MHF4U Grade 12 Advanced Functions Exam Review: Practice Test & Formulas [Ontario]

Grade 12 Advanced Functions (MHF4U) is the decisive prerequisite course for Ontario high school students entering competitive STEM, engineering, commerce, computer science, and health science degree programs. The final examination accounts for 30% of your course grade and synthesizes polynomial behavior, rational functions, radian trigonometry, and logarithmic equations.

Achieving an 85%+ on this exam requires both algebraic speed and rigorous adherence to Ontario assessment criteria. This exam review hub provides the official curriculum formula sheet, K/T/C/A rubric weightings, and unit-by-unit practice problems with step-by-step solutions.

The Complete MHF4U Exam Formula Cheat Sheet

• Polynomial Theorems: Remainder Theorem states that when P(x) is divided by (x - a), remainder R = P(a). Factor Theorem states that (x - a) is a factor of P(x) if and only if P(a) = 0.
• Sum and Difference of Cubes: a³ + b³ = (a + b)(a² - ab + b²); a³ - b³ = (a - b)(a² + ab + b²).
• Rational Root Theorem: Any rational zero p/q of an integer polynomial has p dividing the constant term and q dividing the leading coefficient.
• Rational Function Asymptotes: Vertical Asymptote at x = a where denominator = 0 after common factors are cancelled. Point discontinuity (hole) at x = a where (x - a) cancels from both numerator and denominator. Horizontal Asymptote: y = 0 if degree(numerator) < degree(denominator); y = a/b (ratio of leading coefficients) if degrees are equal; Oblique Asymptote found via polynomial division if degree(num) = degree(den) + 1.
• Radian Conversion & Arc Length: θ (radians) = θ (degrees) · (π / 180°); Arc length s = r·θ; Sector area A = (1/2)r²·θ (where θ is in radians).
• Compound Angle Formulas: sin(A ± B) = sin(A)cos(B) ± cos(A)sin(B); cos(A ± B) = cos(A)cos(B) ∓ sin(A)sin(B); tan(A ± B) = [tan(A) ± tan(B)] / [1 ∓ tan(A)tan(B)].
• Double Angle Formulas: sin(2θ) = 2sin(θ)cos(θ); cos(2θ) = cos²(θ) - sin²(θ) = 2cos²(θ) - 1 = 1 - 2sin²(θ); tan(2θ) = [2tan(θ)] / [1 - tan²(θ)].
• Logarithmic Laws: log_b(xy) = log_b(x) + log_b(y); log_b(x/y) = log_b(x) - log_b(y); log_b(x^k) = k·log_b(x); Change of base: log_b(x) = [log_a(x)] / [log_a(b)] = [ln(x)] / [ln(b)].
• Logarithmic Scales: pH = -log[H⁺]; Earthquake Magnitude: M₂ - M₁ = log(I₂ / I₁); Decibel Sound: L₂ - L₁ = 10 log(I₂ / I₁).
• Rates of Change: Average Rate of Change (AROC) = [f(b) - f(a)] / (b - a); Instantaneous Rate of Change (IROC) estimated using symmetric difference quotient: IROC ≈ [f(a + 0.001) - f(a - 0.001)] / 0.002.

Unit 1: Polynomial Functions & Inequalities Practice

Practice Question 1 (Thinking & Application): Solve the inequality 2x³ - 5x² - 4x + 3 ≥ 0 algebraically. State the solution in interval notation.

Worked Solution 1:

• Step 1: Test factors of constant term 3 (±1, ±3) using the Factor Theorem. Let P(x) = 2x³ - 5x² - 4x + 3. Test x = -1: P(-1) = 2(-1)³ - 5(-1)² - 4(-1) + 3 = -2 - 5 + 4 + 3 = 0. Therefore, (x + 1) is a factor.
• Step 2: Use synthetic or long division to divide P(x) by (x + 1): Quotient = 2x² - 7x + 3.
• Step 3: Factor the quadratic quotient by decomposition: 2x² - 6x - x + 3 = 2x(x - 3) - 1(x - 3) = (2x - 1)(x - 3).
• Step 4: Completely factored inequality: (x + 1)(2x - 1)(x - 3) ≥ 0. Critical numbers: x = -1, x = 1/2, x = 3.
• Step 5: Create a Table of Signs testing intervals (-∞, -1), (-1, 1/2), (1/2, 3), and (3, ∞): For x < -1: (-)·(-)·(-) = - (Negative); For -1 < x < 1/2: (+)·(-)·(-) = + (Positive); For 1/2 < x < 3: (+)·(+)·(-) = - (Negative); For x > 3: (+)·(+)·(+) = + (Positive).
• Final Solution: We require ≥ 0, so include critical roots: x ∈ [-1, 1/2] ∪ [3, ∞).

Unit 2: Rational Functions & Asymptotes Practice

Practice Question 2 (Knowledge & Communication): For the rational function f(x) = (2x² - 8) / (x² - x - 2), state all intercepts, holes, vertical asymptotes, and horizontal asymptotes.

Worked Solution 2:

• Step 1: Factor numerator and denominator: f(x) = [2(x - 2)(x + 2)] / [(x - 2)(x + 1)]. Restrictions before cancellation: x ≠ 2, x ≠ -1.
• Step 2: Cancel common factor (x - 2): Simplified function is f(x) = [2(x + 2)] / (x + 1) for x ≠ 2.
• Step 3: Identify hole: Common factor (x - 2) creates a point discontinuity at x = 2. Substitute x = 2 into simplified function: y = 2(2 + 2)/(2 + 1) = 8/3. Hole coordinates: (2, 8/3).
• Step 4: Vertical Asymptote: Non-cancelling denominator root: x = -1.
• Step 5: Horizontal Asymptote: Degrees of numerator and denominator are equal (degree 2). Ratio of leading coefficients is 2/1: Horizontal Asymptote at y = 2.
• Step 6: Intercepts: y-intercept: f(0) = [2(2)]/1 = 4 ⇒ (0, 4). x-intercept: set simplified numerator to 0: 2(x + 2) = 0 ⇒ x = -2 ⇒ (-2, 0).

Unit 3: Radian Trigonometry & Double Angle Identities Practice

Practice Question 3 (Thinking): Prove the identity: [1 - cos(2θ)] / [sin(2θ)] = tan(θ), and solve the equation 2sin²(x) - 3cos(x) = 0 for x ∈ [0, 2π].

Worked Solution 3:

• Part A Identity Proof: Left Hand Side (LHS) = [1 - cos(2θ)] / sin(2θ). Substitute double angle formulas: cos(2θ) = 1 - 2sin²(θ) and sin(2θ) = 2sin(θ)cos(θ).
• LHS = [1 - (1 - 2sin²(θ))] / [2sin(θ)cos(θ)] = [2sin²(θ)] / [2sin(θ)cos(θ)] = sin(θ) / cos(θ) = tan(θ). Since LHS = RHS, the identity is proven.
• Part B Equation Solving: 2sin²(x) - 3cos(x) = 0. Use Pythagorean identity sin²(x) = 1 - cos²(x): 2(1 - cos²(x)) - 3cos(x) = 0 ⇒ 2 - 2cos²(x) - 3cos(x) = 0 ⇒ 2cos²(x) + 3cos(x) - 2 = 0.
• Factor the quadratic in terms of cos(x): (2cos(x) - 1)(cos(x) + 2) = 0.
• Case 1: cos(x) + 2 = 0 ⇒ cos(x) = -2 (No real solution, since -1 ≤ cos(x) ≤ 1).
• Case 2: 2cos(x) - 1 = 0 ⇒ cos(x) = 1/2. In [0, 2π], cosine is positive in Quadrants I and IV. Reference angle θ_r = π/3. Quadrant I: x = π/3; Quadrant IV: x = 2π - π/3 = 5π/3.
• Final Exact Solutions: x = π/3, 5π/3.

Unit 4: Logarithmic & Exponential Equations Practice

Practice Question 4 (Application): Solve the equation log₄(x + 3) + log₄(x - 3) = 2 algebraically. State restrictions and verify your solution.

Worked Solution 4:

• Step 1: State domain restrictions: Arguments of logs must be strictly positive: x + 3 > 0 ⇒ x > -3, and x - 3 > 0 ⇒ x > 3. Combined restriction is x > 3.
• Step 2: Combine logs using product rule: log₄[(x + 3)(x - 3)] = 2 ⇒ log₄(x² - 9) = 2.
• Step 3: Convert to exponential form: x² - 9 = 4² ⇒ x² - 9 = 16 ⇒ x² = 25 ⇒ x = ±5.
• Step 4: Check against restriction x > 3: x = 5 satisfies x > 3. x = -5 is extraneous because log₄(-5 + 3) = log₄(-2) is undefined.
• Final Unique Solution: x = 5.

Top 5 Costly Exam Traps in MHF4U

• Trap 1: Forgetting to check extraneous roots in logarithmic equations. Every negative argument must be explicitly discarded with justification.
• Trap 2: Calculator in Degree mode instead of Radian mode. MHF4U trig units strictly use radians unless an applied word problem specifies degrees.
• Trap 3: Missing Oblique Asymptotes when degree(numerator) = degree(denominator) + 1. Performing polynomial long division to find y = mx + b is required.
• Trap 4: Forgetting that a hole has BOTH an x and y coordinate. Always substitute the x-value back into the simplified expression to report the (x, y) hole location.
• Trap 5: Confusing Average Rate of Change (secant line slope) with Instantaneous Rate of Change (tangent line slope).