Review of the Basics: Product Rule

Write each with a single, positive exponent, showing your work without using a calculator.

$3^2 × 3^5$ Solution

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= 3⁷

$5^2 × 5^{-2}$ Solution

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= 5^{2 + (-2)}

= 5⁰

= 1

$(-10)^{-5} × (-10)^7$ Solution

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= (-10)^{-5 + 7}

= (-10)²

$\left(\dfrac{2}{3}\right)^4 × \left(\dfrac{2}{3}\right)^3$ Solution

Hint Clear Info

$\bigg( \dfrac{2}{3} \bigg)$

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$= \left(\dfrac{2}{3}\right)^{4+3} \\ \\ \left(\dfrac{2}{3}\right)^7$

Review of the Basics: Quotient Rule

Simplify, showing your work without using a calculator.

$7^7 \div 7^3$ Solution

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= 7^{7 - 3}

= 7⁴

$(-2)^{-3} \div (-2)^8$ Solution

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= (-2) ^-11

$(-5)^{6} \div (-5)^{-4}$ Solution

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= (-5)^{6 - (-4)}

= (-5)¹⁰

Review of the Basics: Power of a Power

Write each with a single, positive exponent, or reduced fraction, showing your work without using a calculator.

$\left(3^2\right)^4$ Solution

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= 3^{2 × 4}

= 3⁸

$\left(2^2\right)^3$ Solution

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= 2^{2 × 3}

= 2⁶

$\left(\left(2^2\right)^3\right)^2$ Solution

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2^{2 × 3 × 2}

= 2¹²

$\left(\cfrac{1}{2}\right)^3$ Solution

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$\begin{align} & = \cfrac{1^3}{2^3} \\ \\ & = \cfrac{1}{2^3} \\ \\ & = \cfrac{1}{8} \\ \\ \end{align}$

$\left(\cfrac{3}{2^2}\right)^3$ Solution

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$\begin{align} & = \cfrac{3^3}{2^{2 × 3}} \\ \\ & = \cfrac{3^3}{2^6} \\ \\ & = \cfrac{27}{64} \\ \\ \end{align}$

Review of the Basics: Negative Powers

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

$3^{-2}$ Solution

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$\begin{align} & = \dfrac{1}{3^2} \\ \\ & = \dfrac{1}{9} \end{align}$

$-4^{-3}$ Solution

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$\begin{align} & = -4^{-3} \\ \\ & = \dfrac{1}{-4^3} \\ \\ & = \dfrac{-1}{64} \\ \\ \end{align}$

$\left( \dfrac{3}{4} \right)^{-1}$ Solution

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$\quad = \left( \dfrac{4}{3} \right)^{+1} = \dfrac{4}{3}$

$\left( \dfrac{2}{1} \right)^{-2}$ Solution

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$\quad \left( \dfrac{1}{2} \right)^{2} = \dfrac{1^2}{2^2} = \dfrac{1}{4}$

$\quad \left( \dfrac{2}{3} \right)^{-3}$ Solution

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$\quad \left( \dfrac{3}{2} \right)^{3} = \dfrac{3^3}{2^3} = \dfrac{27}{8}$

$\quad \cfrac{2}{3^{-2}}$ Solution

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$\begin{align} & = \cfrac{2 × 3^{2}}{1} \\ \\ & = \cfrac{2 × 9}{1} \\ \\ & = 18 \\ \\ \end{align}$

$\quad \left(9^{-7}\right)^0$ Solution

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$\quad \left(9^{-7 × 0}\right) = 9^0 = 1$

New Law: Rational Exponents & Roots/Radicals

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

$\sqrt{x}$ Solution

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$\begin{align} & = x^{\tfrac{1}{2}} \\ \\ \end{align}$

$\sqrt[3]{x}$ Solution

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$\begin{align} & \sqrt[3]{x} \\ \\ & = x^{\tfrac{1}{3}} \\ \\ \end{align}$

$\sqrt{x^5}$ Solution

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$\begin{align} & \sqrt{x^5} \\ \\ & = (x^5)^{\tfrac{1}{2}} \\ \\ & = (x)^{5 × \tfrac{1}{2}} \\ \\ & = x^{\tfrac{5}{2}} \\ \\ \end{align}$

$\sqrt[4]{x^3}$ Solution

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$\begin{align} & \sqrt[4]{x^3} \\ \\ & = (x^3)^{\tfrac{1}{4}} \\ \\ & = (x)^{3×\tfrac{1}{4}} \\ \\ & = x^{\tfrac{3}{4}} \\ \\ \end{align}$

$(-27)^{\tfrac{1}{3}}$ Solution

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$\begin{align} & (-27)^{\tfrac{1}{3}} \\ \\ & = (-3^3)^{\tfrac{1}{3}} \\ \\ & = (-3)^{3×\tfrac{1}{3}} \\ \\ & = (-3)^{\cancel{3}×\tfrac{1}{\cancel{3}}} \\ \\ & = (-3)^1 \\ \\ & = -3 \\ \\ \end{align}$

$(8)^{\tfrac{4}{3}}$ Solution

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$\begin{align} & = (2^3)^{\tfrac{4}{3}} \\ \\ & = (2)^{3×\tfrac{4}{3}} \\ \\ & = (2)^{\cancel{3}×\tfrac{4}{\cancel{3}}} \\ \\ & = (2)^{×\tfrac{4}{1}} \\ \\ & = 2^4 \\ \\ & = 16 \\ \\ \end{align}$

$(64)^{-\tfrac{5}{6}}$ Solution

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$\begin{align} & = (2^6)^{-\tfrac{5}{6}} \\ \\ & = (2)^{\cancel{6}×-\tfrac{5}{\cancel{6}}} \\ \\ & = (2)^{-\tfrac{5}{1}} \\ \\ & = (2)^{-5} \\ \\ & = \left(\dfrac{1}{2}\right)^{5} \\ \\ & = \dfrac{1^5}{2^5} \\ \\ & = \dfrac{1}{2^5} \\ \\ \end{align}$

$\left(\dfrac{16}{25}\right)^{\tfrac{3}{2}}$ Solution

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$\begin{align} & = \left(\dfrac{4^2}{5^2}\right)^{\tfrac{3}{2}} \\ \\ & = \left(\dfrac{4^{2 × \tfrac{3}{2}}}{5^{2 × \tfrac{3}{2}}}\right)^{\tfrac{3}{2}} \\ \\ & = \left(\dfrac{4^{\cancel{2} × \tfrac{3}{\cancel{2}}}}{5^{\cancel{2} × \tfrac{3}{\cancel{2}}}}\right) \\ \\ & = \left(\dfrac{4^{\tfrac{3}{1}}}{5^{\tfrac{3}{1}}}\right) \\ \\ & = \dfrac{4^3}{5^3} \\ \\ \end{align}$

Rational Expressions with Exponent Laws

Simplify fully, using positive exponents.

Click to show Exponent Laws Solution

$\quad (3x^3)(x^4)(2x^{-2})$ Solution

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(3)(2)x^{3 + 4 + (-2)}

= (6)x^{7 - 2}

= 6x⁵

$\dfrac{(x^2y^3)(-2x)}{4x^3}$ Solution

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$\begin{align} & = \dfrac{-2 x^{2 + 1} y^3}{4x^3} \\ \\ & = \dfrac{-\cancel{2} x^{3} y^3}{^2\!\cancel{4}x^3} \\ \\ & = \dfrac{-1 x^{3} y^3}{2x^3} \\ \\ & = \dfrac{-1 \cancel{x^{3}} y^3}{2\cancel{x^{3}}} \\ \\ & = \dfrac{-y^3}{2} \\ \\ \end{align}$

$\cfrac{(4x^3y^2)^4}{-3x^2y^2x^3}$ Solution

$\dfrac{3x^6y}{x^3y\,^{-2}}$ Solution

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$\begin{align} & = 3x^{6 - 3}y^{1 - (- 2)} \\ \\ & = 3x^3y \\ \\ \end{align}$

$\cfrac{(27x^{-1}y^4)^{-2}}{(81y^{-3}x^2)^{-1}}$ Solution

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$\begin{align} & = \dfrac{27^{-2}}{81^{-1}} \, \dfrac{x^{-2 × -1} y^{4 × -2}}{y^{-3 × - 1} x^{2 × -1}} \\ \\ & = \dfrac{3^{3(-2)}}{3^{4(-1)}} \, \dfrac{x^2 y^{-8}}{y^3 x^{-2}} \\ \\ & = \dfrac{3^{-6}}{3^{-4}} x^{2 - (-2)} y^{-8 - 3} \\ \\ & = 3^{-6 - (-4)} x^{0} y^{-11} \\ \\ & = 3^{-2} y^{-11} \\ \\ & = \dfrac{1}{3^2 y^{11}} \\ \\ \end{align}$

$\ \ \ \Bigg( \dfrac{ \Big( \big( x^3 \big)^2 \Big)^1}{x} \Bigg)^2$ Solution

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$\begin{align} & = \dfrac{x^{3 × 2 × 1 × 2}}{x^2} \\ \\ & = \dfrac{x^12}{x^2} \\ \\ & = x^{12 - 2} \\ \\ & = x^{10} \\ \\ \end{align}$

$\ \ \ \Bigg(\!\dfrac{\big(i\,^4\big)\big(i\,^3\big)}{\big(i\,^2\big)}\!\Bigg)^2$ Solution

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$\begin{align} & = \dfrac{i^{2(4 + 3)}}{i^{2 × 2}} \\ \\ & = \dfrac{i^{14}}{i^4} \\ \\ & = i^{14 - 4} \\ \\ & = i^{10} \\ \\ \end{align}$

Roots and Radicals

Simplify given two of the most basic rules for roots.

Simplify the root, $\sqrt{704}$ by dragging the steps into the correct order. Solution

$=8\sqrt{11}$

$=\sqrt{64 × 11}$

$\sqrt{704}$

$=\sqrt{8^{\cancel{2}}} × \sqrt{11}$

$=\sqrt{64} × \sqrt{11}$

Find the greatest perfect square (like 8²) under the root to bring it out.

Simplify $\sqrt{75}$. Solution

Hint Clear Info

$\quad\sqrt{ \quad \ }$

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$\begin{align} & = \sqrt{75} \\ \\ & = \sqrt{25 × 3} \\ \\ & = \sqrt{25} × \sqrt{3} \\ \\ & = \sqrt{5^2} × \sqrt{3} \\ \\ & = 5 \sqrt{3} \\ \\ \end{align}$

Practice with Exponent Laws

Write each with a single positive exponent, or integer, or reduced fraction, showing your work without using a calculator.

$(-27x^3y^6)^{^{\tfrac{1}{3}}}$ Solution Video

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$\begin{align} & = -27^{\tfrac{1}{3}} x^{^{\tfrac{1}{3}} × 3} y^{^{\tfrac{1}{3}} × 6} \\ \\ & = -(3^3)^{\tfrac{1}{3}} x^1 y^2 \\ \\ & = -(3)^{3 × \tfrac{1}{3}} x^1 y^2 \\ \\ & = -(3)^{1} x^1 y^2 \\ \\ & = -3xy^2\\ \\ \end{align}$

$\cfrac{(4)^{\tfrac{1}{8}}}{(4)^{\tfrac{3}{8}} \cdot (4)^{\tfrac{1}{4}}}$ Solution Video

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$\begin{align} & = \cfrac{(4)^{\tfrac{1}{8}}}{(4)^{\tfrac{3}{8} + \tfrac{1}{4}} } \\ \\ & = \cfrac{(4)^{\tfrac{1}{8}}}{(4)^{\tfrac{5}{8} } } \\ \\ & = (4)^{\tfrac{1}{8} - \tfrac{5}{8} } \\ \\ & = (4)^{ \tfrac{-4}{8} } \\ \\ & = (2^2)^{ \tfrac{-1}{2} } \\ \\ & = (2)^{2 × \tfrac{-1}{2} } \\ \\ & = 2^{-1} \\ \\ & = \dfrac{1}{2} \\ \\ \end{align}$

$\cfrac{(a^5b^6)^{-2}}{(a^2b^3)^{-3}}$ Solution

$\begin{align} & = \cfrac{(a^5b^6)^{-2}}{(a^2b^3)^{-3}} \\ \\ & = \cfrac{ a^{5 \,× -2} b^{6 \,× -2} }{ a^{2 \,× -3} b^{3 \,× -3} } \\ \\ & = \cfrac{ a^{-10} b^{-12} }{ a^{-6} b^{-9} } \\ \\ & = a^{-10 - (-6)} b^{-12 - (-9)} \\ \\ & = a^{-4} b^{-3} \\ \\ & = \dfrac{1}{a^{4} b^{3}} \\ \\ \end{align}$

$25^{^{\tfrac{1}{2}}} - 16^{^{\tfrac{3}{4}}} + \sqrt[3]{27} + 9^{^{0.5}}$ Solution

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$\begin{align} & = (5^2)^{^{\tfrac{1}{2}}} - (2^4)^{^{\tfrac{3}{4}}} + \sqrt[3]{3^3} + (3^2)^{^{0.5}} \\ \\ & = (5)^{^ 2 × {\tfrac{1}{2}}} - (2)^{4 × ^{\tfrac{3}{4}}} + (3)^{3 × ^{\tfrac{1}{3}}}+ (3)^{2 × ^{0.5}} \\ \\ & = (5)^1 - (2)^3 + (3)^1 + (3)^1 \\ \\ & = 5 - 8 + 3 + 3 \\ \\ & = 3 \\ \\ \end{align}$

$3^5 \cdot 4^5 \cdot 12^{-4}$ Solution

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$\begin{align} & = 3^5 \cdot 4^5 \cdot (3 × 4)^{-4} \\ \\ & = 3^5 \cdot 4^5 \cdot 3^{-4} \cdot 4^{-4} \\ \\ & = 3^{5 + (-4)} \cdot 4^{5 + (-4)} \\ \\ & = 3^1 \cdot 4^1 \\ \\ & = 12 \\ \\ \end{align}$

$\cfrac{15^2}{5^4\cdot4^1}\cdot\left(\cfrac{10}{3}\right)^2$ Solution Video

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$\left(8^4 + 8^3\right)\left(3^{-4} \cdot 8^{-3}\right)$ Solution

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$\begin{align} & = \left(8^{4 + (-3)} + 8^{3 + (-3)}\right)\left( 3^{-4} \right) \\ \\ & = \left(8^1 + 8^{0}\right)\left( 3^{-4} \right) \\ \\ & = \left(8 + 1\right)\left( 3^{-4} \right) \\ \\ & = \left(9\right)\left( 3^{-4} \right) \\ \\ & = \left(3^2\right)\left( 3^{-4} \right) \\ \\ & = 3^{2 + (-4)}\\ \\ & = 3^{-2}\\ \\ & = \dfrac{1}{3^2} \\ \\ & = \dfrac{1}{9} \\ \\ \end{align}$

$\quad (2^6 - 2^4)(2^{-4} - 2^{-2})$ Solution

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FOIL: $\begin{align} & = (2^6)(2^{-4}) + (2^6)(-2^{-2}) + (- 2^4)(2^{-4}) + (- 2^4)(- 2^{-2}) \\ \\ & = 2^{6 + (-4)} - 2^{6 + (-2)} - 2^{4 + (-4)} + 2^{4 + (-2)} \\ \\ & = 2^{2} - 2^{4} - 2^{0} + 2^{2} \\ \\ & = 4 - 16 - 1 + 4 \\ \\ & = -9 \\ \\ \end{align}$

Function Output

ƒ(x) = x² is the same as y = x² Solution

1. True
2. False

In function notation, ƒ(x) is a way of representing 'y'. Furthermore the function notation ƒ(x) shows the certain 'x' value that is plugged into the function to obtain the value of 'y'.

Input and Output Notation of a Function

It is important to make the distinction between the input and output of a function.

The output of the following functions would be the same. Solution ƒ(2) = x² + 2x                    ƒ(2) = n² + 2n                    y = x² + 2x

1. True
2. False

Any input would result in the same output regardless of the different variables used in the functions: x² + 2x and n² + 2n

Function notation uses ƒ(x) instead of y, but they mean the same thing. Think of ƒ(x) as just, y.

Function notation is useful to show substitutions, for example: $\begin{align} f(x) & = x^2 + 2x \\ \\ f(2) & = (2)^2 + 2(2) \\ \\ f(2) & = 4 + 4 \\ \\ f(2) & = 8 \\ \\ y & = 8 \\ \\ \end{align}$ (2, 8)

Which output is true for the function ƒ(x) = 3x² Solution

1. ƒ(2) = 4
2. ƒ(2) = 7
3. ƒ(2) = 6
4. ƒ(2) = 12

ƒ(2) = 12

• (2) is the input of the independent variable
• 12 is the output of the dependent variable

The input in the following function is: Solution $\begin{align} f(x) & = 4x^2 - 2x + 1 \\ \\ f(3) & = 4(3)^2 - 2(3) + 1 \\ \\ \end{align}$

1. 1
2. 2
3. 3
4. 4

The input of the independent variable is the (3)...

Evaluate. Solution $\begin{align} f(x) & = 2(2x + 1)(x - 3) \\ \\ f(4) & = \ ... \end{align}$

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Using function notation, sub the x-value (x = 4) into the function: $\begin{align} f(4) & = 2(2x + 1)(x - 3) \\ \\ & = 2(2(4) + 1)((4) - 3) \\ \\ & = 2(8 + 1)(4 - 3) \\ \\ & = 2(9)(1) \\ \\ & = 2(9) \\ \\ & = 18 \\ \\ \end{align}$ So, ƒ(4) = 18, or y = 18...

In the function below, x is the dependent variable and y is the independent variable. Solution y = 3x² + 1

1. True
2. False

It is the other way around.

x is the independent variable and y is the dependent variable.

Review from Grade 10: Factoring Trinomials when a = 1

Factor fully.

$x^2 + 4x + 4$ Solution

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$\begin{align} & x^2 + 4x + 4 \\ \\ & = (x + 2)(x + 2) \\ \\ & = (x + 2)^2 \\ \\ \end{align}$

$x^2 - 6x + 5$ Solution

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$\begin{align} & x^2 - 6x + 5 \\ \\ & = (x - 1)(x - 5) \\ \\ \end{align}$

Review from Grade 10: Factoring Trinomials when a ≠ 1

Factor by decomposition, or directly (if you are able).

$2x^2 - 13x + 6$ Solution

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What multiplies to (2)(6) = 12? ==> -12 and -1
What adds to -13? ==> -12 and -1 $\begin{align} & 2x^2 - 13x + 6 \\ \\ & = 2x^2 - 12x - 1x + 6 \\ \\ & = 2x(x - 6) - 1(x - 6) \\ \\ & = (2x - 1)(x - 6) \\ \\ \end{align}$

$10x^2 + 17x + 6$ Solution

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What multiplies to (10)(6) = 60? ==> 5 and 12
What adds to 17? ==> 5 and 12 $\begin{align} & 10x^2 + 17x + 6 \\ \\ & = 10x^2 + 5x + 12x + 6 \\ \\ & = 5x(2x + 1) + 6(2x + 1) \\ \\ & = (5x + 6)(2x + 1) \\ \\ \end{align}$

$3x^2 + 5x - 42$ Solution

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Review from Grade 10: Factoring Trinomials when a = 1, and x⁴

Factor using any method.

$x^4 - 9x^2 - 10 = 0$ Solution

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$x^4 + 4x^2 + 4$ Solution

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$x^4 + 7x^2y^2 + 12y^4$ Solution

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$\begin{align} & x^4 + 7x^2y^2 + 12y^4 \\ \\ & = x^4 + 3x^2y^2 + 4x^2y^2 + 12y^4 \\ \\ & = x^2(x^2 + 3y^2) + 4y^2(x^2 + 3y^2) \\ \\ & = (x^2 + 3y^2)(x^2 + 4y^2) \\ \\ \end{align}$

$16x^4 - 81$ Solution

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There are two rounds of differences of squares in here... $\begin{align} & = 16x^4 - 81 \\ \\ & = (4x^2 + 9)(4x^2 - 9) \\ \\ & = (4x^2 + 9)(2x + 3)(2x - 3) \\ \\ \end{align}$

Factoring: Difference of Squares and Fractions

Factor fully.

$x^2 - 4$ Solution

Hint Clear Info

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$\begin{align} & x^2 - 4 \\ \\ & = (x + 2)(x - 2) \\ \\ \end{align}$

$\dfrac{1}{2}x^2 = 2x - 2$ Solution

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$\begin{align} \dfrac{1}{2}x^2 & = 2x - 2 \\ \\ \dfrac{1}{2}x^2 - 2x + 2 & = 0 \\ \\ \dfrac{1}{2}x^2(2) - 2x(2) + 2(2) & = 0(2) \\ \\ x^2 - 4x + 4 & = 0 \\ \\ (x - 2)(x - 2) & = 0 \\ \\ (x - 2)^2 & = 0 \\ \\ \end{align}$

$\dfrac{900}{x} = \dfrac{900}{x\ +\ 6} + 25$ Solution

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$\begin{align} \dfrac{900}{x} & = \dfrac{900}{x\ +\ 6} + \dfrac{25(x\ +\ 6)}{(x\ +\ 6)} \\ \\ \dfrac{900}{x} & = \dfrac{900\ +\ 25(x\ +\ 6)}{x\ +\ 6} \\ \\ \dfrac{900}{x} & = \dfrac{25x\ +\ 1050}{x\ +\ 6} \\ \\ (900)(x\ +\ 6) & = (25x\ +\ 1050)(x) \\ \\ 900x + 5400 & = 25x^2 + 1050x \\ \\ 0 & = 25x^2 + 150x - 5400 \\ \\ 0 & = 25(x^2 + 6x - 216) \\ \\ 0 & = 25(x + 18)(x - 12) \\ \\ \end{align}$

$3x^2 - 27 = 0$ Solution

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$\begin{align} 3x^2 - 27 & = 0 \\ \\ 3(x^2 - 9) & = 0 \\ \\ 3(x + 3)(x - 3) & = 0 \\ \\ \end{align}$

$x^4 - y^2$ Solution

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$\begin{align} & x^4 - y^2 \\ \\ & = (x^2 + y)(x^2 - y) \\ \\ \end{align}$

$25x^4 - 9y^2$ Solution

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$\begin{align} & 25x^4 - 9y^2 \\ \\ & = (5x^2 + 3y)(5x^2 - 3y) \\ \\ \end{align}$

Tricky Factoring

Factor fully.

x² - y² + 4y - 4 Solution

Hint Clear Info

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See that you can factor the last three terms -(y² - 4y + 4)... $\begin{align} & x^2 - y^2 + 4y - 4 \\ \\ & = x^2 -(y^2 - 4y + 4) \\ \\ & = x^2 - (y - 2)^2 \\ \\ & = [x + (y - 2)][x - (y - 2)] \\ \\ & = (x + y - 2)(x - y + 2) \\ \\ \end{align}$

16x² - 72x + 81 - 9y² + 12y - 4 Solution

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See that you can factor the first 3 terms, and separately factor the last 3 terms. $\begin{align} & = 81 - 72x + 16x^2 - 9y^2 + 12y - 4 \\ \\ & = 81 - 72x + 16x^2 - (9y^2 - 12y + 4) \\ \\ & = (9 - 4x)^2 - (3y - 2)^2 \\ \\ & = [(9 - 4x) + (3y - 2)][(9 - 4x) - (3y - 2)] \\ \\ & = (-4x + 3y + 7)(-4x - 3y + 11) \\ \\ \end{align}$

Inverse Equations

Determine the inverse of the following equations.

$y = -\dfrac{2}{5}x + 1$ Solution

y =

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$\dfrac{}{\quad\quad}x \ + \ \dfrac{}{\quad\quad}$

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$\begin{align} y & = -\dfrac{2}{5}x + 1 \\ \\ x & = -\dfrac{2}{5}y + 1 \\ \\ x - 1 & = -\dfrac{2}{5}y \\ \\ -\dfrac{2}{5}y & = x - 1 \\ \\ -2y & = 5(x - 1) \\ \\ -2y & = 5x - 5 \\ \\ y & = \dfrac{-5}{2}x + \dfrac{5}{2} \\ \\ \end{align}$

$y = (x - 5)^2$ Solution

y =

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$\quad\sqrt{ \quad \ }$

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$\begin{align} y & = (x - 5)^2 \\ \\ x & = (y - 5)^2 \\ \\ \sqrt{x} & = \sqrt{(y - 5)^2} \\ \\ y - 5 & = \sqrt{x} \\ \\ y & = \sqrt{x} + 5 \\ \\ \end{align}$

$y = \sqrt{x + 4}$ Solution

y =

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$\begin{align} y & = \sqrt{x + 4} \\ \\ x & = \sqrt{y + 4} \\ \\ x^2 & = (\sqrt{y + 4})^2 \\ \\ x^2 & = y + 4 \\ \\ y & = x^2 - 4 \\ \\ y & = (x + 2)(x - 2) \\ \\ \end{align}$

$y = (2x - 3)^2 - 9$ Solution

Hint Clear Info

$\dfrac{±\sqrt{x \quad\quad} \quad\quad}{\quad}$

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$\begin{align} y & = (2x - 3)^2 - 9 \\ \\ x & = (2y - 3)^2 - 9 \\ \\ x + 9 & = (2y - 3)^2 \\ \\ \pm \sqrt{x + 9} & = 2y - 3 \\ \\ 2y - 3 & = \pm \sqrt{x + 9} \\ \\ 2y & = \pm \sqrt{x + 9} + 3 \\ \\ y & = \dfrac{\pm \sqrt{x + 9} + 3}{2} \\ \\ \end{align}$

Equations in Vertex Form

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

Reflection in the x-axis, vertical stretch by a factor of 3, vertical translation 5 units up. Solution

1. $y = -⅓(x)^2 - 5$
2. $y = -3(x + 5)^2$
3. $y = ⅓(-x)^2 - 5$
4. $y = -3(x)^2 + 5$
5. $y = 3(x - 5)^2$

$y = -3(x)^2 + 5$

A parabola with vertex at (-1, 3), opening upward, and with a vertical compression by a factor of ⅛. Solution $y = a(x - h)^2 + k$

1. $y = \dfrac{1}{8}(x - 1)^2 + 3$
2. $y = \dfrac{1}{8}(x + 1)^2 + 3$
3. $y = \dfrac{1}{8}(x + 1)^2 - 3$
4. $y = 8(x - 1)^2 + 3$

$y = \dfrac{1}{8}(x + 1)^2 + 3$

Finding the Vertex by Completing the Square

Determine the vertex for each of the following by completing the square.

ƒ(x) = -3x² + 9x - 1 Solution Video

Hint Clear Info

$\bigg( \dfrac{\quad}{\quad}, \dfrac{\quad}{\quad} \bigg)$

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$\begin{align} f(x) & = -3x^2 + 9x - 1 \\ \\ & = -3(x^2 - 3x) - 1 \\ \\ & = -3\left(x^2 - 3x + \frac{9}{4} - \frac{9}{4}\right) - 1 \\ \\ & = -3\left(x^2 - 3x + \frac{9}{4}\right) - 1 + 3\left(\frac{9}{4}\right) \\ \\ & = -3\left(x^2 - 3x + \frac{9}{4}\right) - 1 + \left(\frac{27}{4}\right) \\ \\ & = -3\left(x^2 - 3x + \frac{9}{4}\right) - \frac{4}{4} + \frac{27}{4} \\ \\ & = -3\left(x^2 - 3x + \frac{9}{4}\right) + \frac{23}{4} \\ \\ & = -3\left(x - \frac{3}{2}\right)\left(x - \frac{3}{2}\right) + \frac{23}{4} \\ \\ & = -3\left(x - \frac{3}{2}\right)^2 + \frac{23}{4} \\ \\ \\ \\ \\ \\ & \text{Vertex:} \left(\frac{3}{2}, \frac{23}{4}\right) \end{align}$

ƒ(x) = -2x² + 8x + 4 Solution

Hint Clear Info

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$\begin{align} f(x) & = -2x^2 + 8x + 4 \\ \\ & = -2(x^2 + -4x) + 4 \\ \\ & = -2(x^2 + -4x + 4 - 4) + 4 \\ \\ & = -2(x^2 + -4x + 4) + (-4)(-2) + 4 \\ \\ & = -2(x - 2)(x - 2) + 8 + 4 \\ \\ & = -2(x - 2)^2 + 12 \\ \\ \end{align}$ (2, 12)

Finding the Vertex by Completing the Square (With Fractions!)

Determine the vertex for each of the following by completing the square (without decimals).

y = -3x² + 5x - 2 Solution

Hint Clear Info

$\bigg( \dfrac{\quad}{\quad}, \dfrac{\quad}{\quad} \bigg)$

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$\begin{align} y & = -3\left( x^2 - \frac{5}{3}x + \underline{ \ \quad \ } - \underline{ \ \quad \ } \right) - 2 \\ \\ & = -3\left[ x^2 - \frac{5}{3}x + (\frac{5}{2(3)})^2 - (\frac{5}{2(3)})^2 \right] - 2 \\ \\ & = -3\left( x^2 - \frac{5}{3}x + \frac{25}{36} - \frac{25}{36} \right) - 2 \\ \\ & = -3\left( x^2 - \frac{5}{3}x + \frac{25}{36} \right) - 2 + \left( -3 \right)\left( - \frac{25}{36} \right) \\ \\ & = -3\left( x - \frac{5}{6} \right)^2 - \frac{24}{12} + \frac{25}{12} \\ \\ & = -3\left( x - \frac{5}{6} \right)^2 + \frac{1}{12} \\ \\ \\ \\ & \text{Vertex:} \left(\frac{5}{6}, \frac{1}{12}\right) \end{align}$

ƒ(x) = ⅕x² + 2x + 4 Solution

Hint Clear Info

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$\begin{align} & = \frac{1}{5} \left( x^2 + 10x + \underline{ \ \quad \ } - \underline{ \ \quad \ } \right) + 4 \\ \\ & = \frac{1}{5} \left[ x^2 + 10x + (\frac{10}{2})^2 - (\frac{10}{2})^2 \right] + 4 \\ \\ & = \frac{1}{5} \left( x^2 + 10x + \frac{100}{4} \right) + 4 + (\frac{1}{5})(-\frac{100}{4}) \\ \\ & = \frac{1}{5} \left( x^2 + 10x + 25 \right) + 4 + (\frac{1}{5})(-25) \\ \\ & = \frac{1}{5} \left( x + 5 \right)^2 + 4 - 5 \\ \\ & = \frac{1}{5} \left( x + 5 \right)^2 - 1 \\ \\ \end{align}$ (-5, 1)

Reciprocal Functions (Basics with Inputs and Outputs)

State the domain and range of the reciprocal function: $f(x) = \dfrac{1}{x}$ Solution

Domain: {x | x ≠ 0}

Range: {y | y ≠ 0}

Determine ƒ(5) in the function $f(x) = \dfrac{3x - 5}{2 - x}$ Solution

1. $\dfrac{1}{2}$
2. $\dfrac{2}{5}$
3. $-\dfrac{8}{7}$
4. $-\dfrac{10}{3}$

$\begin{align} f(x) & = \dfrac{3x - 5}{2 - x} \\ \\ f(5) & = \dfrac{3(5) - 5}{2 - (5)} \\ \\ & = \dfrac{15 - 5}{2 - 5} \\ \\ & = \dfrac{10}{-3} \\ \\ & = -\dfrac{10}{3} \\ \\ \end{align}$

Determine ƒ(x + 2) in the function $f(x) = \dfrac{2x - 4}{4 - 2x}$ Solution

Hint Clear Info

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$\begin{align} f(x) & = \dfrac{2x - 4}{4 - 2x} \\ \\ f(x + 2) & = \dfrac{2(x + 2) - 4}{4 - 2(x + 2)} \\ \\ & = \dfrac{2(x) + 2(2) - 4}{4 - 2(x) - 2(2)} \\ \\ & = \dfrac{2x + 4 - 4}{4 - 2x - 4} \\ \\ & = \dfrac{2x}{-2x} \\ \\ & = -1 \\ \\ \end{align}$

Rational Expressions with Multiplication and Division of Monomials

Simplify (and state the restrictions in your notes - answer enter not available).

$\dfrac{3ab}{4a} × \dfrac{8a}{9b}$ Solution

Hint Clear Info

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$\begin{align} & = \dfrac{3\cancel{a}\cancel{b}}{4\cancel{a}} × \dfrac{8a}{9\cancel{b}} \\ \\ & = \dfrac{3}{^1\cancel{4}} × \dfrac{^2\cancel{8}a}{9} \\ \\ & = \dfrac{^1\cancel{3}}{1} × \dfrac{2a}{^3\cancel{9}} \\ \\ & = \dfrac{2a}{3} \\ \\ \end{align}$ a ≠ 0, b ≠ 0

(Even though 'b' cancels, you still have to state the restriction on it. (Any variable that was ever in the denominator has a restriction).

$\dfrac{40a^2b^3}{3b} × \dfrac{21ab}{10a^3}$ Solution

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$\begin{align} & = \dfrac{^4\cancel{40}a^2b^3}{\cancel{3}b^1} × \dfrac{^7\cancel{21}a^1b^1}{\cancel{10}a^3} \\ \\ & = \dfrac{4a^{2+1-3}b^{3-1+1}}{1} × \dfrac{7}{1} \\ \\ & = \dfrac{28a^0b^3}{1} \\ \\ & = 28b^3 \\ \\ \end{align}$ a ≠ 0, b ≠ 0

(Even though 'b' cancels, you still have to state the restriction on it. (Any variable that was ever in the denominator has a restriction).

$\dfrac{16m^2n}{9n} ÷ \dfrac{4m}{3}$ Solution

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Multiply by the reciprocal, simplify the coefficients, and reduce the variables, $\begin{align} & = \dfrac{^4\bcancel{16}m^2n}{^3\cancel{9}n} \times \dfrac{^1\cancel{3}}{^1\bcancel{4}m} \\ \\ & = \dfrac{4m^2n}{3n} \times \dfrac{1}{m} \\ \\ & = \dfrac{4m^2n^1}{3m^1n^1} \\ \\ & = \dfrac{4m^{2-1}n^{1-1}}{3} \\ \\ & = \dfrac{4m^1\cancelto{1}{n^0}}{3} \\ \\ & = \dfrac{4m}{3} \\ \\ \end{align}$ n ≠ 0, m ≠ 0

You state a restriction on 'm' even if it is on the top? → Yes, $÷ \dfrac{m}{1}$ would have become $× \dfrac{1}{m}$ if we took the time to write it and you can see the variable 'm' is in the denominator so the restriction has to be stated.

$\dfrac{14x^2}{9} ÷ \dfrac{49}{3x}$ Solution

Rational Expressions with Multiplication and Division of Monomials with Negative Exponents

Simplify and state the restrictions.

$\dfrac{15a^{-2}b^3}{3ab^{-1}}$ Solution

$\begin{align} & = \dfrac{15}{3}a^{-2 - 1}b^{3 - (- 1)} \\ \\ & = 5a^{-3}b^4 \\ \\ & = \dfrac{5b^4}{a^3} \\ \\ \end{align}$ a ≠ 0, b ≠ 0

$\dfrac{(3a^2b^{-1})^{-2}}{b^{-2}}$ Solution

$\begin{align} & = \dfrac{3^{-2}a^{2(-2)}b^{-1(-2)}}{b^{-2}} \\ \\ & = \dfrac{3^{-2}a^{-4}b^2}{b^{-2}} \\ \\ & = 3^{-2}a^{-4}b^{2 - (-2)} \\ \\ & = 3^{-2}a^{-4}b^4 \\ \\ & = \dfrac{b^4}{3^2a^4} \\ \\ \end{align}$ a ≠ 0, b ≠ 0

$\dfrac{21y^{-2}}{11x^2} × \dfrac{22x^3}{7y(2x^{-1})}$ Solution

$\begin{align} & = \dfrac{21}{7} × \dfrac{22}{11} × \dfrac{1}{2} y^{-2 - (-1)} x^{3 - 2 - (-1)} \\ \\ & = 3 × 2 × \dfrac{1}{2}y^{-3}x^{2} \\ \\ & = \dfrac{3x^2}{y} \\ \\ \end{align}$ x ≠ 0, y ≠ 0

$\dfrac{18x^2y^{-1}}{25xy} ÷ \dfrac{3x^{-2}y}{5x^2y^2}$ Solution Video

$\begin{align} & = \dfrac{18x^2y^{-1}}{25xy} ÷ \dfrac{3x^{-2}y}{5x^2y^2} \\ \\ & = \dfrac{18x^2y^{-1}}{25xy} × \dfrac{5x^2y^2}{3x^{-2}y} \\ \\ & = \dfrac{18}{3} × \dfrac{5}{25} x^{2 + 2 - 1 - (-2)} y^{-1 + 2 - 1 - 1} \\ \\ & = 6 × \dfrac{1}{5} x^5 y^{-1} \\ \\ & = \dfrac{6x^5}{5y} \\ \\ \end{align}$ x ≠ 0, y ≠ 0

Rational Expression Simplification

The following is simplified fully.

$\dfrac{(x + 3)(x - 1) + 2(x - 1)}{3(x - 1)(x + 5)}$ Solution

1. True
2. False

The top has two terms (separated by + or -), so you should to distribute, the collect like terms, then factor... $\begin{align} & = \dfrac{x^2 + 2x - 3 + 2x - 2}{3(x - 1)(x + 5)} \\ \\ & = \dfrac{x^2 + 4x - 5}{3(x - 1)(x + 5)} \\ \\ & = \dfrac{\bcancel{(x + 5)}\cancel{(x - 1)}}{3\cancel{(x - 1)}\bcancel{(x + 5)}} \\ \\ & = \dfrac{1}{3} \end{align}$ (restrictions: x ≠ 1, -5)

Rational Expressions: Restrictions

State the restrictions for the following rational expressions:

$\dfrac{7x\ +\ 1}{(x\ +\ 2)(x\ -\ 3)(4x\ +\ 2)}$ Solution

$x \ne -2, +3, -\dfrac{1}{2}$

$\dfrac{2x}{(x\ +\ 5)^3(x\ -\ 2)^2}$ Solution

$x \ne -5, +2$

$\dfrac{9x\ +\ 2}{(x^2\ -\ y^2)(x\ -\ 2)} × \dfrac{3x\ +\ 2}{x\ (2x\ +\ 3)}$ Solution

$x \ne -y, +y, +2, 0, -\dfrac{3}{2}$

$\dfrac{11\ +\ 5x}{(4x^2\ -\ 25)(3x^2)} \div \dfrac{3\ +\ 3x}{(7x\ -\ 1)}$ Solution Video

$x \ne -\dfrac{5}{2}, +\dfrac{5}{2}, 0, -1, \dfrac{1}{7}$

(Don't forget to state a restriction on anything that was ever a denominator, so even though the 7x-1 gets flipped to the top, it still has to be included in the restriction list!)

Transformations and Translations of Functions with y = af[k(x - d)] + c

Determine the transformations and translations in the following functions, and any asymptotes (if exists).

y = 2(x - 2)² + 1 Solution

y = 7(2(x - 3))² - 3 Solution

y = -2(3x + 6)² + 0 Solution

$y = -\sqrt{x + 4} - 2$ Solution

• Vertical reflection (on the x-axis)
• Horizontal shift 4 units left
• Vertical shift 2 units down

$y = 2\sqrt{4x + 3} - 5$ Solution

$y = 2\sqrt{4(x + \dfrac{3}{4})} - 5$

Transformation: Vertical stretch by 2

Transformation: k = 4, Horizontal stretch (compression) by factor of 1/k = 1/4

Translation: 3/4 units left

Translation: 5 units down

$y = \dfrac{3}{2x\ +\ 1} + 7$ Solution

$y = \dfrac{3}{2(x + \dfrac{1}{2})} + 7$

Vertical stretch by a factor of 3/2

Translation 7 units up

Translation 1/2 units left

Asymptotes at y = 7, and x = -1/2

$y = \dfrac{1}{(x\ +\ 1)(x\ -\ 2)}$ Solution

Asymptotes at y = 0, and x = -1, +2

$y = \dfrac{5}{(2x\ +\ 1)(x\ +\ 4)}$ Solution

Asymptotes at y = 0, and x = -1/2, - 4

$y = \dfrac{5}{2x^2\ +\ 6x\ +\ 4}$ Solution

$\begin{align} y & = \dfrac{5}{2x^2\ +\ 6x\ +\ 4} \\ \\ & = \dfrac{5}{(2x + 2)(x + 2)} \\ \\ & = \dfrac{5}{2(x + 1)(x + 2)} \\ \\ \end{align}$ Asymptotes at y = 0, and x = -1, -2

$3f\Big(-\dfrac{x}{4} - 2\Big) + 3$ Solution

$\begin{align} & = 3f\Big(-\dfrac{x}{4} - 2\Big) + 3 \\ \\ & = 3f\Big(-\dfrac{1}{4}\left(x + 8 \right) \Big) + 3 \\ \\ \end{align}$

• Vertical stretch by a factor of 3
• Horizontal reflection (on y-axis)
• Horizontal stretch by a factor of 4 (it's because 1/k...)
• Horizontal shift 8 units left
• Vertical shift 3 units up

$f(x) = -2(4x + 12)^2 - 3$ Solution

Transformations and Translations of Individual Coordinates with Function Notation y = aƒ[k(x - d)] + c

Determine the new coordinate (image) given function notation and a point:

ƒ(x - 3),   P(1, 2) Solution

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Horizontal shift 3 units right...
Only affects the x-value (1 + 3, 2)
= (4, 2)

ƒ(x) - 3,   P(5, -2) Solution

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This is a vertical shift 3 units down...
Only affects the y-value (5, -2 - 3)
= (5, -5)

2ƒ(x),   P(-3, 4) Solution

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Vertical stretch by a factor of 2...
This affects the y-value because it's vertical. (-3, 4 × 2)
= (-3, 8)

ƒ(0.5x),   P(2, -7) Solution

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Horizontal transformations by a factor of 1/k !
This is a horizontal stretch by factor of 2
This affects the x-value (2 × 2, -7)
= (4, -7)

ƒ(3x),   P(1, 2) Solution

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Horizontal transformations by a factor of 1/k !
This is a horizontal compression by a factor of 1/3
On the x-value (1 × ⅓, 2)
= (0.33, 2)

ƒ(2x - 6) - 3,   P(1, -4) Solution

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Careful!
To get the horizontal shift you have to factor out the horizontal stretch/compression first! $\begin{align} & = f(2x - 6) - 3 \\ \\ & = f(2(x - 3)) - 3 \\ \\ \end{align}$ This is a horizontal compression by 1/2
Horizontal shift by 3 units right
Vertical shift 3 units down (1 × ½ + 3, -4 - 3)
= (3.5, -7)

-2ƒ(¼x + 1)² - 5,   P(2, 3) Solution

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Transformations to 'y'...

• Vertical reflection
• Vertical shift by a factor of 2
• Vertical shift 5 units down

Transformations to 'x'...

• Horizontal stretch by a factor of 4
• Horizontal shift 4 units left

Apply the transformations to 'x' on the 'x' value, and the transformations to 'y' to the 'y' value... $\begin{align} & = \left(2, 3\right) \\ \\ & = \left(\dfrac{1}{\tfrac{1}{4}}(2) - 4, -2(3) - 5\right) \\ \\ & = \left(4(2) - 4, -2(3) - 5\right) \\ \\ & = \left(4, -11\right) \\ \\ \end{align}$

Quadratic Functions Comparing the Discriminant to Zero

For what values of k will the following functions have no zeros, one zero, and two zeros?

ƒ(x) = 4x² - 2x + k Solution Video

k =

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$\begin{align} f(x) & = 4x^2 - 2x + k \\ \\ a & = 4 \\ \\ b & = -2 \\ \\ c & = k \\ \\ \end{align}$ for 1 x-intercept... $\begin{align} b^2 - 4ac & = 0 \\ \\ (-2)^2 - 4(4)(k) & = 0 \\ \\ 4 - 16k & = 0 \\ \\ -16k & = -4 \\ \\ k & = \dfrac{1}{4} \\ \\ \end{align}$ for 2 x-intercepts... $\begin{align} b^2 - 4ac & > 0 \\ \\ (-2)^2 - 4(4)(k) & > 0 \\ \\ 4 - 16k & > 0 \\ \\ - 16k & > -4 \\ \\ k & < \dfrac{1}{4} \\ \\ \end{align}$ for 0 x-intercept... $\begin{align} b^2 - 4ac & < 0 \\ \\ (-2)^2 - 4(4)(k) & < 0 \\ \\ 4 - 16k & < 0 \\ \\ - 16k & < -4 \\ \\ k & > \dfrac{1}{4} \\ \\ \end{align}$

ƒ(x) = kx² + x + k Solution Video

k =

Hint Clear Info

± ━━

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Sine Law and Cosine Law (Grade 10 Review)

Given the adjacent side lengths, angle-angle-side (AAS) in an acute triangle which of the following will solve the triangle? Solution

1. Sine Law
2. Cosine Law
3. Pythagorean Theorem
4. S^OH C^AH T^OA

Use Sine Law for AAS or SSA

Given adjacent side lengths, side-side-angle (SSA) in an acute triangle, which of the following will solve the triangle? Solution

1. Sine Law
2. Cosine Law
3. Pythagorean Theorem
4. S^OH C^AH T^OA

Use Sine Law for AAS or SSA

Given adjacent side lengths, side-angle-side (SAS) in an acute triangle, which of the following will solve the triangle? Solution

1. Sine Law
2. Cosine Law
3. Pythagorean Theorem
4. S^OH C^AH T^OA

Use Cosine Law for SSS or SAS

The Location of Angles in the Quadrant System

Determine the location of each angle.

Which of the following angles is in quadrant 1? Solution

1. 90˚
2. 0˚
3. 360˚
4. -300˚

The negative angle -300˚ starts from 0˚ and rotates clockwise to the first quadrant.

θ = 120˚ Solution

1. Quadrant 1
2. Quadrant 2
3. Quadrant 3
4. Quadrant 4

Start from 0˚ on the x-axis and rotate counter-clockwise, past 90˚ to 120˚ in quadrant 2, before 180˚.

θ = -135˚ Solution

1. Quadrant 1
2. Quadrant 2
3. Quadrant 3
4. Quadrant 4

Start from 0˚ and rotate in the negative direction, clockwise by 135˚. This puts the terminal arm in Quadrant 3 in the normal sense.

θ = 420˚ = 60˚ + 360˚ Solution

1. Quadrant 1
2. Quadrant 2
3. Quadrant 3
4. Quadrant 4

420˚ is a coterminal angle (greater than 360˚).

= 420˚
= 60˚ + 360˚
≈ 60˚

This is located in quadrant 1

θ = 700˚ Solution

1. Quadrant 1
2. Quadrant 2
3. Quadrant 3
4. Quadrant 4

700˚ is a coterminal angle.
= 340˚ + 360˚
≈ 340˚

This is in Quadrant 4.

Signs of Trig

Sin$\,\theta$ is positive in which of the following quadrants? Solution

1. Quadrant 1 and Quadrant 2
2. Quadrant 1 and Quadrant 4
3. Quadrant 2 and Quadrant 3
4. Quadrant 1 and Quadrant 3

Use C.A.S.T. Sin is positive in 'A' and 'S', which corresponds to Quadrant 1 and Quadrant 2.

Predict which of the following will be negative. Solution

1. sin60˚
2. sin135˚
3. cos60˚
4. cos180˚

Using your understanding of a graph of cos180˚ or using your calculator, you can see the value of this is -1.

If tan$\,\theta$ is negative, then the terminal arm (ray) can only be located in Quadrant 2. Solution

1. True
2. False

Tan$\,\theta$ is negative in quadrant 2 and quadrant 4.

Trig and Points on Terminal Arms

Given that a point on the terminal arm makes a right angle triangle with the reflex angle...

A point on the terminal arm in quadrant 2, using the ratio below, could be (-1, 1). Solution $\text{tan}\,\theta = \dfrac{1}{-1}$

1. True
2. False

The point (-1, 1) gives the side lengths (x, y) of the triangle formed with the reflex angle. $\begin{align} \text{tan}\,\theta & = \dfrac{Opposite}{Adjacent} \\ \\ \end{align}$ Opposite = y = 1

Adjacent = x = -1

Point: (-1, 1)

A point on the terminal arm in quadrant 4, using the ratio below, could be (1, -1). Solution $\text{sin}\,\theta = \dfrac{-1}{1}$

1. True
2. False

$\begin{align} \text{sin}\,\theta & = \dfrac{Opposite}{hypotenuse} \\ \\ \end{align}$ Opposite = y = -1

Find the hypotenuse using the Pythagorean Theorem:

$\begin{align} (hypotenuse)^2 & = b^2 + c^2 \\ \\ hypotenuse & = \sqrt{b^2 + c^2} \\ \\ hypotenuse & = \sqrt{(-1)^2 + 1^2} \\ \\ hypotenuse & = \sqrt{2} \\ \\ \end{align}$ Point = $(\sqrt{2}, -1)$

Exact Values

Determine the exact value of each of the following:

cos30˚ Solution

Hint Clear Info

$\dfrac{\sqrt{ \quad \ }}{\quad}$

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Hint Unavailable

Reference angle = 30˚ in quadrant 1
Use special triangle side lengths: $\begin{align} \text{cos}\,\theta & = \dfrac{Adjacent}{hypotenuse} \\ \\ \text{cos}\,30˚ & = \dfrac{\sqrt{3}}{2} \\ \\ & = \dfrac{\sqrt{3}}{2} \\ \\ \end{align}$

sin135˚ Solution

Hint Clear Info

$\dfrac{\sqrt{ \quad \ }}{\quad}$

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Use 45-45-90 special triangle...

Reference angle = 45˚ in quadrant 2
Use special triangle side lengths: $\begin{align} \text{sin}\,\theta & = \dfrac{Opposite}{hypotenuse} \\ \\ \text{sin}\,45˚ & = \dfrac{1}{\sqrt{2}} \\ \\ & = \dfrac{1}{\sqrt{2}} × \dfrac{\sqrt{2}}{\sqrt{2}} \quad derationalize \\ \\ & = \dfrac{\sqrt{2}}{2} \\ \\ \end{align}$

tan660˚ Solution

Hint Clear Info

$\quad\sqrt{ \quad \ }$

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Coterminal angle with 660˚ - 360˚ = 300˚
Reference angle = 60˚ in quadrant 4
Use special triangle side lengths: $\begin{align} \text{tan}\,\theta & = \dfrac{Opposite}{Adjacent} \\ \\ \text{tan}\,60˚ & = \dfrac{-\sqrt{3}}{1} \\ \\ & = \dfrac{-\sqrt{3}}{1} \\ \\ & = -\sqrt{3} \\ \\ \end{align}$

cos(-120˚) Solution

Hint Clear Info

$\quad\sqrt{ \quad \ }$

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Hint Unavailable

Reference angle = 60˚ in quadrant 3
Use special triangle side lengths: $\begin{align} \text{cos}\,\theta & = \dfrac{Adjacent}{hypotenuse} \\ \\ \text{cos}\,60˚ & = \dfrac{-\sqrt{3}}{-1} \\ \\ & = +\dfrac{\sqrt{3}}{1} \\ \\ & = \sqrt{3} \\ \\ \end{align}$

Converting Degrees and Radians

Convert the following angles into radians.

60˚ Solution

Hint Clear Info

$\dfrac{\pi}{\quad\quad\quad}$

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radians

Hint Unavailable

$\begin{align} Radians & = Degrees\ × \dfrac{\pi\ radians}{180˚} \\ \\ Radians & = 60˚\ × \dfrac{\pi\ radians}{180˚} \\ \\ Radians & = 60\cancel{˚}\ × \dfrac{\pi\ radians}{180\cancel{˚}} \\ \\ & = \dfrac{60 \pi\ radians}{180} \\ \\ & = \dfrac{1 \pi\ radians}{3} \\ \\ & = \dfrac{\pi}{3} \ radians \\ \\ \end{align}$

120˚ Solution

Hint Clear Info

$\dfrac{\quad\quad \ \pi}{\quad\quad\quad}$

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radians

Hint Unavailable

$\begin{align} Radians & = Degrees\ × \dfrac{\pi}{180˚} \\ \\ & = 120˚\ × \dfrac{\pi}{180˚} \\ \\ & = \dfrac{2\pi}{3} \ radians \\ \\ \end{align}$

180˚ Solution

Hint Clear Info

$\pi$

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radians

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$\begin{align} Radians & = Degrees\ × \dfrac{\pi}{180˚} \\ \\ & = 180˚\ × \dfrac{\pi}{180˚} \\ \\ & = \pi \ radians \\ \\ \end{align}$

270˚ Solution

Hint Clear Info

$\dfrac{\quad\quad \ \pi}{\quad\quad\quad}$

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radians

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$\begin{align} Radians & = Degrees\ × \dfrac{\pi}{180˚} \\ \\ & = 270˚\ × \dfrac{\pi}{180˚} \\ \\ & = \dfrac{3\pi}{2} \ radians \\ \\ \end{align}$

Converting Degrees and Radians

Convert the following angles into degrees.

$\dfrac{\pi}{3}$ Solution

Hint Clear Info

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degrees

Hint Unavailable

$\begin{align} Degrees & = \dfrac{\pi}{3}\ radians\ × \dfrac{180˚}{\pi\ radians} \\ \\ Degrees & = \dfrac{\cancel{\pi}}{3}\ \cancel{radians}\ × \dfrac{180˚}{\cancel{\pi}\ \cancel{radians}} \\ \\ & = \dfrac{180˚}{3} \\ \\ & = 60˚ \\ \\ \end{align}$

$\dfrac{2\pi}{3}$ Solution

Hint Clear Info

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degrees

Hint Unavailable

$\begin{align} Degrees & = \dfrac{2\cancel{\pi}}{3}\ \cancel{radians}\ × \dfrac{180˚}{\cancel{\pi}\ \cancel{radians}} \\ \\ & = 120˚ \\ \\ \end{align}$

$2\pi$ Solution

Hint Clear Info

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degrees

Hint Unavailable

$\begin{align} Degrees & =2\cancel{\pi}\ \cancel{radians}\ × \dfrac{180˚}{\cancel{\pi}\ \cancel{radians}} \\ \\ & = 360˚ \\ \\ \end{align}$

$\dfrac{3\pi}{4}$ Solution

Hint Clear Info

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degrees

Hint Unavailable

$\begin{align} Degrees & = \dfrac{3\cancel{\pi}}{4}\ \cancel{radians}\ × \dfrac{180˚}{\cancel{\pi}\ \cancel{radians}} \\ \\ & = 135˚ \\ \\ \end{align}$

2.1 radians Solution

Hint Clear Info

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degrees

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$\begin{align} Degrees & = 2.1\ \cancel{radians}\ × \dfrac{180˚}{\pi \ \cancel{radians}} \\ \\ & = 120.3˚ \\ \\ \end{align}$

Exact Values with Radians

Determine the exact value (y-value) of each of the following, given the angles. (A calculator will not work for this. Hint: draw a sketch of the principle angle first).

$\text{tan}\,\dfrac{2\pi}{3}$ Solution

Hint Clear Info

$-\sqrt{ \begin{matrix} \quad \\ \quad \end{matrix} }$

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Draw a sketch with $\dfrac{2\pi}{3}$ is the principal angle in Quadrant 2.
REFERENCE ANGLE: $\begin{align} & = \pi - \dfrac{2\pi}{3} \\ \\ & = \dfrac{3\pi}{3} - \dfrac{2\pi}{3} \\ \\ & = \dfrac{\pi}{3}\ rad\ or\ 60˚ \\ \\ \end{align}$ Draw a sketch of the reference angle in quadrant 2 to determine the following:
x = -1, y = $\sqrt{3}$, r = 2 $\begin{align} tan & = \dfrac{Opp}{Adj} \\ \\ & = \dfrac{\sqrt{3}}{-1} \\ \\ & = -\sqrt{3} \\ \\ \end{align}$

$\text{sin}\,\dfrac{5\pi}{4}$

Hint Clear Info

$- \dfrac{\sqrt{ \quad}}{\quad}$

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$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

$\text{cos}\,\dfrac{11\pi}{6}$

Hint Clear Info

$\dfrac{\sqrt{ \quad}}{\quad}$

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$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

$\text{sin}\,\dfrac{-\pi}{3}$

Hint Clear Info

$- \dfrac{\sqrt{ \quad}}{\quad}$

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Hint Unavailable

$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

$\text{csc}\,\dfrac{7\pi}{4}$

Hint Clear Info

$-\sqrt{ \begin{matrix} \quad \\ \quad \end{matrix} }$

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$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

$\text{sin}\,\dfrac{\pi}{3} + \text{cos}\,\dfrac{\pi}{6}$

Hint Clear Info

$\quad\sqrt{ \quad \ }$

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Hint Unavailable

$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

Exact Values with Radians

Determine the exact value of the following. Note that we can use a calculator to find the exact value only when the angles are a multiple of $n \cdot \dfrac{\pi}{2}$.

$\text{sin}\,\dfrac{\pi}{2}$ Solution

Hint Clear Info

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$\begin{align} \text{sin}\,\dfrac{\pi}{2} & = 1 \\ \\ \end{align}$

$\text{cos}\,\pi$

Hint Clear Info

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$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

$\text{sin}\,\dfrac{3\pi}{2}$

Hint Clear Info

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$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

$\text{cos}\,2\pi$

Hint Clear Info

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$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

$\text{sin}(-2\pi)$

Hint Clear Info

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$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

$\text{cos}\,\dfrac{5\pi}{2}$

Hint Clear Info

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Hint Unavailable

$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

Determining Radian Angles Given Exact Values

Using the CAST system, determine two angles between 0 and $2\pi$ $(0 \le \theta \le 2\pi)$ that have the following values:

$\text{tan}\,\theta = -\dfrac{1}{1}$ Solution Video

Hint Clear Info

$\dfrac{\quad\quad\pi}{\quad\quad\quad\quad}, \dfrac{\quad\quad\pi}{\quad\quad\quad\quad}$

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radians

Hint Unavailable

CAST: tan is negative in Quadrants 2 and 4.

Based on the side lengths, -1 and 1, or 1 and -1:

Reference angle = 45˚

The angles ($\theta$) in your answer will be the two principle angles:

Quadrant 2 = 180˚ - 45˚ = 135˚ $= \dfrac{3\pi}{4}\ rad$

Quadrant 4 = 360˚ - 45˚ = 315˚ $= \dfrac{7\pi}{4}\ rad$

$\text{sin}\,\theta = \dfrac{\sqrt{3}}{2}$

Hint Clear Info

$\dfrac{\pi}{\quad\quad\quad}, \dfrac{\quad\quad\pi}{\quad\quad\quad\quad}$

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radians

Hint Unavailable

$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

$\text{cos}\,\theta = -\dfrac{\sqrt{3}}{2}$

Hint Clear Info

$\dfrac{\quad\quad\pi}{\quad\quad\quad\quad}, \dfrac{\quad\quad\pi}{\quad\quad\quad\quad}$

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radians

Hint Unavailable

$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

$\text{tan}\,\theta = \dfrac{\sqrt{3}}{1}$

Hint Clear Info

$\dfrac{\pi}{\quad\quad\quad}, \dfrac{\quad\quad\pi}{\quad\quad\quad\quad}$

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radians

Hint Unavailable

$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

Solving Trig Equations

Solve the following between: $0 \le x \le 2\pi$. Order your answers from lowest to highest.

$\quad \Bigg(\text{tan}\,\theta + 1\Bigg)\Bigg(\text{tan}\,\theta + \dfrac{1}{\sqrt{3}}\Bigg) = 0$ Solution

Hint Clear Info

$\dfrac{\quad\quad\pi}{\quad\quad\quad}, \dfrac{\quad\quad\pi}{\quad\quad\quad}, \dfrac{\quad\quad\pi}{\quad\quad\quad}, \dfrac{\quad\quad\pi}{\quad\quad\quad}$

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radians

Hint Unavailable

CAST: tan is negative in Quadrants 2 and 4. $\begin{align} \text{tan}\,\theta = -1 \quad & and \quad \text{tan}\,\theta = \dfrac{-1}{\sqrt{3}} \\ \\ \end{align}$ For $\text{tan}\,\theta = -\dfrac{1}{1} \ $ determine that the REFERENCE ANGLE = 45˚ or $\dfrac{\pi}{4}\ rad$ and then find the principle angles on the CAST plane: $\begin{align} \theta_1 & = \pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4} \\ \\ \theta_2 & = 2\pi - \dfrac{\pi}{4} = \dfrac{7\pi}{4} \\ \\ \end{align}$ For $\text{tan}\,\theta = \dfrac{-1}{\sqrt{3}} \ $ determine that the REFERENCE ANGLE = 30˚ or $\dfrac{\pi}{6}\ rad$ and then find the principle angles on the CAST plane: $\begin{align} \theta_3 & = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6} \\ \\ \theta_4 & = 2\pi - \dfrac{\pi}{6} = \dfrac{11\pi}{6} \\ \\ \end{align}$

$\quad 2\text{sin}^2\,\theta - \text{sin}\,\theta - 1 = 0$

Hint Clear Info

$\dfrac{\pi}{\quad\quad\quad}, \dfrac{\quad\quad\pi}{\quad\quad\quad}, \dfrac{\quad\quad\pi}{\quad\quad\quad}, \dfrac{\quad\quad\pi}{\quad\quad\quad}$

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radians

Hint Unavailable

$\begin{align} 2\text{sin}^2\,\theta - \text{sin}\,\theta - 1 & = 0 \\ \\ (2\sin\,\theta + 1)(\sin\,\theta - 1) & = 0 \\ \\ \end{align}$ $\sin\,\theta = \dfrac{-1}{2} \quad and \quad \sin\,\theta = 1$ For $\text{sin}\,\theta = -\dfrac{1}{2} \ $ determine that the REFERENCE ANGLE = 30˚ or $\dfrac{\pi}{6}\ rad$ and then find the principle angles on the CAST plane: $\begin{align} \theta_1 & = \pi + \dfrac{\pi}{6} = \dfrac{7\pi}{6} \\ \\ \theta_2 & = 2\pi - \dfrac{\pi}{6} = \dfrac{11\pi}{6} \\ \\ \end{align}$ For $\text{sin}\,\theta = \dfrac{1}{1} \ $ determine that the REFERENCE ANGLE = 45˚ or $\dfrac{\pi}{4}\ rad$ and then find the principle angles on the CAST plane: $\begin{align} \theta_3 & = \dfrac{\pi}{4} \\ \\ \theta_4 & = \pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4} \\ \\ \end{align}$

Solving Trig Equations

Given the trig equation,

How many answers will the following equation have, between the interval: $0 \le x \le 2\pi$ ? Solution

1. 2
2. 4
3. 6
4. 8

This resembles 4x² - 4x + 1 = 0

Factor: $\begin{align} 4\text{cos}^2\,\theta - 4\text{cos}\,\theta + 1 & = 0 \\ \\ (2\text{cos}\theta - 1)(2\text{cos}\theta - 1) & = 0 \\ \\ (2\text{cos}\theta - 1)^2 & = 0 \\ \\ \end{align}$ Cos is positive in quadrant I and IV... there are 2 answers.

Solve for the principle angles, in degrees. Solution

Hint Clear Info

$\theta_1$ = $\theta_2$ =

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degrees

Hint Unavailable

Solve for θ: $\begin{align} 2\text{cos}\theta - 1 & = 0 \\ \\ 2\text{cos}\theta & = 1 \\ \\ \text{cos}\theta & = \dfrac{1}{2} \\ \\ \theta & = \text{cos}^{-1}\left(\dfrac{1}{2}\right) \\ \\ \theta & = 60˚ \ or \ \tfrac{\pi}{3} \\ \\ \end{align}$ Cos is positive in quadrant I and IV... Principle angles with 60˚ reflex angles are: 60˚, 300˚.