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# Principles of Math MPM2D

This course enables students to broaden their understanding of relationships and extend their problem-solving and algebraic skills through investigation, the effective use of technology, and abstract reasoning. Students will explore quadratic relations and their applications; solve and apply linear systems; verify properties of geometric figures using analytic geometry; and investigate the trigonometry of right and acute triangles. Students will reason mathematically and communicate their thinking as they solve multi-step problems. Prerequisite: Grade 9 Principles of Math MPM1D

# Analytic Geometry

## Given the equation for a circle with a center at (h, k)...

\begin{align} (x - h)^2 + (y - k)^2 = r^2 \\ \\ \end{align}

## A volcano in Iceland erupts with center C(10, 14), the ash cloud formed has radius 3 km and can be modeled with the circle equation below. A weather drone plane flies on route y = 0.8x + 9.

\begin{align} (x - 10)^2 + (y - 14)^2 = 3^2 \\ \\ \end{align} # Geometric Properties

## Given the following points, determine the circumcenter of the triangle. (Hint: a circumcenter is where all the right bisectors intersect.)

L(2, 2)     M(1, 1)     N(0, -3)

A (5, 9)
B (-10, -9)
C (15, -10)

xa × xb = xa + b

xa ÷ xb = xa - b

(xa)b = xa × b

## Simplify fully to an integer or fraction using positive exponents, showing your work without using a calculator.

$a^{-n} = \left( \cfrac{1}{a^{\ +n}} \right) \quad\quad or \quad\quad \left( \cfrac{a}{b} \right)^{\!\!-n} = \left( \cfrac{b}{a} \right)^{\!\!+n}$

## Solve the following exponential equations, similar to the steps shown in the example below. Simplify fully.

\begin{align} 3^{(3x + 2)} & = 243 \\ 3^{(3x + 2)} & = 3^{(5)} \\ 3x + 2 & = 5 \\ 3x & = 5 - 2 \\ 3x & = 3 \\ x & = 1 \\ \\ \end{align}

## The table below shows the flight path of a ball, with distance and height in meters.

 Distance (x) Height (y) 0 1 1 8 2 13 3 16 4 17 5 16 6 13 7 8 8 1

## A basketball shot is taken from a horizontal distance of 5 m from the hoop. The height of the ball can be modeled by the equation below. Where h is height in meters, and t is time in seconds since the ball was released.

$h(t) = -7.3t^2 + 8.25t + 2.1$

## Given the general, vertex form of a quadratic.

$y = ±\,a(x - h)^2 + k$

## Given the general form of a quadratic, determine the equation of a quadratic function with the transformations listed below....

$y = a(x - h)^2 + k$

## Based on the value 'n' determine some of the horizontal transformations below.

$y = a\left[ n\left(x - h\right)\right]^2 + k$

## Find the vertex using each of the following methods: ## A quadratic function passes through the points (-1, 1), (0, -3), and (5, 1).

### Determine the equation in vertex form. Solution Hint Clear Info $\dfrac{\quad\quad}{\quad\quad} \Bigg(x \quad\quad\quad \Bigg)^2 - \dfrac{\quad\quad}{\quad\quad}$ Incorrect Attempts: CHECK Hint Unavailable First determine the Axis of Symmetry (AOS) of the quadratic. The AOS goes through the vertex and is the midpoint of two points on the same horizontal line: (-1, 1) and (5, 1). \begin{align} \text{AOS} & = \text{Midpoint of x} \\ \\ x & = \dfrac{x_1 + x_2}{2} \\ \\ x & = \dfrac{-1 + 5}{2} \\ \\ x & = 2 \\ \\ \end{align} Substitute AOS and the easiest point (0, -3) into the vertex form... This AOS represents 'h' in the equation... \begin{align} y & = a(x - h)^2 + k \\ \\ -3 & = a(0 - 2)^2 + k \\ \\ -3 & = 4a + k \\ \\ \end{align} Substitute AOS and any other point, (-1, 1)... \begin{align} 1 & = a(-1 - 2)^2 + k \\ \\ 1 & = 9a + k \\ \\ \end{align} Solve with elimination... \begin{align} & -3 = 4a + k \\ (-) \ & \underline {1 = 9a + k} \\ \\ & -4 = -5a \\ \\ & a = \dfrac{4}{5} \\ \\ \end{align} Now find the last value, 'k by subbing in the last point (5, 1)... \begin{align} y & = a(x - h)^2 + k \\ \\ y & = \dfrac{4}{5}(x - 2)^2 + k \\ \\ 1 & = \dfrac{4}{5}(5 - 2)^2 + k \\ \\ 1 & = \dfrac{36}{5} + k \\ \\ k & = -\dfrac{31}{5} \\ \\ \end{align} Here is your equation in vertex form, \begin{align} y & = \dfrac{4}{5}(x - 2)^2 - \dfrac{31}{5} \\ \\ \end{align}

Specific Topic General Topic School Date
The Correct Way to Expand a Squared Binomial Expanding Branksome Sep 2013
Factoring a Perfect Square Trinomial Quadratics Branksome Sep 2013
Factoring Unordered Trinomials Example 1 Quadratics Branksome Sep 2013
Factoring Unordered Trinomials Example 2 Quadratics Branksome Sep 2013

## Now that you understand more about factoring it's time to take the training wheels off - no more decomposition (aka splitting and grouping). Factor straight into the parenthesis like the example shown below.

$6x^2 + 11x + 3$ What multiplies to +18 and adds to +11? = +9 and +2 \begin{align} & = (2x \quad\quad )(3x \quad\quad ) \\ \\ & = ( \ \underbrace{2x + \underbrace{3 \ )( \ 3x}_{+9} + 1 }_{+2} \ ) \\ \\ \end{align}

## Factor the following perfect squaresfully, using any method (direct is encouraged). The general formulas for perfect squares are given below.

\begin{align} & a^2 + 2ab + b^2 \\ \\ & = (a)^2 + 2ab + (b)^2 \\ \\ & = (a + b)^2 \\ \\ \end{align} \begin{align} & a^2 - 2ab + b^2 \\ \\ & = (a)^2 - 2ab + (b)^2 \\ \\ & = (a - b)^2 \\ \\ \end{align}

## Given the general formula for a perfect square quadratic...

E.g. 1) (a)2x2 + 2(a)(b)x + (b)2
= (ax + b)2
E.g. 2) x2 + 6x + 9
= (a)2x2 + 2(a)(b)x + (b)2
= (1)2x2 + 2(1)(3)x + (3)2
= (1x + 3)2

## Factor the difference of squares fully (meaning common factor whenever possible). Show as much work in your notes as you need.

\begin{align} & 1x^2 - 9 \\ \\ & = (1x + 3)(1x - 3) \\ \\ \end{align} \begin{align} & 75a^2 - 27 \\ \\ & = 3(25a^2 - 9) \\ \\ & = 3\left[ (5a)^2 - (3)^2 \right] \\ \\ & = 3(5a + 3)(5a - 3) \\ \\ \end{align}

## A triangular sandbox is surrounded by a rectangular field of grass. The dimensions of the sandbox and field are listed below, in terms of x.

 Base (cm) Height (cm) Length (cm) Width (cm) Triangle 2x x + 4 Rectangle x + 1 4x - 4

### Determine the most simplified expression for the total surface area of a cube with side lengths equal to (2x - 3). Solution ⁰¹²³⁴⁵⁶⁷⁸⁹⁻⁺⁽⁾₀₁₂₃₄₅₆₇₈₉₋₊₍₎ Incorrect Attempts: CHECK Hint Unavailable There are 6 equal faces on a cube.The area of each face is (2x - 3)(2x - 3) = (2x - 3)2 The total surface area = 6(area) = 6(2x - 3)2

Specific Topic General Topic School Date
Solving for X by Completing the Square Factoring Quadratics Branksome Sep 2013

## Given the equation already in factored form.

\begin{align} y = x(2x - 3) \\ \\ \end{align}

## Determine the solutions, hence solve. Use the zero-product property or complete the square where most applicable. Order solutions from lowest to highest... ## The revenue, expense, and profit functions are given below, where x is the number of items sold.

\begin{align} & Revenue = 5x^2 + 200 \\ \\ & Expense = 35x^2 - 130x + 100 \\ \\ & Profit = Revenue \ - \ Expense \end{align}

## The cost of operating a machine, in dollars, is given by the formula below where t is time, in hours, that the machine operates.

C(t) = 2t2 - 16t + 682

# Trigonometry with Right Triangles

## The interior angles of a triangle are given in terms of x in the diagram below. Base = x
Height = x + 7
Hypotenuse = 17

## A very tall ladder is placed 20 m from the base of a building of unknown height, at an angle of elevation of 60˚ to the top of the building. ## Gold coins are at the bottom of a diving pool, which is 14 m deep. A diver is underwater on the bottom of the pool, 20 m away from the coins and a swimmer is at the edge of the pool. The angle of depression from the swimmer to the coins is 45 degrees. ## Some people are trying to measure the height of a super-tall statue. Point B is S 45˚ W from the bottom of the statue, and point C is E 45˚ S from the bottom. Let point A be the top of the statue and point D the bottom. ∆ABD and ∆ACD are in the vertical plane while ∆BDC is in the horizontal plane. The distance between point B and C is measured at 68 m. # Trigonometry with Acute and Obtuse

## Given the complex shape, determine the missing quantities stated below. ### Explain how the Cosine law is based on the Pythagorean theorem.  Solution Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable Pythagorean Theorem $a^2 + b^2 = c^2$ Cosine law when angle (A˚) = 90˚ \begin{align} a^2 & = b^2 + c^2 - 2bc \cos A˚ \\ \\ a^2 & = b^2 + c^2 - 2bc \cos (90˚) \\ \\ a^2 & = b^2 + c^2 - 2bc \cancelto{0}{\cos (90˚)} \\ \\ a^2 & = b^2 + c^2 - \cancelto{0}{2bc (0)} \\ \\ a^2 & = b^2 + c^2 \\ \\ \end{align} The Cosine law is the Pythagorean theorem when the angle is 90˚... Percent complete:
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