes. | Hole or vertical asymptote, worked example 2, Parts A and B |
Sources and places to read more
- The text of HSF.IF.C.7d at the Common Core State Standards Initiative, which is the wording this page is built against.
- The AP Precalculus Course and Exam Description from the College Board. Unit 1 covers polynomial and rational functions and carries 30 to 40 percent of the exam, and topics 1.7 through 1.10 are the rational function material on this page.
- Stitz and Zeager, Graphs of Rational Functions, an open textbook section on LibreTexts with a longer procedure and full worked solutions.
- Wolfram MathWorld on rational functions, for the formal definition and where the topic sits in the wider subject.
- The California Algebra II content standards from the California Department of Education, useful if your student is in an FUSD, Newark USD, or New Haven USD course and you want to see where this sits in the year.
- The Desmos graphing calculator. Graph your function and your reduced function on the same axes. They agree everywhere except at the excluded values, which is the clearest picture of what a cancelled factor does to the domain.
Related worksheets on this site
- Simplifying rational expressions worksheetThe cancelling step from this page, practiced on its own.
- Rational equations worksheet and extraneous solution guideSolving rather than graphing, with the two list method for excluded values.
- Multiplying and dividing rational expressions worksheetThe restriction ledger, and restrictions that vanish when factors cancel.
- Adding and subtracting rational expressions worksheetBuilding an LCD from a factor inventory.
- Factoring polynomials guide and flowchartEvery factoring pattern the numerators and denominators here need.
- Polynomial division worksheetThe long division that produces a slant asymptote.
- Polynomial inequalities worksheetThe same sign chart, applied to inequalities.
- Graphing polynomial functions worksheetEnd behavior and multiplicity without the denominator.
- All polynomials worksheets and guidesThe hub page for the whole polynomials unit.
- All quadratics worksheets and guidesFactoring, the quadratic formula, and graphing parabolas.
About the author
Berke Sahbazoglu holds a BA in Biochemistry from Washington University in St. Louis and has tutored Algebra 1, Algebra 2, Precalculus, and AP science for more than ten years. He owns Burke Tutoring in Fremont, an in-home tutoring practice serving Fremont, Newark, and Union City, California.
The problems on this page were prepared from previous class notes and assigned homework. The worked examples follow the order used in sessions with students taking Algebra 2 and Precalculus at FUSD, Newark USD, and New Haven USD schools.
Working through rational functions with a tutor in the Tri City area
If you can find the asymptotes but your sketch still feels uncertain, make a sign chart. One test value in each interval tells you whether that part of the graph sits above or below the x-axis. That step is usually where our sessions on this topic go.
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