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MCR3U Quadratic Functions
Grade 11 Unit 2 — practice questions with full solutions

Completing the square, maximum and minimum problems, the discriminant, and quadratics with irrational or no real roots.

Free MCR3U quadratic functions practice for Grade 11 students in Ontario. These 5 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 4 core 1 challenge
1Core
Complete the square for y = 2x² + 12x + 7 and state the minimum value.
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  1. Factor the leading coefficient from the first two terms only: y = 2(x² + 6x) + 7.
  2. Half of 6 is 3, and 3² = 9. Add and subtract 9 inside the bracket.
  3. y = 2(x² + 6x + 9 − 9) + 7 = 2((x+3)² − 9) + 7.
  4. Distribute the 2: y = 2(x+3)² − 18 + 7 = 2(x+3)² − 11.
  5. Since a = 2 > 0 the parabola opens up, so the vertex value is the minimum.
Final answery = 2(x + 3)² − 11; minimum value −11 at x = −3
2Core
Solve exactly, leaving answers in radical form:   x² − 6x + 4 = 0
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  1. Use the quadratic formula with a = 1, b = −6, c = 4.
  2. Discriminant: 36 − 16 = 20.
  3. x = (6 ± √20) ÷ 2.
  4. Simplify the radical: √20 = 2√5, so x = (6 ± 2√5) ÷ 2.
  5. Divide every term by 2.
Final answerx = 3 ± √5
3Core
A company's profit is P(x) = −2x² + 60x − 250, where x is the selling price in dollars. Find the price that maximises profit and the maximum profit.
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  1. The graph is a downward parabola, so the vertex gives the maximum.
  2. x = −b ÷ (2a) = −60 ÷ (2 × −2) = 15.
  3. Substitute: P(15) = −2(225) + 60(15) − 250.
  4. That is −450 + 900 − 250 = 200.
Final answerprice $15, maximum profit $200
4Core
Determine the number of x-intercepts of y = 3x² − 4x + 5 without graphing, and explain what this means about the graph's position.
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  1. Use the discriminant: D = b² − 4ac = (−4)² − 4(3)(5).
  2. D = 16 − 60 = −44, which is negative.
  3. A negative discriminant means no real roots, so there are no x-intercepts.
  4. Since a = 3 > 0 the parabola opens upward and never reaches the axis, so it lies entirely above the x-axis.
Final answerNo x-intercepts; the parabola lies entirely above the x-axis
5Challenge
The sum of two numbers is 20. Find the two numbers that make the sum of their squares as small as possible.
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  1. Let the numbers be x and 20 − x.
  2. Sum of squares: S = x² + (20 − x)².
  3. Expand: S = x² + 400 − 40x + x² = 2x² − 40x + 400.
  4. This opens upward, so the minimum is at x = 40 ÷ (2 × 2) = 10.
  5. Both numbers are 10, giving S = 100 + 100 = 200.
Final answerBoth numbers are 10; minimum sum of squares is 200

MCR3U Quadratic Functions — common questions

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

How do you complete the square when a is not 1?

Factor a out of the first two terms only, complete the square inside the bracket, then multiply the correction term by a as you bring it outside.

What does a negative discriminant mean for the graph?

There are no x-intercepts, so the parabola sits entirely above the x-axis if a is positive, or entirely below it if a is negative.

More MCR3U practice — grade 11 functions

MCR3U Functions & TransformationsMCR3U Exponential FunctionsMCR3U Sequences & SeriesMCR3U Financial ApplicationsAll MCR3U units
← Unit 1: Functions & TransformationsUnit 3: Exponential Functions →
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