MHF4U Advanced Functions: Topics & Worked Examples

MHF4U is Ontario’s Grade 12 university-preparation Advanced Functions course. It extends function concepts and equips students to connect algebraic manipulation, graphical properties, and real-world modeling. The official prerequisite listed by the Ontario Ministry of Education is MCR3U (Grade 11 Functions) or MCT4C (Mathematics for College Technology).

What Does MHF4U Cover?

The MHF4U curriculum is organized into four main strands: Exponential and Logarithmic Functions, Trigonometric Functions, Polynomial and Rational Functions, and Characteristics of Functions. Throughout each strand, students analyze rates of change, solve multi-step equations, and combine distinct functions using addition, subtraction, multiplication, division, and composition.

MHF4U vs MCV4U: How the Grade 12 Math Courses Connect

A common point of confusion for families is how MHF4U relates to MCV4U (Calculus and Vectors). MHF4U focuses strictly on function behavior—understanding transformations, domain restrictions, asymptotic limits, and logarithmic/trigonometric identities. MCV4U takes those function foundations and applies them to instantaneous rates of change, derivative rules, optimization problems, and 3D spatial vector geometry.

MHF4U must be completed before, or taken concurrently with, MCV4U. Postsecondary mathematics requirements vary by program and admission year. Check the published course codes for the exact program rather than assuming one rule covers all business, science or engineering degrees.

A Crucial Concept: Checking Algebra and Domain Restrictions

Consider the rational function g(x) = (x² − 9) / (x − 3). Factoring the numerator yields g(x) = [(x − 3)(x + 3)] / (x − 3). The expression simplifies to y = x + 3 only under the explicit restriction x ≠ 3. The original denominator is zero at x = 3, meaning the input is permanently excluded from the domain.

On a graph, this function does not have a vertical asymptote at x = 3; instead, it is the linear line y = x + 3 with a point discontinuity (a 'hole') at (3, 6). Establishing the discipline to write down restrictions prior to algebraic cancellation prevents fatal errors on senior assessments.

Worked Example 1: Polynomial End Behaviour, Roots & Multiplicity

Problem: Analyze and sketch the polynomial P(x) = -(x - 1)² (x + 2)(x - 3) without technology.

• Degree and Leading Coefficient: The sum of powers is 2 + 1 + 1 = 4 (even degree). The leading coefficient is negative (-1). Therefore, as x → ∞, P(x) → -∞, and as x → -∞, P(x) → -∞ (both tails point downward in Quadrants III and IV).
• x-Intercepts (Zeros): x = 1 has multiplicity 2 (even multiplicity: the curve turns around and touches/is tangent to the x-axis at (1, 0)). x = -2 has multiplicity 1 (odd multiplicity: passes straight through the x-axis). x = 3 has multiplicity 1 (passes straight through the x-axis).
• y-Intercept: Evaluate P(0) = -(0 - 1)² (0 + 2)(0 - 3) = -(1)(2)(-3) = +6. The curve crosses the y-axis at (0, 6).
• Sketching: Start from bottom-left (Quadrant III), cross upward through (-2, 0), rise to a local maximum, pass through (0, 6), descend to touch the x-axis at (1, 0), bounce upward to a local maximum, then cross downward through (3, 0) into Quadrant IV.

Worked Example 2: Logarithmic Equations & Extraneous Roots

Problem: Solve the equation log₂(x + 2) + log₂(x - 2) = 5 for all real values of x.

• Step 1: State restrictions. Logarithm arguments must be strictly positive: x + 2 > 0 ⇒ x > -2, and x - 2 > 0 ⇒ x > 2. The combined restriction is x > 2.
• Step 2: Apply the product rule of logarithms. log₂[(x + 2)(x - 2)] = 5 ⇒ log₂(x² - 4) = 5.
• Step 3: Convert to exponential form. x² - 4 = 2⁵ ⇒ x² - 4 = 32.
• Step 4: Solve the quadratic. x² = 36 ⇒ x = 6 or x = -6.
• Step 5: Check against restrictions. x = 6 satisfies x > 2. However, x = -6 violates the domain restriction (log₂(-4) is undefined in real numbers). Therefore, x = -6 is an extraneous root. The unique valid solution is x = 6.

Worked Example 3: Trigonometric Transformations & Graphing

Problem: State the amplitude, period, phase shift, vertical translation, and five key points for f(x) = 3 cos(2(x - π/4)) + 1 over the domain [0, π].

• Amplitude |a|: |3| = 3. The vertical stretch is 3 units from the sinusoidal axis.
• Vertical Shift (Sinusoidal Axis): c = 1, so the centerline is the line y = 1. Maximum value = 1 + 3 = 4; Minimum value = 1 - 3 = -2.
• Period: T = 2π / k = 2π / 2 = π radians.
• Phase Shift: d = π/4 units to the right.
• Five key points within [0, π]: (0, 1), (π/4, 4), (π/2, 1), (3π/4, −2), and (π, 1). These are spaced by one quarter-period, π/4.

5-Question Diagnostic Readiness Check

Before advancing into midterm exams or Calculus, test your conceptual understanding on these five diagnostic questions:

• Question 1 (Logarithmic Laws): Evaluate log₃(81) - log₃(1/9). Solution: log₃(3⁴) - log₃(3⁻²) = 4 - (-2) = 6.
• Question 2 (Rational Discontinuities): State all discontinuities of f(x) = (x - 2) / (x² - 4). Solution: Factoring denominator yields (x - 2)/[(x - 2)(x + 2)]. Point discontinuity (hole) occurs at x = 2 with coordinates (2, 1/4). Vertical asymptote occurs at x = -2.
• Question 3 (Exact Trig Ratios): Determine the exact value of sin(5π/6). Solution: The reference angle is π/6. In Quadrant II, sine is positive: sin(5π/6) = sin(π/6) = 1/2.
• Question 4 (Polynomial Remainder Theorem): Find the remainder when P(x) = 2x³ - 3x² + 4x - 5 is divided by (x - 2). Solution: Evaluate P(2) = 2(2)³ - 3(2)² + 4(2) - 5 = 16 - 12 + 8 - 5 = 7.
• Question 5 (Exponential Substitution): Solve 2^(2x) - 5(2^x) + 4 = 0. Solution: Let u = 2^x. Then u² - 5u + 4 = 0 ⇒ (u - 1)(u - 4) = 0 ⇒ u = 1 or u = 4. Substituting back: 2^x = 1 ⇒ x = 0; 2^x = 4 ⇒ x = 2. Solutions are x = 0 and x = 2.

Build a Practical Study Routine for MHF4U

• Start with a short algebra review. Use errors to decide what to revisit rather than repeating questions you already find easy.
• Sketch a function by hand before relying on graphing tools like Desmos, then use software to verify your critical points and intercepts.
• Always write domain restrictions beside original expressions before applying algebraic cancellation or logarithmic compression.
• Sort assessment mistakes into concept gaps, algebraic slips, and misread questions to focus your next revision session efficiently.

Check a rational expression before cancelling

For r(x) = (x² - 4)/(x - 2), factoring gives r(x) = x + 2 only when x ≠ 2. The original expression is undefined at 2, so its graph has a hole at (2, 4). Cancelling a factor simplifies a formula; it does not restore an excluded input. Compare this with 1/(x - 2), which has a vertical asymptote at x = 2. State restrictions before simplifying.

A readiness check by function family

• Polynomial: for (x − 2)²(x + 1), identify zeros 2 and −1 and explain why the graph touches at 2 and crosses at −1.
• Rational: for 1/(x − 3), state x ≠ 3 before describing the vertical asymptote.
• Exponential: solve 2^(x + 1) = 16 by matching powers, giving x = 3.
• Logarithmic: log₂(x) = 3 means x = 8, with x > 0.
• Trigonometric: for y = 2 sin(x), give amplitude 2 and period 2π in radians.

Use the question you could not explain as your next practice target. These are brief checks, not a complete assessment of the course.