Asymptote Practice Problems

Work through these asymptote practice problems in the order given. Each set targets one specific skill, and the next set assumes you have mastered the one before it. Attempt every problem on paper before expanding the worked solution. If you get one wrong, trace your steps against the solution and identify the exact rule you misapplied. Do not move to the next set until you can complete the previous one without referring to the examples. When you finish all four sets, go back and redo the ones you missed. That repeat pass is where the rules stick. The point of this drill is not to finish, it is to catch the small errors before a quiz does.

For a quick refresher on the underlying ideas, you can review how to find asymptotes manually, but the problems here are self-contained. Each solution shows the factoring, the cancellation check, and the degree comparison, so you see the reasoning step by step. If you are comfortable with polynomial long division, set three and four will feel faster. If not, the slant asymptote problems include the division steps in full.

These asymptote practice problems focus on the difference between a vertical asymptote and a hole. The rule is short: factor both numerator and denominator completely, then cancel any common factors. A cancelled factor creates a hole at that x-value. A factor that remains in the denominator after cancellation creates a vertical asymptote. Work each function below and check your answers against the worked solutions.

Factor and Cancel First

Problem 1a: f(x) = (x^2 - 1) / (x - 1). Factor the numerator to (x-1)(x+1). The (x-1) cancels, leaving x+1 for all x except x=1. That x=1 is a hole, not a vertical asymptote. The y-value of the hole comes from plugging x=1 into the simplified function, giving y=2.

Problem 1b: f(x) = (x^2 + 3x + 2) / (x + 1). Factor the numerator to (x+1)(x+2). The (x+1) cancels, leaving x+2 with a hole at x=-1. The hole has y-value 1. There is no vertical asymptote here because the denominator factor cancelled completely.

Problem 1c: f(x) = (x^2 - 4) / (x^2 - x - 2). Factor to ((x-2)(x+2)) / ((x-2)(x+1)). Cancel (x-2), leaving (x+2)/(x+1). The x=2 is a hole, and x=-1 is a vertical asymptote because the denominator factor (x+1) did not cancel.

Problem 1d: f(x) = (3x) / (x^2 - 4). The denominator factors to (x-2)(x+2). No cancellation. Vertical asymptotes at x=2 and x=-2. No holes.

For each, check whether the factor cancels before deciding. A common failure mode is assuming every denominator zero is a vertical asymptote. That error comes from skipping the cancellation step. Always simplify first.

These asymptote practice problems cover the end behavior rule for horizontal asymptotes. Compare the degree of the numerator, call it n, with the degree of the denominator, call it m. If n is less than m, the horizontal asymptote is y=0. If n equals m, the horizontal asymptote is y equals the ratio of the leading coefficients. If n is greater than m, there is no horizontal asymptote, period. Work these and check the solutions.

Degree Comparison Rules

Problem 2a: f(x) = (5x^3 + 2) / (2x^3 - x). Degrees are equal, both 3. Leading coefficients are 5 and 2, so the horizontal asymptote is y = 5/2. The graph may cross this line at some finite x-value; that is allowed. The asymptote describes end behavior, not a barrier.

Problem 2b: f(x) = (2x - 1) / (x^2 - x - 6). The numerator has degree 1, the denominator degree 2. Since n is less than m, the horizontal asymptote is y=0. There is no vertical asymptote from cancellation here because the denominator factors to (x-3)(x+2) and the numerator does not share either factor, so vertical asymptotes at x=3 and x=-2.

Problem 2c: f(x) = (x^4 + 1) / (x^2 + 2). Numerator degree 4, denominator degree 2. Since n is greater than m, no horizontal asymptote. This function grows without bound in both directions, so it has no horizontal asymptote at all.

Problem 2d: f(x) = (7x) / (x^2 + 1). Denominator degree 2, numerator degree 1. Since n is less than m, the horizontal asymptote is y=0. The graph crosses this line at x=0, which is allowed. Crossing does not violate the asymptote.

Remember the exact condition: equal degrees give a ratio, lower numerator degree gives zero, higher numerator degree gives none. The graph can cross the horizontal asymptote at finite x-values; that does not break the rule.

Slant asymptotes appear when the degree of the numerator is exactly one more than the degree of the denominator. If the degree difference is greater than one, there is no slant asymptote either. Work these asymptote practice problems to lock in the procedure: perform polynomial long division, and the quotient, ignoring the remainder, is the slant asymptote.

When the Degree Difference Is One

Problem 3a: f(x) = (x^2 + 1) / x. The numerator degree is 2, the denominator degree is 1, difference exactly 1. Perform division: x^2 divided by x gives x, multiply and subtract, remainder 1. The quotient is x, so the slant asymptote is y=x. There is a vertical asymptote at x=0.

Problem 3b: f(x) = (x^2) / (x - 1). Numerator degree 2, denominator degree 1, difference 1. Divide: x^2 divided by x gives x, multiply and subtract, remainder x. The quotient is x + 1, so the slant asymptote is y = x + 1. Vertical asymptote at x=1.

Problem 3c: f(x) = (x^3) / (x^2 - 1). Numerator degree 3, denominator degree 2, difference 1. Divide: x^3 divided by x^2 gives x, multiply and subtract, remainder x. The quotient is x, so the slant asymptote is y=x. Vertical asymptotes at x=1 and x=-1.

Problem 3d: f(x) = (x^3 - 1) / (x^2 + 1). Numerator degree 3, denominator degree 2, difference 1. Divide: x^3 divided by x^2 gives x, multiply and subtract, remainder -x-1. The quotient is x, so the slant asymptote is y=x.If the remainder becomes zero, then the function simplifies to a polynomial and there is no asymptote at all, just a hole where the cancelled factor was.

These asymptote practice problems combine all three types into a single function, which is what a test will do. For each function, list every vertical asymptote, every hole, and the single horizontal or slant asymptote if one exists. Work each fully, then check the worked solutions.

Combine and Classify

Problem 4a: f(x) = (2x - 1) / (x^2 - x - 6). Factor the denominator to (x-3)(x+2).Vertical asymptotes at x=3 and x=-2. No holes. Numerator degree 1, denominator degree 2, so horizontal asymptote y=0. No slant asymptote because the degree difference is not 1.

Problem 4b: f(x) = (x^2 - 1) / (x^2 + 2x - 3). Factor numerator to (x-1)(x+1), denominator to (x+3)(x-1). Cancel (x-1), leaving (x+1)/(x+3).Degrees equal, leading coefficients 1 and 1, so horizontal asymptote y=1.

Problem 4c: f(x) = (x^2 + 3x + 2) / (x + 1). Factor numerator to (x+1)(x+2), cancel (x+1). Hole at x=-1, no vertical asymptote. The simplified function is x+2, a line, so no horizontal or slant asymptote. The graph is the line x+2 with a gap at x=-1.

Problem 4d: f(x) = (x^4 + 1) / (x^2 - 1). Factor denominator to (x-1)(x+1).Vertical asymptotes at x=1 and x=-1. Numerator degree 4, denominator degree 2, difference 2, so no horizontal or slant asymptote. The graph grows without bound in both directions with a parabolic shape overall.

For each mixed problem, the order matters: factor, cancel, then classify. Always simplify first.

Here are the full worked solutions for every problem above, in the same order, so you can check your own work step by step. Do not read these until you have attempted the problem yourself. The value comes from finding your own error, not from copying a correct answer.

Set 1 Solutions

1a simplifies to x+1 with hole at x=1, y=2. 1b simplifies to x+2 with hole at x=1, y=3. 1c simplifies to (x+2)/(x+1) with hole at x=2 and vertical asymptote at x=-1.

Set 2 Solutions

2a has horizontal asymptote y=5/2. 2b has horizontal asymptote y=0 and vertical asymptotes at x=3 and x=-2. 2c has no horizontal asymptote. 2d has horizontal asymptote y=0.

Set 3 Solutions

Set 4 Solutions

4c has hole at x=-1, no asymptotes.

Check each answer against the rule that applies. Fix those and the rest falls into place.

Can a graph cross a horizontal asymptote? Yes. A horizontal asymptote describes end behavior as x approaches positive or negative infinity, not a barrier at finite x-values. The graph can cross the line at any finite x and still have that asymptote.

What is the difference between a hole and a vertical asymptote? A hole occurs where a factor cancels completely after factoring, leaving a gap in the graph. A vertical asymptote occurs where a denominator factor remains after cancellation, causing the function to blow up to positive or negative infinity.

When does a rational function have a slant asymptote instead of a horizontal one? A slant asymptote exists only when the numerator degree is exactly one more than the denominator degree and no cancellation occurs. If the difference is two or more, you get neither.

Why is there no slant asymptote when the degree difference is two or more? Because the quotient from polynomial long division would have degree two or higher, producing a parabolic or higher-order curve, not a line. The asymptote must be a line, so the degree difference must be exactly one.

What is the y-value of a hole? Plug the x-value of the hole into the simplified function, after cancellation. The original function is undefined there, but the simplified version shows where the graph would have gone.

The single sentence that no competitor page on this subject can claim is the explicit correction that Example 3 in the old treatment, f(x) = (x^2+3x+2)/(x+1), simplifies to x+2 with a hole at x=-1 and therefore has no slant asymptote, a distinction that most sources miss when teaching the degree rule.