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MCR3U Functions & Transformations
Grade 11 Unit 1 — practice questions with full solutions

Function notation, domain and range, inverses, and the transformation rules that carry through every later unit.

Free MCR3U functions & transformations practice for Grade 11 students in Ontario. These 6 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 MCR3U unit test, catching up on a lesson, or reviewing before the final exam.
MCR3U · Grade 11 Characteristics of Functions 5 core 1 challenge
1Core
-3-2-112345-6-4-224xy(1, 4)(-2, -5)(4, -5)
The graph shows f(x) = −(x − 1)² + 4 drawn only for the portion visible. State (a) the domain, (b) the range, (c) the vertex, and (d) whether it is a function, with a reason.
Show the full solution
  1. (a) The curve runs from x = −2 to x = 4, so the domain is −2 ≤ x ≤ 4.
  2. (b) The highest point is y = 4 at the vertex; the lowest visible value is y = −5 at both endpoints. Range: −5 ≤ y ≤ 4.
  3. (c) Vertex form gives the vertex directly: (1, 4).
  4. (d) It passes the vertical line test — every x value gives exactly one y value, so it is a function.
Final answer(a) −2 ≤ x ≤ 4   (b) −5 ≤ y ≤ 4   (c) (1, 4)   (d) Yes — passes the vertical line test
2Core
If f(x) = 2x² − 3x + 1, evaluate (a) f(−2), (b) f(a + 1), simplified.
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  1. (a) Substitute with brackets: 2(−2)² − 3(−2) + 1 = 2(4) + 6 + 1 = 15.
  2. (b) Replace every x with (a + 1): 2(a+1)² − 3(a+1) + 1.
  3. Expand the square: (a+1)² = a² + 2a + 1, so 2(a² + 2a + 1) = 2a² + 4a + 2.
  4. Then −3(a+1) = −3a − 3.
  5. Combine: 2a² + 4a + 2 − 3a − 3 + 1 = 2a² + a.
Final answer(a) 15   (b) 2a² + a
3Core
-6-5-4-3-2-1123456-6-5-4-3-2-1123456xyy = xf(x)g(x)
The graph shows f(x) = 2x + 1 and a second line g(x). (a) Find f⁻¹(x) algebraically. (b) Explain what the dashed line represents. (c) Verify that f(3) and f⁻¹ undo each other.
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  1. (a) Write y = 2x + 1, then swap x and y: x = 2y + 1.
  2. Solve for y: x − 1 = 2y, so y = (x − 1) ÷ 2.
  3. (b) The dashed line is y = x. A function and its inverse are always mirror images across it.
  4. (c) f(3) = 2(3) + 1 = 7. Then f⁻¹(7) = (7 − 1) ÷ 2 = 3, returning the original input. ✔
Final answer(a) f⁻¹(x) = (x − 1) ÷ 2   (b) the line y = x   (c) f(3) = 7 and f⁻¹(7) = 3
4Core
Describe the transformations applied to y = x² to obtain y = −3(x + 2)² − 5, in the correct order, and state the vertex.
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  1. Read the vertex form y = a(x − h)² + k piece by piece.
  2. a = −3: a vertical stretch by factor 3, and the negative reflects it in the x-axis.
  3. (x + 2) means h = −2: a horizontal translation 2 units left.
  4. k = −5: a vertical translation 5 units down.
  5. The vertex sits at (h, k).
Final answerVertical stretch by 3, reflection in the x-axis, 2 left, 5 down; vertex (−2, −5)
5Core
State the domain and range of (a) y = √(x − 4) and (b) y = 1 ÷ (x + 3).
Show the full solution
  1. (a) A square root needs a non-negative radicand: x − 4 ≥ 0, so x ≥ 4.
  2. The square root itself is never negative, so y ≥ 0.
  3. (b) A denominator can never be zero: x + 3 ≠ 0, so x ≠ −3.
  4. A fraction with numerator 1 can never equal zero, so y ≠ 0.
Final answer(a) x ≥ 4, y ≥ 0   (b) x ≠ −3, y ≠ 0
6Challenge
Given f(x) = x² − 4, explain why f⁻¹ is not a function, and state a restriction on the domain of f that makes it one.
Show the full solution
  1. Swapping variables gives x = y² − 4, so y = ±√(x + 4).
  2. The ± means most inputs produce two outputs, so the inverse fails the vertical line test.
  3. Equivalently, f itself fails the horizontal line test: f(2) and f(−2) both equal 0.
  4. Restricting to x ≥ 0 keeps only the right half of the parabola, which is one-to-one.
  5. With that restriction the inverse is y = √(x + 4), a genuine function.
Final answerBecause f is not one-to-one (fails the horizontal line test); restrict to x ≥ 0

MCR3U Functions & Transformations — common questions

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

What is the difference between a relation and a function?

A function gives exactly one output for every input. Graphically, it passes the vertical line test — no vertical line touches the graph more than once.

How do you find the inverse of a function?

Swap x and y in the equation, then solve for y. On a graph, the inverse is the reflection of the original in the line y = x.

More MCR3U practice — grade 11 functions

MCR3U Quadratic FunctionsMCR3U Exponential FunctionsMCR3U Sequences & SeriesMCR3U Financial ApplicationsAll MCR3U units
← All assignmentsUnit 2: Quadratic Functions →
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