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## 12

More homework, harder tests, and tough assignments, tutoring is all about an individualized plan that builds the academic skills, good habits and positive attitudes needed to succeed in high school and beyond.

This course extends students' experience with functions. Students will investigate the properties of polynomial, rational, logarithmic, and trigonometric functions; develop techniques for combining functions; broaden their understanding of rates of change; and develop facility in applying these concepts and skills. Students will also refine their use of the mathematical processes necessary for success in senior mathematics. This course is intended both for students taking the Calculus and Vectors course as a prerequisite for a university program and for those wishing to consolidate their understanding of mathematics before proceeding to any one of a variety of university programs. Prerequisite: Grade 11 Functions and Relations MCR3U

# Functions

## Given the function,

ƒ(x) = (3x2 + 8x + 4)(x2 - 9)

## Given the piecewise function,

$f(x) = \begin{cases} 4x + 1 & x < 1 \\ 3 & x \ge 1 \\ \end{cases}$

# Polynomial Functions

Specific Topic General Topic School Date
Solving a Volume Word Problem Factoring Polynomials North Toronto Sep 2013

## The degree of a polynomial equals the number of real plus unreal roots. This rule does not work well for linear polynomials. Anyways, determine the number of roots in each...

Degree = Real Roots + Unreal Roots

## Given the following formula for finite differences of a polynomial, where 'n' is the degree of the function,

Finite Difference = (Leading Coefficient) × (n!)

## The remainder theorem equation, in corresponding form is:

ƒ(x) = d(x)·q(x) + r(x)
dividend = (divisor)(quotient) + remainder
And in quotient form is: $\dfrac{ \text{dividend} }{ \text{divisor} } \ = \ \text{quotient} \ + \ \dfrac{ \text{remainder} }{ \text{divisor} }$

## Solve the following inequalities with interval notation. Remember for interval notation:

±∞ ... uses ... ( )
>, < ... uses ... ( )
≥, ≤ ... uses ... [ ]

# Rational Functions

## Solve the rational equation word problems, and remember the general form equation for problems on 'working together'.

$\dfrac{\text{Time Together}}{\text{Time for 'A'}} \ + \ \dfrac{\text{Time Together}}{\text{Time for 'B'}} \ = \ 1$

## Determine the 'x' and 'y' intercepts in coordinate form (x, y). Round your answers to one decimal place and list in increasing order, where applicable.

1) x-intercept(s): solve for the numerator
2) y-intercept: set x = 0 and solve for 'y'

## Determine the horizontal and vertical asymptotes of the following functions. (Order from low to high, where applicable)

The horizontal asymptote (HA) is always y = 0 + 'd' when the degree of the numerator is less than the degree of the denominator. (where 'd' is the vertical translation, not shown here)

## Determine the horizontal and vertical asymptotes of the following functions.

The horizontal asymptote (HA) is always $= \cfrac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} + d$ when the degree of the numerator is equal to the degree of the denominator (where 'd' is the vertical translation, not shown here)

## Determine the oblique (slant) and vertical asymptotes of the following functions.

There is no horizontal asymptote (HA) when the degree of the numerator is one more than the degree of the denominator.
The slant/oblique asymptote is the quotient of $=\cfrac{numerator}{denominator}$

## Determine where holes exist for the following functions.

Holes exist for all values where the numerator cancels with the denominator...

# Trigonometric Functions

## Given the function,

$f(x) = -\dfrac{1}{3} \cos \left(\dfrac{4}{3}\left(x - \dfrac{\pi}{4} \right)\right) - 2$

## Given the function below...

$5 = 13\sin \left( \dfrac{2\pi}{5} x + 1 \right) - 1.5$

## Determine the features of the function below. Reduce your answers fully.

$P(t) = 75\sin\left(\dfrac{5\pi}{12}t - 6\right) + 10$

## A function modelling the average temperature T, in degrees Celsius, in a certain ountry over the course of one year is given below where t is the time in days. Determine the following

$T(t) = 16\sin\left(\frac{2\pi}{365}(t - 99)\right) + 13$

# Trigonometric Equations and Identities

## Given the cofunction identity,

\begin{align} \sin\theta & = \cos(90˚ - \theta) \\ \sin\theta & = \cos\left(\dfrac{\pi}{2} - \theta \right) \end{align}

## Evaluate each of the following with an exact value, using the compound angle formulas.

 Angle (radians) Angle (degrees) sinx (exact value) cosx (exact value) $0 \pi$ 0˚ = 0 = 1 $\dfrac{\pi}{6}$ 30˚ = $\dfrac{1}{2}$ = $\dfrac{\sqrt{3}}{2}$ $\dfrac{\pi}{4}$ 45˚ = $\dfrac{1}{\sqrt{2}}$ = $\dfrac{1}{\sqrt{2}}$ $\dfrac{\pi}{3}$ 60˚ = $\dfrac{\sqrt{3}}{2}$ = $\dfrac{1}{2}$ $\dfrac{\pi}{2}$ 90˚ = 1 = 0 $\pi$ 180˚ = 0 = -1 $\dfrac{3\pi}{2}$ 270˚ = -1 = 0 $2\pi$ 360˚ = 0 = 1

## Evaluate each of the following with an exact value, using the compound angle formulas. These questions use the principle angles of related acute angles that are special triangle angles.

 Angle (degrees) Related Special Acute Angle sinx (exact value) cosx (exact value) 150˚ 30˚ = $\dfrac{1}{2}$ = $\dfrac{-\sqrt{3}}{2}$ 210˚ 30˚ = $\dfrac{-1}{2}$ = $\dfrac{-\sqrt{3}}{2}$ 330˚ 30˚ = $\dfrac{-1}{2}$ = $\dfrac{\sqrt{3}}{2}$ 135˚ 45˚ = $\dfrac{1}{2}$ = $-\dfrac{1}{2}$ 225˚ 45˚ = $\dfrac{-1}{2}$ = $\dfrac{-1}{\sqrt{2}}$ 315˚ 45˚ = $\dfrac{-1}{\sqrt{2}}$ = $\dfrac{1}{2}$ 120˚ 60˚ = $\dfrac{\sqrt{3}}{2}$ = $\dfrac{-1}{2}$ 240˚ 60˚ = $\dfrac{-\sqrt{3}}{2}$ = $\dfrac{-1}{\sqrt{2}}$ 300˚ 60˚ = $\dfrac{-\sqrt{3}}{2}$ = $\dfrac{1}{2}$

## Solve the trig equation on the interval: 0 ≤ x ≤ $\pi$

$\cos\left(\dfrac{\pi}{6}x\right) = \dfrac{1}{2}$

# Exponential & Logarithmic Functions

## Evaluate.

 Log Laws loga(1) = 0 loga(a) = 1 a(logab) = b loga(ab) = b -log(a) = log(1/a) log(ab) = log(a) + log(b) log(a/b) = log(a) - log(b)

## Given the change of base formula,

$\log_m(n) = \dfrac{\log_{b}(n)}{\log_{b}(m)}$

## Sound intensity is measured in units of decibels (dB), where 'I' is sound intensity and 'I0' is a reference intensity of the threshold of our hearing ability.

$dB = 10\log\left(\dfrac{\text{I}}{\text{I}_0}\right)$

# Combinations & Compositions of Functions

f(x) = 3x + 2
g(x) = -5x2 - 2x

## Determine each of the following, using the table given below.

 x g(x) -4 -30 -2 -10 -1 -3 0 0 1 -5 2 30 3 -135
 x ƒ(x) -3 5 -2 0 -1 -3 0 -4 1 -3 2 0 3 5

## For the functions,

\begin{align} g(x) = \dfrac{1}{2x + 5} \\ \\ f(x) = \sqrt{2x + 1} \end{align}

## Two functions are defined:

\begin{align} f(x) & = \sqrt{2x + 13} \\ \\ \\ g(x) & = 2x^2 \end{align}

## Given the equation

ƒ(x) = 3x + 4

### The two points (-1, -24), (-3, 0) lie on the combined function of, (n × m)(x). Determine the values, 'a' and 'b'. Solution n(x) = 2x2 + ax - 3 m(x) = bx2 - 5x + 3 Hint Clear Info a = b = Incorrect Attempts: CHECK Hint Unavailable The domain of the combined function is shared (the same) for both n(x) and m(x). This allows us to use the 'x' value from the combined function, '-1' and '-3', and substitute into both n(x) and m(x). We do this to get the 'y' values for n(x) and m(x). For the combined function, (n × m)(x), the 'y' values multiply so that: yn × ym = ycombined. We show this below. Substitute the 'x = -1' into the equations for n(x), \begin{align} n(x) & = 2(x)^2 + a(x) - 3 \\ \\ n(-1) & = 2(-1)^2 + a(-1) - 3 \\ \\ y_1 & = -a - 1 \end{align} \begin{align} n(x) & = 2(x)^2 + a(x) - 3 \\ \\ n(-3) & = 2(-3)^2 + a(-3) - 3 \\ \\ y_3 & = -3a + 15 \end{align} And the 'x = -3', into m(x), \begin{align} m(x) & = b(x)^2 - 5(x) + 3 \\ \\ m(-1) & = b(-1)^2 - 5(-1) + 3 \\ \\ y_2 & = b + 8 \end{align} \begin{align} m(x) & = b(x)^2 - 5(x) + 3 \\ \\ m(-3) & = b(-3)^2 - 5(-3) + 3 \\ \\ y_4 & = 9b + 18 \end{align} The combination of functions, (n × m)(x), is the product of the 'y' values of n(x) and m(x), yn × ym = ycombined: \begin{align} (n × m)(x) & = y_{c\ 1} \\ \\ y_1 × y_2 & = -24 \\ \\ (-a - 1)(b + 8) & = -24 \\ \\ -ab - 8a - b - 8 & = -24 \\ \\ -ab - 8a - b + 16 & = 0 \end{align} \begin{align} (n × m)(x) & = y_{c\ 2} \\ \\ y_3 × y_4 & = 0 \\ \\ (-3a + 15)(9b + 18) & = 0 \\ \\ -27ab - 54a + 135b + 270 & = 0 \end{align} Substitute one equation for another. You can isolate for 'a' in one equation \begin{align} -27ab - 54a + 135b + 270 & = 0 \\ \\ a(-27b - 54) & = -135b - 270 \\ \\ a & = \dfrac{-135b - 270}{-27b - 54} \\ \\ a & = \dfrac{135b + 270}{27b + 54} \end{align} And substitute it into the other, \begin{align} -ab - 8a - b + 16 & = 0 \\ \\ -b\left(\dfrac{135b + 270}{27b + 54}\right) - 8\left(\dfrac{135b + 270}{27b + 54}\right) - b + 16 & = 0 \\ \\ -b\left(135b + 270\right) - 8\left(135b + 270\right) - b\left(27b + 54\right) + 16\left(27b + 54\right) & = 0\left(27b + 54\right) \\ \\ 135b^2 + 270b + 1080b + 2160 + 27b^2 + 54b - 432b - 864 & = 0 \\ \\ 162b^2 + 972b + 1296 & = 0 \\ \\ 162(b^2 + 6b + 8) & = 0 \\ \\ 162(b + 4)(b + 2) & = 0 \\ \\ b & = -4,\ -2 \end{align} Plug in each 'b' value into one of the previous equations to solve for 'a' \begin{align} a & = \dfrac{135b + 270}{27b + 54} \\ \\ a & = \dfrac{135(-4) + 270}{27(-4) + 54} \\ \\ a & = \dfrac{270}{54} \\ \\ a & = 5 \end{align} \begin{align} a & = \dfrac{135b + 270}{27b + 54} \\ \\ a & = \dfrac{135(-2) + 270}{27(-2) + 54} \\ \\ a & = \dfrac{0}{0} \\ \\ a & = \text{UNDEFINED\DNE} \end{align} Therefore, b = -4, and a = 5 [b ≠ -2 since 'a' is undefined for that value].

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