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MHF4U Rational Functions
Grade 12 Unit 3 — practice questions with full solutions

Vertical, horizontal and oblique asymptotes, holes, and solving rational equations and inequalities.

Free MHF4U rational functions practice for Grade 12 students in Ontario. These 4 questions cover the same material as a typical unit test or exam review on this topic, and every one comes with a complete worked solution — not just an answer key. Useful whether you are preparing for a MHF4U unit test, catching up on a lesson, or reviewing before the final exam.
MHF4U · Grade 12 Rational Functions 3 core 1 challenge
1Core
-4-3-2-112345678910-8-6-4-224681012xyx = 3y = 2
The graph shows f(x) = (2x + 1) ÷ (x − 3). (a) State both asymptotes and justify each algebraically. (b) Find the intercepts. (c) State the domain and range.
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  1. (a) The vertical asymptote occurs where the denominator is zero: x − 3 = 0, so x = 3.
  2. For the horizontal asymptote the degrees of numerator and denominator are equal, so divide leading coefficients: y = 2 ÷ 1 = 2.
  3. (b) y-intercept: set x = 0 to get (1) ÷ (−3) = −1⁄3.
  4. x-intercept: set the numerator to zero, 2x + 1 = 0, giving x = −1⁄2.
  5. (c) Domain excludes the vertical asymptote: x ≠ 3. Range excludes the horizontal asymptote: y ≠ 2.
Final answer(a) x = 3 and y = 2   (b) (−½, 0) and (0, −⅓)   (c) x ≠ 3, y ≠ 2
2Core
Identify all asymptotes and any holes for f(x) = (x² − 9) ÷ (x² − x − 6).
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  1. Factor both parts: numerator (x−3)(x+3), denominator (x−3)(x+2).
  2. The common factor (x−3) cancels, which creates a hole rather than an asymptote at x = 3.
  3. Find the hole's height using the reduced function (x+3)÷(x+2) at x = 3: 6 ÷ 5 = 1.2.
  4. The remaining denominator factor gives a vertical asymptote at x = −2.
  5. Degrees are equal, so the horizontal asymptote is the ratio of leading coefficients: y = 1.
Final answerhole at (3, 1.2); vertical asymptote x = −2; horizontal asymptote y = 1
3Core
Solve:   3⁄x − 2 = x⁄4
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  1. Cross-multiply: 3(4) = x(x − 2), so 12 = x² − 2x.
  2. Rearrange: x² − 2x − 12 = 0.
  3. This does not factor, so use the quadratic formula: x = (2 ± √(4 + 48)) ÷ 2.
  4. √52 = 2√13, so x = (2 ± 2√13) ÷ 2 = 1 ± √13.
  5. Check neither value makes the original denominator zero — both are fine since x ≠ 2.
Final answerx = 1 ± √13
4Challenge
Solve the inequality:   (x + 1)⁄(x − 4) ≥ 0
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  1. A quotient is zero or positive when numerator and denominator share a sign, or the numerator is zero.
  2. Critical values are x = −1 (numerator zero) and x = 4 (denominator zero).
  3. Test x = −2: (−1)÷(−6) = positive. ✔
  4. Test x = 0: (1)÷(−4) = negative. ✘ Test x = 5: (6)÷(1) = positive. ✔
  5. Include x = −1 because the expression equals 0 there, but exclude x = 4 since the function is undefined.
Final answerx ≤ −1 or x > 4

MHF4U Rational Functions — common questions

Short answers to the things students ask most about this unit.

How do you find a horizontal asymptote?

Compare degrees: if the numerator's degree is smaller, y = 0; if equal, divide the leading coefficients; if larger, there is no horizontal asymptote.

What is the difference between a hole and a vertical asymptote?

A factor that cancels from both numerator and denominator creates a hole. A factor that remains only in the denominator creates a vertical asymptote.

More MHF4U practice — grade 12 advanced functions

MHF4U Polynomial FunctionsMHF4U Polynomial Equations & InequalitiesMHF4U Exponential & Logarithmic FunctionsMHF4U Trigonometric FunctionsAll MHF4U units
← Unit 2: Polynomial Equations & InequalitiesUnit 4: Exponential & Logarithmic Functions →
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