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# Math: Functions and Relations MCR3U

This course introduces some financial applications of mathematics, extends students' experiences with functions, and introduces second degree relations. Students will solve problems in personal finance involving applications of sequences and series; investigate properties and applications of trigonometric functions; develop facility in operating with polynomials, rational expressions, and exponential expressions; develop and understanding of inverses and transformations of functions; and develop facility in using function notation and in communicating mathematical reasoning. Students will also investigate loci and the properties and applications of conics. MCR3U has 25% to 30% more content than MCF3M, requiring that the pace be much faster than in MCF3M. MCR3U requires a very high degree of commitment on the part of the student. Prerequisite: Grade 10 Principles of Math MPM2D (Academic)

# Exponent Laws

Specific Topic General Topic School Date
Exponent Laws Fractions, Negatives, Roots Bayview Nov 2013
Exponent Laws Decimals, Negatives Bayview Nov 2013

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.

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

## Write each with a single positive exponent, or an integer, or a reduced fraction, where applicable. (Without using a calculator)

$\sqrt[n]{a^m} = (\sqrt[n]{a})^m = a^{\frac{m}{n}}$

## Simplify given two of the most basic rules for roots.

\begin{align} & \sqrt{a^2} = a \\ \\ & \sqrt{ab} = \sqrt{a} × \sqrt{b} \end{align}

# Functions

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

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

# Rational Expressions

## When a soccer ball is kicked the parabolic path of its vertical height, h, in meters is given by the equation below, where the time, t, is in seconds.

$h(t) = -5t^2 + 15t + 0.1$

# Trigonometry

## Given the trig equation,

$4\text{cos}^2\,\theta - 4\text{cos}\,\theta + 1 = 0$

# Periodic Trig Functions

## Consider a graph of the parent function between the interval: $0 \le x \le 3\pi$

$y = \text{cos}\,\theta$

## Given the trig function below:

ƒ(x) = -4cos 2(θ + $\pi$) - 4

## Determine the features below given the function:

$P(t) = 4\sin\left(x - \pi\right) + 10$

## A periodic function expresses the period as a horizontal stretch or compression, with the value k according to the formula:

$k = \dfrac{2\pi}{period}$

## The average monthly maximum temperature of a certain city can be modeled by the periodic function below where T(t) is the temperature in ˚Celsius, and t is the time in months, where t = 0 represents January 1, t = 1 represents February 1, etc.

$T(t) = 7\,\text{sin}\dfrac{\pi}{6}(t - 3) + 25$

## A ferris wheel starts rotating from a point at ground level, represented by the equation below where H(t) is height above the ground, in meters, and t is time, in seconds.

$H(t) = -20\,\text{cos}\dfrac{\pi}{30}(t) + 20$

# Exponential Functions

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

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

## Given the function for compounding below, where 'i' is initial amount, 'r' is rate of interest, 't' is time, and 'n' is compounding period:

$A = i\, \left(1 + \dfrac{\left(\tfrac{r}{100}\right)}{n}\right)^{n\cdot t}$

# Financial Mathematics

### Calculate the monthly payments on a $300,000 mortgage over 25 years at 4% compounded monthly. Solution$ Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable Givens: PV = 300,000 t = 25 years n = 12 i = 0.04 \begin{align} PV & = \dfrac{R[1 - (1 + \tfrac{i}{n})^{-n\,×\,t}]}{\tfrac{i}{n}} \\ \\ 300,000 & = \dfrac{R[1 - (1 + \tfrac{0.04}{12})^{-12\,×\,25}]}{\tfrac{0.04}{12}} \\ \\ 300,000(\tfrac{0.04}{12}) & = R[1 - (\tfrac{301}{300})^{-300}] \\ \\ R & = \dfrac{1000}{0.63151} \\ \\ R & = 1,583.51 \\ \\ \end{align} Monthly payments are1,583.51

### If the driver put $3540 down on the car, calculate the original price of the car. Solution Video$ Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable Loans are present value (PV)... PV = Original Price - Down Payment = OP - 3540 \begin{align} PV & = \dfrac{R[1 - (1 + \tfrac{i}{n})^{-n\,×\,t}]}{\tfrac{i}{n}} \\ \\ OP - 3540 & = \dfrac{500[1 - (1 + \tfrac{0.029}{12})^{-12\,×\,4}]}{\tfrac{0.029}{12}} \\ \\ OP & = \dfrac{500[1 - (1 + \tfrac{0.029}{12})^{-12\,×\,4}]}{\tfrac{0.029}{12}} + 3540 \\ \\ OP & = 22,634.51 + 3540 \\ \\ OP & = 26,174.51 \\ \\ \end{align} The value of the loan was22,634.51 and the original selling price of the car including the down payment was $26,174.51 ### Calculate the interest paid on the loan. Solution Video$ Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable Interest Paid = Sum of Monthly Payments - Present Value of Loan \begin{align} & = (500/mo)(48\ mo) - 22,634.51 \\ \\ & = 24,000 - 22,634.51 \\ \\ & = 1,365.49 \\ \\ \end{align} The interest paid to use the car for 48 months is $1,365.49 (Aside: part of this can be written off on taxes when the car is used only for your business) ##### Mortgage Precision of Different Compound Periods ### A mortgage on a home is$250,000 at 3.5% compounded semiannually, with a 25 year amortization period. Determine the monthly payments. Solution \$ Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable Mortgages typically have semi-annual compounding, while the payments are monthly. This is an exception that is allowed. First determine the interest rate on a monthly basis (for monthly payments). A 3.5% rate is 1.75% every 6 months. Convert this rate to a monthly rate after 6 months: \begin{align} 1 + \tfrac{i}{n} & = (1 + i)^6 \\ \\ 1 + \tfrac{0.035}{2} & = (1 + i)^6 \\ \\ 1.0175 & = (1 + i)^6 \\ \\ \sqrt[6]{1.0175} & = \sqrt[6]{(1 + i)^6 } \\ \\ \sqrt[6]{1.0175} & = 1 + i \\ \\ i & = \sqrt[6]{1.0175} - 1 \\ \\ i & = 0.0028956 \\ \\ \end{align} n = 12 (monthly), t = 25, Solve for R... We have determined i = 0.0028956 so we use a slightly different equation... \begin{align} PV & = \dfrac{R[1 - (1 + i)^{-n\,×\,t}]}{i} \\ \\ PV(i) & = R[1 - (1 + i)^{-n\,×\,t}] \\ \\ R & = \dfrac{PV(i)}{[1 - (1 + i)^{-n\,×\,t}]} \\ \\ R & = \dfrac{250,000(0.0028956)}{[1 - (1 + 0.0028956)^{-12\,×\,25}]} \\ \\ R & = \dfrac{723.906}{[1 - (1.0028956)^{-300}]} \\ \\ R & = 1,248.18 \\ \\ \end{align}

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