MHF4U Rational Functions Practice: Restrictions, Holes and Asymptotes

For a rational function, find restrictions from the original denominator before simplifying. A fully cancelled denominator factor can produce a hole; a denominator zero that remains after cancellation can produce a vertical asymptote. Use the simplified expression to analyse behaviour while keeping every original excluded input.

This original MHF4U practice set covers restrictions, discontinuities, intercepts, asymptotes and an inequality. It is not a complete Advanced Functions assessment.

Why does cancelling not restore an excluded input?

Consider f(x) = (x² − 4)/(x − 2). The original expression is undefined at x = 2. Factoring gives (x − 2)(x + 2)/(x − 2), which simplifies to x + 2 only when x ≠ 2. The graph is the line y = x + 2 with the point (2, 4) removed.

The simplified formula describes the same outputs on the original domain. It does not change where the original expression was defined. Write that restriction beside the first line of your solution so it is not lost during algebra.

A checklist before sketching

• 1. Factor numerator and denominator where possible.
• 2. Record all original denominator zeros as restrictions.
• 3. Cancel common factors while retaining the restrictions.
• 4. Identify holes, remaining vertical asymptotes and valid intercepts.
• 5. Analyse end behaviour and check signs on intervals.

For a polynomial ratio, comparing degrees helps with horizontal asymptotes. If the numerator degree is lower, y = 0 is the horizontal asymptote. If the degrees match, use the ratio of leading coefficients. A higher numerator degree needs further analysis, such as division.

Eight questions with worked solutions

1. Simplify and locate a hole

For f(x) = (x² − 9)/(x − 3), state the restriction and hole.

Factor the numerator: (x − 3)(x + 3). The simplified rule is x + 3 with x ≠ 3. The missing point is (3, 6). There is no vertical asymptote at x = 3 because the denominator factor cancels completely.

2. Distinguish a hole from an asymptote

Analyse g(x) = (x − 2)/(x² − 4).

The original restrictions are x ≠ 2 and x ≠ −2. Factor the denominator and cancel x − 2 to get 1/(x + 2), retaining both restrictions. There is a hole at (2, 1/4) and a vertical asymptote at x = −2. The two excluded inputs have different graph behaviour.

3. Find horizontal and vertical asymptotes

For h(x) = (2x + 1)/(x − 4), identify the asymptotes.

The uncancelled denominator zero gives x = 4. The degrees match, so the horizontal asymptote is y = 2/1 = 2. Rewriting h(x) = 2 + 9/(x − 4) also shows why the output approaches 2 as |x| grows.

4. Check whether a numerator zero is allowed

Find the intercepts of r(x) = (x − 1)/[(x − 1)(x + 2)].

The restrictions are x ≠ 1 and −2. The only numerator zero is excluded, so there is no x-intercept. At x = 0, r(0) = 1/2, so the y-intercept is (0, 1/2). The excluded x = 1 gives a hole at (1, 1/3).

5. Solve a rational equation

Solve 1/(x − 1) = 2/(x + 2).

First require x ≠ 1 and −2. Multiplying by the nonzero denominators gives x + 2 = 2(x − 1). Hence x = 4, which satisfies both restrictions. Substitution gives 1/3 on each side.

6. Solve a rational inequality

Solve (x − 1)/(x + 2) ≥ 0.

The critical values are −2 and 1. Test x = −3, 0 and 2: the signs are positive, negative and positive. Include x = 1 because it makes the expression zero; exclude x = −2 because it is undefined. The solution is (−∞, −2) ∪ [1, ∞).

7. Use division for an oblique asymptote

Find the asymptotes of s(x) = (x² + 1)/(x − 1).

Division gives s(x) = x + 1 + 2/(x − 1). Thus the oblique asymptote is y = x + 1 and the vertical asymptote is x = 1. There is no horizontal asymptote. The remainder term approaches zero at the ends of the graph.

8. Can a graph cross its horizontal asymptote?

For t(x) = x/(x² + 1), identify the horizontal asymptote and determine whether the graph crosses it.

The denominator has no real zero, so the domain is all real numbers. The horizontal asymptote is y = 0, and t(0) = 0. The graph crosses that line at (0, 0). A horizontal asymptote describes end behaviour, not a boundary the graph can never cross.

What to do when a graph and your algebra disagree

Check restrictions and scale before assuming the calculator is wrong. A graphing window may not make a tiny hole visible. Test an ordinary input and inspect the factored expression. For an inequality, check one number in each interval; do not multiply by an expression of unknown sign without considering its sign.

Continue with Advanced Functions

Read the MHF4U guide, explore Grade 12 math tutoring, or book a free one-hour assessment.

Curriculum reference: Ontario Grades 11–12 Mathematics. All examples are original.