Unit 3 Practice Test

Rational Equations Work Application 1 Rational Equations Work Application Gloria can clean a house in 4 hours and Joe can clean the house in 6 hours. About how many hours will it take them to clean the house, if they work together? Gloria’s Rate Joe’s Rate Let x = Number of hours for both to clean house

Rational Equations Work Application 1 Rational Equations Work Application Let x = Number of hours for both to clean house Gloria’s portion of house cleaned Joe’s portion of house cleaned Total house cleaned + = Gloria’s Rate  Hours Worked Joe’s Rate  Hours Worked + = 1 LCD = 12

Rational Equations Work Application 1 Rational Equations Work Application Let x = Number of hours for both to clean house LCD = 12 + = 1 12  + 12  = 12  1 + = 12 3x + 2x = 12 5x = 12 Answer 2.4 hours x = 2.4

2 1. y = x2 + 5 0 = x2 + x – 2 (7 – x) = x2 + 5 0 = (x + 2)(x – 1) What is the solution of the system of equations? Method: Use substitution Step 1: Substitute equation 2 into equation 1. Then solve. 1. y = x2 + 5 0 = x2 + x – 2 (7 – x) = x2 + 5 0 = (x + 2)(x – 1) 7 – x = x2 + 5 + x + x x + 2 = 0 x – 1 = 0 7 = x2 + x + 5 x = –2 x = 1 –7 –7 0 = x2 + x – 2

2 x = –2 x = 1 What is the solution of the system of equations? Step 2: Find y-values, by substituting x-values into either equation. Use equation 2. y = 7 – x x = –2: y = 7 – (–2) x = 1: y = 7 – (1) y = 7 + 2 y = 6 y = 9 (1, 6) (–2, 9)

3 x + 3 > 0 2x – 5 > 0 –3 –3 +5 +5 x > –3 2x > 5 Find the domain of each function. x + 3 > 0 2x – 5 > 0 –3 –3 +5 +5 x > –3 2x > 5

x2 + 8 = y – 6x y-intercept 4 x-intercept Let y = 0 x2 + 8 = y – 6x Find the intercepts of the graphs of each equation. x2 + 8 = y – 6x x-intercept y-intercept Let y = 0 x2 + 8 = y – 6x Let x = 0 x2 + 8 = 0 – 6x x2 + 8 = y – 6x x2 + 8 = –6x (0)2 + 8 = y – 6(0) +6x +6x 0 + 8 = y – 0 x2 + 6x + 8 = 0 (x + 2)(x + 4) = 0 8 = y x + 2 = 0 x + 4 = 0 x = –2 x = –4

4x + 2y = 12 x-intercept y-intercept 4 Let y = 0 Let x = 0 Find the intercepts of the graphs of each equation. 4x + 2y = 12 x-intercept y-intercept Let y = 0 Let x = 0 4x + 2y = 12 4x + 2y = 12 4x + 2(0) = 12 4(0) + 2y = 12 4x + 0 = 12 0 + 2y = 12 4x = 12 2y = 12 x = 3 y = 6

Horizontal compression by ¼ 5 Describe the transformation for each function. Shift down 7 units Horizontal compression by ¼

Reflection across y-axis 5 Describe the transformation for each function. Reflection across y-axis Vertical compression by ⅓

Reflection across x-axis 5 Describe the transformation for each function. Shift 5 units right Reflection across x-axis Vertical stretch by 6

x2 – 1 = 0 (x + 1)(x – 1) = 0 x + 1 = 0 x – 1 = 0 x = –1 x = 1 6 Vertical Asymptote x2 – 1 = 0 Let denominator equal 0 Solve for x. (x + 1)(x – 1) = 0 x + 1 = 0 x – 1 = 0 Domain Range x = –1 x = 1 x  1 y  0 x  –1 Horizontal Asymptote y = 0

Let denominator equal 0 x – 1 = 0 +1 +1 Solve for x. x = 1 x  1 y  4 6 Vertical Asymptote Let denominator equal 0 x – 1 = 0 +1 +1 Solve for x. x = 1 Domain Range Horizontal Asymptote x  1 y  4 y = 4

Let denominator equal 0 x – 1 = 0 +1 +1 Solve for x. x = 1 x  1 6 Vertical Asymptote Let denominator equal 0 x – 1 = 0 +1 +1 Solve for x. x = 1 Domain Range Horizontal Asymptote x  1 All real numbers None

x2 – 9 = 0 (x + 3)(x – 3) = 0 x + 3 = 0 x – 3 = 0 x = –3 x = 3 6 Vertical Asymptote Let denominator equal 0 x2 – 9 = 0 Solve for x. (x + 3)(x – 3) = 0 x + 3 = 0 x – 3 = 0 Domain Range x = –3 x = 3 x  3 y  3 x  –3 Horizontal Asymptote y = 3

7 Sketch the graph of each function.

7 Sketch the graph of each function.

Sketch the graph of x – 2 = 0 x = 2 y = 1 8 Vertical Asymptote Horizontal Asymptote y = 1

Sketch the graph of x – 2 = 0 x = 2 y = 1 x  2 y  1 8 Vertical Asymptote x – 2 = 0 x = 2 Horizontal Asymptote y = 1 Domain Range x  2 y  1

Sketch the graph of x + 3 = 0 x = –3 y = 0 8 Vertical Asymptote Horizontal Asymptote y = 0

Sketch the graph of x + 3 = 0 x = –3 y = 0 x  –3 y  0 8 Vertical Asymptote x + 3 = 0 x = –3 Horizontal Asymptote y = 0 Domain Range x  –3 y  0

a2 = a2 +3a + 9 0 = 3a + 9 –9 = 3a –3 = a 9 Square both sides Subtract a2 on both sides 0 = 3a + 9 –9 = 3a Subtract 9 on both sides –3 = a Final Answer: No Solution

9 Square both sides Subtract a2 on both sides Final Answer: No Solution

5 – x = (x + 1)(x + 1) 5 – x = x2 + 2x + 1 –5 + x –5 + x 10 5 – x = (x + 1)(x + 1) 5 – x = x2 + 2x + 1 –5 + x –5 + x 0 = x2 + 3x – 4

0 = x2 + 3x – 4 0 = (x – 1)(x + 4) x – 1 = 0 x + 4 = 0 x = 1 x = –4 10 0 = x2 + 3x – 4 0 = (x – 1)(x + 4) x – 1 = 0 x + 4 = 0 x = 1 x = –4 Answer: x = 1

11 LCD = 3x 10 4 Solve the equation = + 2. 3 x (3x) = (3x) + 2(3x) 10 3 4 x Multiply each term by the LCD, 3x. Simplify. 10x = 12 + 6x Combine like terms. 4x = 12 Solve for x. x = 3

12 LCD = (x – 4)(x + 4) 16 x2 – 16 2 x – 4 = Solve the equation. Multiply each term by the LCD, (x – 4)(x +4). 16 = 2(x + 4) Simplify. 16 = 2x + 8 Solve for x. 8 = 2x 4 = x No solution

x – 5 = 0 x = 5 3(x – 5) = (12)(1) 3x – 15 = 12 x = 5 3x = 27 x = 9 13 Solve Step 1A Step 1B Let denominator = 0 x – 5 = 0 x = 5 3(x – 5) = (12)(1) Endpoints 3x – 15 = 12 x = 5 3x = 27 x = 9 x = 9

5 < x < 9 2 –4 > 3 F 6 12 > 3 T 11 2 > 3 F Solve A B C 13 Solve Step 2 A B C 5 9 Step 3 Interval Test Number Test of Inequality True / False A B C 2 –4 > 3 F 6 12 > 3 T 11 2 > 3 F Step 4 5 < x < 9

x – 8 = 0 x = 8 3(x – 8) = (6)(1) 3x – 24 = 6 x = 8 3x = 30 x = 10 14 Solve Step 1A Step 1B Let denominator = 0 x – 8 = 0 x = 8 3(x – 8) = (6)(1) Endpoints 3x – 24 = 6 x = 8 3x = 30 x = 10 x = 10

x < 8 or x > 10 7 –6 < 3 T 9 6 < 3 F 11 2 < 3 T Solve A 14 Solve Step 2 A B C 8 10 Step 3 Interval Test Number Test of Inequality True / False A B C 7 –6 < 3 T 9 6 < 3 F 11 2 < 3 T Step 4 x < 8 or x > 10

15 Subtract . 2x2 + 64 x2 – 64 – x + 8 x – 4 2x2 + 64 (x – 8)(x + 8) – Factor the denominators. The LCD is (x – 8)(x + 8), so multiply by . x – 4 x + 8 (x – 8) 2x2 + 64 (x – 8)(x + 8) – x + 8 x – 4 x – 8 2x2 + 64 – (x – 4)(x – 8) (x – 8)(x + 8) Subtract the numerators. 2x2 + 64 – (x2 – 12x + 32) (x – 8)(x + 8) Multiply the binomials in the numerator.

15 2x2 + 64 – x2 + 12x – 32 Distribute the negative sign. Write the numerator in standard form. (x + 8)(x + 4) (x – 8)(x + 8) Factor the numerator. x + 4 x – 8 Divide out common factors.

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x2 + 3x + 2 = 0 (x + 1)(x + 2) = 0 x + 1 = 0 x + 2 = 0 x = –1 x = –2 18 Identify any x-values for which the expression is undefined. Let denominator = 0 x2 + 3x + 2 = 0 (x + 1)(x + 2) = 0 x + 1 = 0 x + 2 = 0 –1 –1 –2 –2 x = –1 x = –2

19 Simplify. LCD = 2x Multiply every fraction of the complex fraction by the LCD.

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Substitute 30 for y and -6 for x. 20 Given: y varies directly as x, and y = 30 when x = –6. Write the direct variation function. y = kx y varies directly as x. 30 = k(–6) Substitute 30 for y and -6 for x. 30 k(-6) = -6 -6 Solve for the constant of variation k. –5 = k y = –5x Write the variation function by using the value of k.

Substitute 8 for y and 4 for x. 21 Given: y varies inversely as x, and y = 8 when x = 4. Write the inverse variation function. y varies inversely as x. Substitute 8 for y and 4 for x. k = 32 Solve for k. Write the variation formula.