Unit 4 · Ontario MPM2D

Ontario Grade 10 Quadratic Relations Practice Questions

The shape of a parabola. Vertex form y=a(x−h)²+k tells you the vertex, axis and direction right away. Practise switching forms, spotting quadratics with second differences, completing the square, and reading a max or min.

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What you need to know

Vertex form — vertex (h,k), axis x=h, a sets stretch and direction.
Second differences — constant second differences = quadratic.
Completing the square — halve b, add the square, keep it balanced.
Optimization — the vertex gives the max or min directly.

Practice questions with solutions

8 questions. Try each one on paper, then open the worked solution.

Question 1 · More Parabola Properties

For the quadratic y=13(x+1)2−4y = \frac{1}{3}(x + 1)^2 - 4y=31​(x+1)2−4, state the vertex, axis of symmetry, stretch/compression, and direction of opening.

Show worked solution

In vertex form y = a(x - h)² + k:
h = -1, k = -4, a = 1/3.
Vertex: (-1, -4)
Axis of symmetry: x = -1
Since |a| = 1/3 < 1, vertically compressed by factor 1/3.
Since a > 0, opens up.

Answer: vertex=(-1,-4), axis=x=-1, compressed by 1/3, opens up

Question 2 · More Parabola Properties

For the quadratic y=2(x+5)(x+1)y = 2(x + 5)(x + 1)y=2(x+5)(x+1), state the x-intercepts, axis of symmetry, and vertex.

Show worked solution

x-intercepts: x = -5 and x = -1.
Axis of symmetry: x = (-5 + (-1))/2 = -3.
Substitute x = -3: y = 2(-3 + 5)(-3 + 1) = 2(2)(-2) = -8.
Vertex: (-3, -8).

Answer: x-intercepts=-5 and -1, axis=x=-3, vertex=(-3,-8)

Question 3 · Properties of Parabolas

For the quadratic y = (x - 5)² + 4, state the vertex, axis of symmetry, and direction of opening.

Show worked solution

In vertex form y = a(x - h)² + k:
h = 5, k = 4
Vertex: (5, 4)
Axis of symmetry: x = 5
Since a = 1 > 0, the parabola opens upward.

Answer: vertex=(5,4), axis of symmetry=x=5, opens up

Question 4 · Properties of Parabolas

For the quadratic y = -3(x + 1)² - 3, state the vertex, axis of symmetry, stretch/compression, and direction of opening.

Show worked solution

In vertex form y = a(x - h)² + k:
h = -1, k = -3, a = -3
Vertex: (-1, -3)
Axis of symmetry: x = -1
Since |a| = 3 > 1, vertically stretched by factor of 3
Since a = -3 < 0, opens downward.

Answer: vertex=(-1,-3), axis of symmetry=x=-1, stretched by 3, opens down

Question 5 · Properties of Parabolas

A parabola has x-intercepts -8 and 2, and passes through the point (0, -8). Determine the equation in factored form y = a(x - r)(x - s).

Show worked solution

x-intercepts are -8 and 2, so:
y = a(x - (-8))(x - 2) = a(x + 8)(x - 2)

Use point (0, -8):
-8 = a(0 + 8)(0 - 2)
-8 = a(8)(-2)
-8 = -16a
a = 1/2

y = 1/2(x + 8)(x - 2)

Answer: y = 1/2(x + 8)(x - 2)

Question 6 · Quadratic Optimization Applications

The path of a rocket is given by h = -5(x - 2)(x - 12), where x is horizontal distance in meters and h is height in meters.

Show worked solution

Let x be the horizontal distance in meters, and h be the height in meters.
Formula/model:
h = -5(x - 2)(x - 12)
Solve:
a) The x-intercepts are 2 and 12. The axis of symmetry is x = (2 + 12)/2 = 7.
b) Substitute x = 7: h = -5(7 - 2)(7 - 12) = 125.
c) The rocket hits the ground at the positive x-intercept, which is x = 12.
Therefore, a) 7 meters, b) 125 meters, c) 12 meters

Answer: a) 7 meters, b) 125 meters, c) 12 meters

Question 7 · Quadratic Optimization Applications

A farmer wants to enclose a rectangular field adjacent to a straight river using 120 m of fencing. No fencing is required along the river. Find the maximum area of the field in square metres.

Show worked solution

Let w be the width of the field perpendicular to the river, and l be the length parallel to the river.
Formula/model:
2w + l = 120
A = w · l
Solve:
From the perimeter equation, l = 120 - 2w. Substitute to get A = w(120 - 2w) = -2w² + 120w.
Complete the square: A = -2(w² - 60w) = -2(w² - 60w + 900 - 900) = -2(w - 30)² + 1800.
The vertex is (30, 1800), giving a maximum area of 1800.
Therefore, 1800

Answer: 1800

Question 8 · Quadratic Optimization Applications

A bus company carries 400 passengers a day at a ticket price of 2.00. A survey shows that for each 0.10 increase in ticket price, the company will lose 10 passengers. What ticket price will maximize daily revenue in dollars?

Show worked solution

Let x be the number of $0.10 price increases.
Formula/model:
P(x) = 2.00 + 0.10x
N(x) = 400 - 10x
R(x) = P(x) · N(x)
Solve:
R(x) = (2.00 + 0.10x)(400 - 10x) = -x² + 20x + 800.
Complete the square: R(x) = -(x² - 20x) + 800 = -(x - 10)² + 900.
The vertex is at (10, 900), so x = 10 increases maximize revenue.
The optimal price is P(10) = 2.00 + 0.10(10) = 3.00.
Therefore, 3.00

Answer: 3.00

Common mistakes

Reading the vertex sign wrong: y=(x−3)² has vertex at +3.
Not factoring out a when completing the square with a≠1.
Mixing first and second differences.
Using factoring when completing the square is quicker.

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Ontario Grade 10 Quadratic Relations Practice Questions

What you need to know

Practice questions with solutions

Common mistakes

Quadratic Relations & Transformations Unit Test

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