Home / Assignments / MPM2D Unit 4
✎ TruMath Assignment

MPM2D Quadratic Relations & Graphing
Grade 10 Unit 4 — practice questions with full solutions

Vertex form, standard form, completing the square, axis of symmetry, and reading everything you need straight off a parabola.

Free MPM2D quadratic relations & graphing practice for Grade 10 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 MPM2D unit test, catching up on a lesson, or reviewing before the final exam.
MPM2D · Grade 10 Quadratic Relations 5 core 1 challenge
1Core
-112345-2-1123456xyvertex(0, 3)y = f(x)
From the parabola shown, state: (a) the x-intercepts, (b) the vertex, (c) the axis of symmetry, (d) the y-intercept, and (e) the equation in factored form.
Show the full solution
  1. (a) The curve crosses the x-axis at x = 1 and x = 3.
  2. (b) The lowest point of the curve is (2, −1).
  3. (c) The axis of symmetry is the vertical line through the vertex: x = 2.
  4. (d) The curve meets the y-axis at (0, 3).
  5. (e) Factored form uses the zeros: y = a(x − 1)(x − 3). Substitute (0, 3): 3 = a(−1)(−3) = 3a, so a = 1.
Final answer(a) 1 and 3   (b) (2, −1)   (c) x = 2   (d) (0, 3)   (e) y = (x−1)(x−3)
2Core
Write y = x² − 6x + 11 in vertex form by completing the square, and state the vertex.
Show the full solution
  1. Take half the coefficient of x and square it: half of −6 is −3, and (−3)² = 9.
  2. Add and subtract 9 inside: y = (x² − 6x + 9) − 9 + 11.
  3. The bracket is a perfect square: y = (x − 3)² + 2.
  4. In vertex form y = a(x−h)² + k, the vertex is (h, k).
Final answery = (x − 3)² + 2; vertex (3, 2)
3Core
For y = −2(x + 1)² + 8, state the vertex, the direction of opening, the maximum or minimum value, and the x-intercepts.
Show the full solution
  1. Vertex form gives the vertex directly: h = −1 and k = 8, so the vertex is (−1, 8).
  2. Since a = −2 is negative, the parabola opens downward, so the vertex is a maximum.
  3. Maximum value is 8.
  4. For intercepts set y = 0: −2(x+1)² + 8 = 0 gives (x+1)² = 4, so x+1 = ±2.
  5. That gives x = 1 and x = −3.
Final answervertex (−1, 8); opens down; maximum 8; x-intercepts −3 and 1
4Core
12344812162024xymaximum(0, 1)Time (seconds)Height (metres)
The graph shows a ball's height h = −5t² + 20t + 1, where t is in seconds. (a) State the launch height. (b) Find the time and value of the maximum height algebraically. (c) Explain why only part of the parabola makes sense here.
Show the full solution
  1. (a) Launch height is the value when t = 0: h = 1 metre.
  2. (b) The t-coordinate of the vertex is t = −b÷(2a) = −20 ÷ (2 × −5) = 2 seconds.
  3. Substitute: h = −5(2)² + 20(2) + 1 = −20 + 40 + 1 = 21 metres.
  4. (c) Negative time has no meaning, and the ball stops when it hits the ground, so only t ≥ 0 up to the landing point is realistic.
  5. This is the difference between the mathematical model and the physical situation it represents.
Final answer(a) 1 m   (b) 21 m at t = 2 s   (c) Only t ≥ 0 until landing is physically meaningful
5Core
A rectangular field is fenced on three sides using 60 m of fencing, with a river forming the fourth side. Find the dimensions giving maximum area.
Show the full solution
  1. Let the two equal sides perpendicular to the river be x, so the parallel side is 60 − 2x.
  2. Area: A = x(60 − 2x) = −2x² + 60x.
  3. This is a downward parabola, so the maximum is at the vertex: x = −60 ÷ (2 × −2) = 15.
  4. The parallel side is 60 − 2(15) = 30.
  5. Maximum area = 15 × 30 = 450 m².
Final answer15 m by 30 m, giving an area of 450 m²
6Challenge
A parabola has zeros at x = −2 and x = 6 and passes through (1, −18). Find its equation in standard form.
Show the full solution
  1. Start in factored form using the zeros: y = a(x + 2)(x − 6).
  2. Substitute (1, −18): −18 = a(3)(−5) = −15a.
  3. So a = 1.2, giving y = 1.2(x + 2)(x − 6).
  4. Expand: (x+2)(x−6) = x² − 4x − 12.
  5. Multiply through by 1.2: y = 1.2x² − 4.8x − 14.4.
Final answery = 1.2x² − 4.8x − 14.4

MPM2D Quadratic Relations & Graphing — common questions

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

How do you find the vertex from standard form?

Use x = −b ÷ 2a to get the x-coordinate, then substitute it back into the equation to get the y-coordinate. Completing the square gives the same result.

What does the value of a tell you about a parabola?

A positive a opens upward with a minimum; a negative a opens downward with a maximum. The larger the size of a, the narrower the parabola.

More MPM2D practice — grade 10 academic

MPM2D Linear SystemsMPM2D Analytic GeometryMPM2D Factoring QuadraticsMPM2D Solving Quadratic EquationsAll MPM2D units
← Unit 3: Factoring QuadraticsUnit 5: Solving Quadratic Equations →
Practice is step one

The classes teach the method behind every one of these.

The solutions above show the steps. The classes teach how to think about the problem in the first place — interactive, and worked through at the student’s own pace. Try 3 complete classes free — no payment required to begin.

Try 3 Classes Free →