Grade 10 Principles of Mathematics (MPM2D)
MPM2D is the gateway academic mathematics course in Ontario's secondary system. Success here unlocks MCR3U in Grade 11. The course develops three major mathematical strands: quadratic relations — including the parabola in all its forms — analytic geometry, and the foundations of trigonometry. Every topic builds the algebraic fluency needed for the advanced courses ahead.
1. Quadratic Relations: Vertex Form
A quadratic relation has the form $y = ax^2 + bx + c$ where $a \neq 0$. Its graph is a parabola. The most useful form for understanding the geometry of a parabola is the vertex form.
Vertex Form of a Quadratic
$$y = a(x - h)^2 + k$$
- The vertex of the parabola is at $(h, k)$.
- The axis of symmetry is the vertical line $x = h$.
- If $a > 0$, the parabola opens upward (minimum at vertex).
- If $a < 0$, the parabola opens downward (maximum at vertex).
- $|a|$ controls the width: larger $|a|$ means narrower parabola.
Converting to Vertex Form by Completing the Square
Worked Example 1.1 — Completing the Square
Write $y = 2x^2 - 12x + 11$ in vertex form and state the vertex.
Step 1 Factor out the leading coefficient from the $x$-terms: $$y = 2(x^2 - 6x) + 11$$
Step 2 Complete the square inside the bracket. Half of $-6$ is $-3$; $(-3)^2 = 9$. Add and subtract 9 inside: $$y = 2(x^2 - 6x + 9 - 9) + 11 = 2\bigl[(x-3)^2 - 9\bigr] + 11$$
Step 3 Distribute the 2: $$y = 2(x-3)^2 - 18 + 11 = 2(x-3)^2 - 7$$
The vertex is at $(3, -7)$, the axis of symmetry is $x = 3$, and the parabola opens upward (minimum value of $-7$).
Transformations of $y = x^2$
Starting from the basic parabola $y = x^2$, vertex form shows us how transformations work:
- $y = x^2 + k$: vertical shift up $k$ units (or down if $k < 0$)
- $y = (x - h)^2$: horizontal shift right $h$ units (or left if $h < 0$)
- $y = ax^2$: vertical stretch by factor $|a|$ (also reflects over $x$-axis if $a < 0$)
2. Factoring Quadratic Expressions
Factoring converts a quadratic from standard form into a product of factors. This directly reveals the $x$-intercepts (roots or zeros) of the parabola.
Common Factoring Strategies
- Common factor: Remove the greatest common factor (GCF) first.
- Difference of squares: $a^2 - b^2 = (a+b)(a-b)$
- Simple trinomials ($a=1$): $x^2 + bx + c = (x + p)(x + q)$ where $p + q = b$ and $pq = c$.
- Complex trinomials ($a \neq 1$): Use decomposition (also called the $ac$-method).
- Perfect square trinomial: $a^2 + 2ab + b^2 = (a+b)^2$
Worked Example 2.1 — Simple Trinomial
Factor $x^2 - 3x - 28$.
We need two numbers that multiply to $-28$ and add to $-3$. The pair is $-7$ and $4$.
$$x^2 - 3x - 28 = (x - 7)(x + 4)$$
The zeros are $x = 7$ and $x = -4$, so the parabola crosses the $x$-axis at these points.
Worked Example 2.2 — Decomposition ($a \neq 1$)
Factor $6x^2 + 7x - 3$.
Step 1 Multiply $a \cdot c = 6 \times (-3) = -18$. Find two numbers that multiply to $-18$ and add to $7$: that's $9$ and $-2$.
Step 2 Rewrite the middle term using these numbers: $$6x^2 + 9x - 2x - 3$$
Step 3 Group and factor: $$= 3x(2x + 3) - 1(2x + 3) = (3x - 1)(2x + 3)$$
Worked Example 2.3 — Solving a Quadratic by Factoring
Solve $x^2 + 2x - 15 = 0$.
Step 1 Factor: $(x + 5)(x - 3) = 0$.
Step 2 Apply the zero product property: $x + 5 = 0$ or $x - 3 = 0$.
$$x = -5 \quad \text{or} \quad x = 3$$
3. The Quadratic Formula and Discriminant
Not all quadratics factor over the integers. The quadratic formula gives the exact roots for any quadratic equation $ax^2 + bx + c = 0$ ($a \neq 0$).
The Quadratic Formula
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
The expression $D = b^2 - 4ac$ is called the discriminant. It tells us about the nature of the roots without solving:
- $D > 0$: two distinct real roots (parabola crosses $x$-axis twice)
- $D = 0$: one repeated real root (parabola touches $x$-axis once — vertex on axis)
- $D < 0$: no real roots (parabola does not cross $x$-axis)
Worked Example 3.1 — Using the Quadratic Formula
Solve $3x^2 - 5x - 2 = 0$ using the quadratic formula.
Step 1 Identify: $a = 3$, $b = -5$, $c = -2$.
Step 2 Compute the discriminant: $D = (-5)^2 - 4(3)(-2) = 25 + 24 = 49$.
Step 3 Apply the formula: $$x = \frac{-(-5) \pm \sqrt{49}}{2(3)} = \frac{5 \pm 7}{6}$$
Step 4 Two solutions: $x = \dfrac{5 + 7}{6} = 2$ or $x = \dfrac{5 - 7}{6} = -\dfrac{1}{3}$.
Worked Example 3.2 — Applying the Discriminant
Determine the nature of the roots of $2x^2 - 4x + 5 = 0$.
$D = (-4)^2 - 4(2)(5) = 16 - 40 = -24 < 0$.
Since $D < 0$, there are no real roots. The parabola does not intersect the $x$-axis.
4. Analytic Geometry: Distance, Midpoint, and Circles
Distance and Midpoint Formulas
For two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$:
Distance: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Midpoint: $M = \left(\dfrac{x_1 + x_2}{2},\, \dfrac{y_1 + y_2}{2}\right)$
Worked Example 4.1 — Median of a Triangle
Triangle $ABC$ has vertices $A(2, 1)$, $B(8, 3)$, and $C(4, 9)$. Find the length of the median from $A$ to side $BC$.
Step 1 Find the midpoint $M$ of $BC$: $M = \left(\dfrac{8+4}{2}, \dfrac{3+9}{2}\right) = (6, 6)$.
Step 2 Find the distance $AM$: $$AM = \sqrt{(6-2)^2 + (6-1)^2} = \sqrt{16 + 25} = \sqrt{41} \approx 6.40 \text{ units}$$
Equation of a Circle
Circle Centred at the Origin
The equation of a circle with centre $(0, 0)$ and radius $r$ is: $$x^2 + y^2 = r^2$$
For a circle centred at $(h, k)$: $$(x - h)^2 + (y - k)^2 = r^2$$
Worked Example 4.2 — Circle Equation
A circle has centre $(3, -2)$ and passes through the point $(7, 1)$. Find its equation.
Step 1 Find the radius using the distance formula: $$r = \sqrt{(7-3)^2 + (1-(-2))^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$
Step 2 Write the equation: $(x - 3)^2 + (y + 2)^2 = 25$.
Right Bisector (Perpendicular Bisector)
The right bisector of a line segment passes through the midpoint and is perpendicular to the segment. Key steps: find the midpoint, find the slope of the segment, then use the negative reciprocal slope with the midpoint to write the equation.
5. Trigonometry: SOHCAHTOA and the Sine and Cosine Rules
Primary Trigonometric Ratios (Right Triangles)
SOHCAHTOA
For an acute angle $\theta$ in a right triangle:
$$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} \qquad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan\theta = \frac{\text{opposite}}{\text{adjacent}}$$
To find an unknown angle from a ratio, use the inverse trig functions: $\theta = \sin^{-1}$, $\cos^{-1}$, or $\tan^{-1}$.
Worked Example 5.1 — Right Triangle Problem
In right triangle $PQR$ with the right angle at $Q$, $PQ = 8$ cm and $PR = 13$ cm. Find angle $P$ and the length of $QR$.
Step 1 Find $QR$ using the Pythagorean theorem: $QR = \sqrt{13^2 - 8^2} = \sqrt{169 - 64} = \sqrt{105} \approx 10.25$ cm.
Step 2 Find angle $P$: $\cos P = \dfrac{PQ}{PR} = \dfrac{8}{13}$, so $P = \cos^{-1}\!\left(\dfrac{8}{13}\right) \approx 52.0°$.
Sine Rule (Law of Sines)
The Sine Rule
For any triangle with sides $a$, $b$, $c$ opposite to angles $A$, $B$, $C$ respectively: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
Use the sine rule when you know: two angles and one side (AAS or ASA), or two sides and a non-included angle (SSA — watch for the ambiguous case).
Worked Example 5.2 — Applying the Sine Rule
In triangle $ABC$, $\angle A = 42°$, $\angle B = 73°$, and $a = 15$ cm. Find side $b$.
Step 1 Note $\angle C = 180° - 42° - 73° = 65°$ (not needed here).
Step 2 Apply the sine rule: $$\frac{b}{\sin B} = \frac{a}{\sin A} \Rightarrow \frac{b}{\sin 73°} = \frac{15}{\sin 42°}$$ $$b = \frac{15 \sin 73°}{\sin 42°} = \frac{15 \times 0.9563}{0.6691} \approx 21.4 \text{ cm}$$
Cosine Rule (Law of Cosines)
The Cosine Rule
$$c^2 = a^2 + b^2 - 2ab\cos C$$
Equivalently, $\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$.
Use the cosine rule when you know: two sides and the included angle (SAS), or all three sides (SSS).
Worked Example 5.3 — Applying the Cosine Rule
Two sides of a triangle are 10 cm and 14 cm and the included angle is 58°. Find the third side.
Solution $$c^2 = 10^2 + 14^2 - 2(10)(14)\cos 58°$$ $$= 100 + 196 - 280(0.5299)$$ $$= 296 - 148.37 = 147.63$$ $$c = \sqrt{147.63} \approx 12.2 \text{ cm}$$
6. Practice Problems
Problem 1 — Vertex Form
Write $y = -x^2 + 6x - 5$ in vertex form. State the vertex and whether it is a maximum or minimum.
Show Solution
Factor out $-1$: $y = -(x^2 - 6x) - 5$.
Complete the square: $y = -(x^2 - 6x + 9 - 9) - 5 = -[(x-3)^2 - 9] - 5$.
$y = -(x-3)^2 + 9 - 5 = -(x-3)^2 + 4$.
Vertex: $(3, 4)$. Since $a = -1 < 0$, this is a maximum.
Problem 2 — Factoring
Factor completely: $4x^2 - 36$.
Show Solution
First remove the common factor: $4x^2 - 36 = 4(x^2 - 9)$.
Then apply difference of squares: $4(x-3)(x+3)$.
Problem 3 — Quadratic Formula
Solve $2x^2 + 3x - 7 = 0$. Leave your answer in exact form (simplified surd).
Show Solution
$a = 2, b = 3, c = -7$. Discriminant: $D = 9 + 56 = 65$.
$$x = \frac{-3 \pm \sqrt{65}}{4}$$
Problem 4 — Discriminant
For what value(s) of $k$ does $x^2 - 4x + k = 0$ have exactly one real root?
Show Solution
Set the discriminant equal to zero: $D = (-4)^2 - 4(1)(k) = 0$.
$16 - 4k = 0 \Rightarrow k = 4$.
Problem 5 — Analytic Geometry
Show that the triangle with vertices $P(1, 2)$, $Q(5, 6)$, and $R(9, 2)$ is isosceles.
Show Solution
$PQ = \sqrt{(5-1)^2 + (6-2)^2} = \sqrt{16+16} = 4\sqrt{2}$
$QR = \sqrt{(9-5)^2 + (2-6)^2} = \sqrt{16+16} = 4\sqrt{2}$
$PR = \sqrt{(9-1)^2 + 0^2} = 8$
Since $PQ = QR = 4\sqrt{2}$, the triangle is isosceles.
Problem 6 — Trigonometry
In a triangle, sides $a = 7$ m, $b = 9$ m, and $c = 11$ m. Find angle $C$ to the nearest degree.
Show Solution
$$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{49 + 81 - 121}{2(7)(9)} = \frac{9}{126} = \frac{1}{14}$$
$C = \cos^{-1}\!\left(\dfrac{1}{14}\right) \approx 86°$.
Problem 7 — Quadratic Word Problem
A ball is launched upward and its height in metres after $t$ seconds is $h(t) = -5t^2 + 20t + 2$. What is the maximum height and when is it reached?
Show Solution
Complete the square: $h = -5(t^2 - 4t) + 2 = -5(t^2 - 4t + 4 - 4) + 2 = -5(t-2)^2 + 20 + 2 = -5(t-2)^2 + 22$.
The vertex is $(2, 22)$: maximum height of 22 m reached at $t = 2$ seconds.
Problem 8 — Sine Rule
In triangle $XYZ$, $\angle X = 35°$, $x = 8.4$ cm, $y = 11.2$ cm. Find angle $Y$ (consider the ambiguous case).
Show Solution
Using the sine rule: $\dfrac{\sin Y}{y} = \dfrac{\sin X}{x}$.
$\sin Y = \dfrac{11.2 \sin 35°}{8.4} = \dfrac{11.2 \times 0.5736}{8.4} \approx 0.7648$.
$Y = \sin^{-1}(0.7648) \approx 49.9°$ or $Y \approx 130.1°$ (both are potentially valid — check by verifying the angle sum is less than 180°).
For $Y \approx 50°$: $\angle Z = 180 - 35 - 50 = 95°$ ✓. For $Y \approx 130°$: $\angle Z = 180 - 35 - 130 = 15°$ ✓. Both triangles exist.