Chapter 8 - Rational and Radical Functions Algebra 2.

Chapter 8 - Rational and Radical Functions Algebra 2

Table of Contents 8.1 – Variation Functions 8.1 8.2 - Multiplying and Dividing Rational Expressions 8.3 - Adding and Subtracting Rational Expressions 8.4 - Rational Functions 8.5 – Solving Rational Equations and Inequalities 8.6 – Radical Expressions and Rational Exponents 8.7 –Radical Functions 8.8 - Solving Radical Equations and Inequalities

8.1 - Variation Functions Algebra 2

A direct variation is a relationship between two variables x and y that can be written in the form y = kx, where k ≠ 0. In this relationship, k is the constant of variation. For the equation y = kx, y varies directly as x. Example: y = mx, is a direct variation (linear) Example: Number of songs downloaded and total cost of songs, number of workers and time needed to build a highway. 8.1 Algebra 2 (bell work)

Given: y varies directly as x, and y = 27 when x = 6. Write and graph the direct variation function. y varies directly as x. Substitute 27 for y and 6 for x. Solve for the constant of variation k. Write the variation function by using the value of k. y = kx 27 = k(6) k = 4.5 y = 4.5x 8.1Writing and Graphing Direct VariationExample 1

y = 4.5x 8.1

Check Substitute the original values of x and y into the equation. y = 4.5x 27 4.5(6) 27 Graph the direct variation function. The y-intercept is 0, and the slope is 4.5. 8.1

Given: y varies directly as x, and y = 6.5 when x = 13. Write and graph the direct variation function. y varies directly as x. Substitute 6.5 for y and 13 for x. Solve for the constant of variation k. Write the variation function by using the value of k. y = kx 6.5 = k(13) k = 0.5 y = 0.5x 8.1

y = 0.5x 8.1

Check Substitute the original values of x and y into the equation. y = 0.5x 6.5 0.5(13) 6.5 Graph the direct variation function. The y-intercept is 0, and the slope is 0.5. 8.1

The cost of an item in euros e varies directly as the cost of the item in dollars d, and e = 3.85 euros when d = $5.00. Find d when e = 10.00 euros. Substitute. Solve for k. Use 0.77 for k. Substitute 10.00 for e. Solve for d. e = kdMethod 1 Find k. 3.85 = k(5.00) 0.77 = k Write the variation function. 10.00 = 0.77d 12.99 ≈ d e = 0.77d 8.1Example 2Solving Direct Variation Problems

When you want to find specific values in a direct variation problem, you can solve for k and then use substitution or you can use the proportion derived below. 8.1

Substitute. Find the cross products. Method 2 Use a proportion. e1e1 = d1d1 d2d2 e2e2 3.85 = 5.00d 10.00 3.85d = 50.00 Solve for d. 12.99 ≈ d 8.1 Euro Dollar

Math Joke Q: What’s reverse variation? A: The relationship between the length of a driveway and the chance of backing into a mailbox 8.1

The perimeter P of a regular dodecagon varies directly as the side length s, and P = 18 in. when s = 1.5 in. Find s when P = 75 in. Substitute. Solve for k. Use 12 for k. Substitute 75 for P. Solve for s. P = ksMethod 1 Find k. 18 = k(1.5) 12 = k Write the variation function. 75 = 12s 6.25 ≈ s P = 12s 8.1

Substitute. Find the cross products. Method 2 Use a proportion. P1P1 = s1s1 s2s2 P2P2 18 = 1.5s 75 18s = 112.5 Solve for s. 6.25 = s 8.1

A joint variation is a relationship among three variables that can be written in the form y = kxz, where k is the constant of variation. For the equation y = kxz, y varies jointly as x and z. The phrases “y varies directly as x” and “y is directly proportional to x” have the same meaning. Reading Math 8.1Day 2

The volume V of a cone varies jointly as the area of the base B and the height h, and V = 12  ft 3 when B = 9  ft 2 and h = 4 ft. Find b when V = 24  ft 3 and h = 9 ft. Step 1 Find k. The base is 8  ft 2. Substitute. Solve for k. V = kBh 12  = k(9  )(4) = k 1 3 Substitute. Solve for B. Step 2 Use the variation function. 24  = B(9) 8  = B V = Bh 1 3 1 3 Use for k. 1 3 8.1Example 3Solving Joint Variation Problems

The lateral surface area L of a cone varies jointly as the area of the base radius r and the slant height l, and L = 63  m 2 when r = 3.5 m and l = 18 m. Find r to the nearest tenth when L = 8  m 2 and l = 5 m. Step 1 Find k. Substitute. Solve for k. L = krl 63  = k(3.5)(18)  = k Substitute. Solve for r. Step 2 Use the variation function. 8  =  r(5) 1.6 = r L =  rl Use  for k. 8.1

A third type of variation describes a situation in which one quantity increases and the other decreases. For example, the table shows that the time needed to drive 600 miles decreases as speed increases. An inverse variation is a relationship between two variables x and y that can be written in the form y =,where k ≠ 0. For the equation y =, y varies inversely as x. k x k x 8.1

Given: y varies inversely as x, and y = 4 when x = 5. Write and graph the inverse variation function. Substitute 4 for y and 5 for x. Solve for k. 4 =4 = k = 20 y = k 5 y varies inversely as x. k x Write the variation formula. y =y = 20 x 8.1Example 4Writing and Graphing Inverse Variation

To graph, make a table of values for both positive and negative values of x. Plot the points, and connect them with two smooth curves. Because division by 0 is undefined, the function is undefined when x = 0. xy –2–10 –4–5 –6– –8– xy 210 45 6 8 3 5 2 5 2 3 8.1

Given: y varies inversely as x, and y = 4 when x = 10. Write and graph the inverse variation function. Substitute 4 for y and 10 for x. Solve for k. 4 =4 = k = 40 y = k 10 y varies inversely as x. k x Write the variation formula. y =y = 40 x 8.1

To graph, make a table of values for both positive and negative values of x. Plot the points, and connect them with two smooth curves. Because division by 0 is undefined, the function is undefined when x = 0. xy –2–20 –4–10 –6– –8–5 xy 220 410 6 85 20 3 3 8.1

When you want to find specific values in an inverse variation problem, you can solve for k and then use substitution or you can use the equation derived below. 8.1

You can use algebra to rewrite variation functions in terms of k. 8.1

Determine whether each data set represents a direct variation, an inverse variation, or neither. A. x6.513104 y840.5 In each case xy = 52. The product is constant, so this represents an inverse variation. B. x5812 y304872 In each case = 6. The ratio is constant, so this represents a direct variation. y x 8.1Example 6Indentifying Direct and Inverse Variation

The time t needed to complete a certain race varies inversely as the runner’s average speed s. If a runner with an average speed of 8.82 mi/h completes the race in 2.97 h, What is the average speed of a runner who completes the race in 3.5 h? Method 1 Find k. Substitute. Solve for k. 2.97 = k = 26.1954 t = k 8.82 k s Use 26.1954 for k. t = 26.1954 s Substitute 3.5 for t. 3.5 = 26.1954 s Solve for s. s ≈ 7.48 8.1Example 5ApplicationOptional

Method Use t 1 s 1 = t 2 s 2. t 1 s 1 = t 2 s 2 7.48 ≈ s 26.1954 = 3.5s (2.97)(8.82) = 3.5s Simplify. Solve for s. Substitute. So the average speed of a runner who completes the race in 3.5 h is approximately 7.48 mi/h. 8.1

HW pg. 573 8.1- – Day 1: 2-6, 20, 21, 55, 56 – Day 2: 7-16, 22, 23, 27 – Ch:37-39, 42-44 – Take out 12-16 if example 5/6 are not covered.

8.2 - Multiplying and Dividing Rational Expressions Algebra 2

When identifying values for which a rational expression is undefined, identify the values of the variable that make the original denominator equal to 0. Caution! Because rational expressions are ratios of polynomials, you can simplify them the same way as you simplify fractions. Recall that to write a fraction in simplest form, you can divide out common factors in the numerator and denominator. 8-2Algebra 2 (bell work) Just Read

Simplify. Identify any x-values for which the expression is undefined. 10x 8 6x46x4 5 10x 8 – 4 3 6x 4 5 3 x4x4 = The expression is undefined at x = 0 because this value of x makes 6x 4 equal 0. 8-2Example 1Simplifying Rational Expressions x 2 + x – 2 x 2 + 2x – 3 (x + 2)(x – 1) (x – 1)(x + 3) (x + 2) (x + 3) The expression is undefined at x = 1 and x = –3 because these values of x make the factors (x – 1) and (x + 3) equal 0.

Simplify. Identify any x-values for which the expression is undefined. 6x 2 + 7x + 2 6x 2 – 5x – 6 (2x + 1)(3x + 2) (3x + 2)(2x – 3) = (2x + 1) (2x – 3) The expression is undefined at x =– and x = because these values of x make the factors (3x + 2) and (2x – 3) equal 0. 3 2 2 3 8-2

Simplify. Identify any x values for which the expression is undefined. Factor out –1 in the numerator so that x 2 is positive, and reorder the terms. Factor the numerator and denominator. Divide out common factors. The expression is undefined at x = –2 and x = 4. 4x – x 2 x 2 – 2x – 8 –1(x 2 – 4x) x 2 – 2x – 8 –1(x)(x – 4) (x – 4)(x + 2) –x–x (x + 2 ) Simplify. 8-2Example 2Simplifying by Factoring -1

Simplify Identify any x values for which the expression is undefined. Factor out –1 in the numerator so that x is positive, and reorder the terms. Factor the numerator and denominator. Divide out common factors. The expression is undefined at x = 5. 10 – 2x x – 5 –1(2x – 10) x – 5 –2 1 Simplify. –1(2)(x – 5) (x – 5) 8-2

Math Joke Q: Why did the doctor send the expression to the psychiatrist A: Because it wasn’t rational 8-2

You can multiply rational expressions the same way that you multiply fractions. 8-2

Multiply. Assume that all expressions are defined. A. 3x 5 y 3 2x3y72x3y7  10x 3 y 4 9x2y59x2y5 3x 5 y 3 2x3y72x3y7  10x 3 y 4 9x2y59x2y5 5 3 3 5x 3 3y 5 B. x – 3 4x + 20  x + 5 x 2 – 9 x – 3 4(x + 5)  x + 5 (x – 3)(x + 3) 1 4(x + 3) 8-2Example 3Multiplying Rational Expressions

Multiply. Assume that all expressions are defined. A. x 15  20 x4x4  2x2x x 7 x 15  20 x4x4  2x2x x7 x7 3 2 2 2x 3 3 B. 10x – 40 x 2 – 6x + 8  x + 3 5x + 15 10(x – 4) (x – 4)(x – 2)  x + 3 5(x + 3) 2 (x – 2) 2 8-2

You can also divide rational expressions. Recall that to divide by a fraction, you multiply by its reciprocal. 1 2 3 4 ÷ = 1 2 4 3  2 2 3 = 8-2Day 2

Divide. Assume that all expressions are defined. 5x 4 8x2y28x2y2 ÷ 8y58y5 15 5x 4 8x2y28x2y2  15 8y58y5 5x4 5x4 8x2y28x2y2  8y58y5 3 23 x 2 y 3 3 8-2Dividing Rational ExpressionsExample 4

x 4 – 9x 2 x 2 – 4x + 3 ÷ x 4 + 2x 3 – 8x 2 x 2 – 16 Divide. Assume that all expressions are defined. x 4 – 9x 2 x 2 – 4x + 3  x 2 – 16 x 4 + 2x 3 – 8x 2 Rewrite as multiplication by the reciprocal. x 2 (x 2 – 9) x 2 – 4x + 3  x 2 – 16 x 2 (x 2 + 2x – 8) x 2 (x – 3)(x + 3) (x – 3)(x – 1)  (x + 4)(x – 4) x 2 (x – 2)(x + 4) (x + 3)(x – 4) (x – 1)(x – 2) 8-2

2x 2 – 7x – 4 x 2 – 9 ÷ 4x 2 – 1 8x 2 – 28x +12 Divide. Assume that all expressions are defined. (2x + 1)(x – 4) (x + 3)(x – 3)  4(2x 2 – 7x + 3) (2x + 1)(2x – 1) (2x + 1)(x – 4) (x + 3)(x – 3)  4(2x – 1)(x – 3) (2x + 1)(2x – 1) 4(x – 4) (x +3) 2x 2 – 7x – 4 x 2 – 9  8x 2 – 28x +12 4x 2 – 1 8-2

Solve. Check your solution. Note that x ≠ 2. x 2 + 3x – 10 x – 2 = 7 (x + 5)(x – 2) (x – 2) = 7 x + 5 = 7 x = 2 Because the left side of the original equation is undefined when x = 2, there is no solution. 8-2Example 5Solving Simple Rational Equations

Solve. Check your solution. Note that x ≠ –4. x 2 + x – 12 x + 4 = –7 (x – 3)(x + 4) (x + 4) = –7 x – 3 = –7 x = –4 Because the left side of the original equation is undefined when x = –4, there is no solution. 8-2

Solve. Check your solution. 4x 2 – 9 2x + 3 = 5 (2x + 3)(2x – 3) (2x + 3) = 5 2x – 3 = 5 x = 4 Note that x ≠ –. 3 2 8-2

HW pg. 580 8.2- – Day 1: 1-10, 54-56, 58, 59 – Day 2: 11-17, 29-33 (Odd) 8-2

8.3 Adding and Subtracting Rational Expressions Algebra II 8-3

Adding and subtracting rational expressions is similar to adding and subtracting fractions. To add or subtract rational expressions with like denominators, add or subtract the numerators and use the same denominator. 584 8-3

Add or subtract. Identify any x-values for which the expression is undefined. The expression is undefined at x = –4 because this value makes x + 4 equal 0. x – 3 x + 4 + x – 2 x + 4 x – 3 + x + 4 x – 2 2x – 5 x + 4 8-3

Add or subtract. Identify any x-values for which the expression is undefined. There is no real value of x for which x 2 + 1 = 0; the expression is always defined. 3x – 4 x 2 + 1 – 6x + 1 x 2 + 1 3x – 4 – x 2 + 1 (6x + 1) –3x – 5 x 2 + 1 3x – 4 – x 2 + 1 6x – 1 8-3

Find the least common multiple for each pair. A. 4x 2 y 3 and 6x 4 y 5 4x 2 y 3 = 2  2  x 2  y 3 6x 4 y 5 = 3  2  x 4  y 5 The LCM is 2  2  3  x 4  y 5, or 12x 4 y 5. B. x 2 – 2x – 3 and x 2 – x – 6 x 2 – 2x – 3 = (x – 3)(x + 1) x 2 – x – 6 = (x – 3)(x + 2) The LCM is (x – 3)(x + 1)(x + 2). 8-3

Find the least common multiple for each pair. a. 4x 3 y 7 and 3x 5 y 4 4x 3 y 7 = 2  2  x 3  y 7 3x 5 y 4 = 3  x 5  y 4 The LCM is 2  2  3  x 5  y 7, or 12x 5 y 7. b. x 2 – 4 and x 2 + 5x + 6 x 2 – 4 = (x – 2)(x + 2) x 2 + 5x + 6 = (x + 2)(x + 3) The LCM is (x – 2)(x + 2)(x + 3). 8-3

To add rational expressions with unlike denominators, rewrite both expressions with the LCD. This process is similar to adding fractions. 8-3

Add. Identify any x-values for which the expression is undefined. x – 3 x 2 + 3x – 4 + 2x2x x + 4 x – 3 (x + 4)(x – 1) + 2x2x x + 4 x – 3 (x + 4)(x – 1) + 2x2x x + 4 x – 1 x – 3 + 2x(x – 1) (x + 4)(x – 1) 2x 2 – x – 3 (x + 4)(x – 1) 2x 2 – x – 3 (x + 4)(x – 1) 2x 2 – x – 3 x 2 + 3x – 4 or The expression is undefined at x = –4 and x = 1 because these values of x make the factors (x + 4) and (x – 1) equal 0. 8-3

Add. Identify any x-values for which the expression is undefined. x x + 2 + –8 x 2 – 4 x x + 2 + –8 (x + 2)(x – 2) x – 2 x x + 2 + –8 (x + 2)(x – 2) x(x – 2) + (–8) (x + 2)(x – 2) x 2 – 2x – 8 (x + 2)(x – 2) (x + 2)(x – 4) (x + 2)(x – 2) x – 4 x – 2 The expression is undefined at x = –2 and x = 2 because these values of x make the factors (x + 2) and (x – 2) equal 0. 8-3

Add. Identify any x-values for which the expression is undefined. 3x 2x – 2 + 3x – 2 3x – 3 3x 2(x – 1) + 3x – 2 3(x – 1) 3 3 3x 2(x – 1) + 3x – 2 3(x – 1) 2 2 15x – 4 6(x – 1) 9x + 6x – 4 6(x – 1) The expression is undefined at x = 1 because this value of x make the factor (x – 1) equal 0. 8-3

Subtract Identify any x-values for which the expression is undefined. 2x 2 – 30 x 2 – 9 – x + 3 x + 5 2x 2 – 30 (x – 3)(x + 3) – x + 3 x + 5 2x 2 – 30 (x – 3)(x + 3) – x + 3 x + 5x – 3 2x 2 – 30 – (x + 5)(x – 3) (x – 3)(x + 3) 2x 2 – 30 – (x 2 + 2x – 15) (x – 3)(x + 3) 2x 2 – 30 – x 2 – 2x + 15 (x – 3)(x + 3) x 2 – 2x – 15 (x – 3)(x + 3) (x + 3)(x – 5) (x – 3)(x + 3) x – 5 x – 3 The expression is undefined at x = 3 and x = –3 because these values of x make the factors (x + 3) and (x – 3) equal 0. 8-3

Subtract. Identify any x-values for which the expression is undefined. 3x – 2 2x + 52x + 5 – 5x – 2 2 (3x – 2)(5x – 2) – 2(2x + 5) (2x + 5)(5x – 2) 15x 2 – 16x + 4 – (4x + 10) (2x + 5)(5x – 2) 3x – 2 2x + 5 – 5x – 2 22x + 5 5x – 2 The expression is undefined at x = – and x = because these values of x make the factors (2x + 5) and (5x – 2) equal 0. 15x 2 – 16x + 4 – 4x – 10 (2x + 5)(5x – 2) 5 2 2 5 8-3

Some rational expressions are complex fractions. A complex fraction contains one or more fractions in its numerator, its denominator, or both. Examples of complex fractions are shown below. Recall that the bar in a fraction represents division. Therefore, you can rewrite a complex fraction as a division problem and then simplify. You can also simplify complex fractions by using the LCD of the fractions in the numerator and denominator. 8-3

Math Joke Teacher: Why didn’t you simplify the fraction Student: It was to Complex X X + 2 X 2 X - 3

Simplify. Assume that all expressions are defined. x + 2 x – 1 x – 3 x + 5 Write the complex fraction as division. x + 2 x – 1 x – 3 x + 5 ÷ x + 2 x – 1 x + 5 x – 3  x 2 + 7x + 10 x 2 – 4x + 3 (x + 2)(x + 5) (x – 1)(x – 3) or 8-3 Method #1

Simplify. Assume that all expressions are defined. x – 1 x x 2 3 x + x (2x) x 2 3 x + x 2 + 6 2(x – 1) (3)(2) + (x)(x) (x – 1)(2) x 2 + 6 2x – 2 or 8-3 Method #1 Method #2

Simplify. Assume that all expressions are defined. x + 4 x – 2 1 2x2x 1 x + x + 4 x – 2 (2x)(x – 2) 1 2x2x 1 x + 3x – 6 (x + 4)(2x) (2)(x – 2) + (x – 2) (x + 4)(2x) 3(x – 2) 2x(x + 4) or 8-3 Method #2Method #1

Simplify. Assume that all expressions are defined. x + 1 x 2 – 1 x x – 1 Write the complex fraction as division. x + 1 x 2 – 1 x x – 1 ÷ x + 1 x 2 – 1 x – 1 x  1 x x + 1 (x – 1)(x + 1) x – 1 x  8-3

Simplify. Assume that all expressions are defined. 20 x – 1 6 3x – 3 Write the complex fraction as division. 20 x – 1 6 3x – 3 ÷ 20 x – 1 3x – 3 6  20 x – 1 3(x – 1) 6  10 8-3

HW Section 8.3 #2-15, 28-30, 61 (Use Table) Challenge: 32, 42, 46, 8-3

A rational function is a function whose rule can be written as a ratio of two polynomials. The parent rational function is f(x) =. Its graph is a hyperbola, which has two separate branches. You will learn more about hyperbolas in Chapter 10. 1 x Like logarithmic and exponential functions, rational functions may have asymptotes. The function f(x) = has a vertical asymptote at x = 0 and a horizontal asymptote at y = 0. 1 x Algebra II (Bell work) 1)Read the following information Below 2)Copy the Know it Notes pg. 592/593 8-4

8.4 Rational Functions Algebra II

Using the graph of f(x) = as a guide, describe the transformation and graph each function. A. g(x) = Because h = –2, translate f 2 units left. 1 x + 2 1 x B. g(x) = Because k = –3, translate f 3 units down. 1 x – 3 8-4

a. g(x) = Because h = –4, translate f 4 units left. 1 x + 4 b. g(x) = Because k = 1, translate f 1 unit up. 1 x + 1 Using the graph of f(x) = as a guide, describe the transformation and graph each function. 1 x 8-4

The values of h and k affect the locations of the asymptotes the domain and the range of rational functions whose graphs are hyperbolas. Pg. 593 8-4

Identify the asymptotes, domain, and range of the function g(x) = – 2. Vertical asymptote: x = –3 Domain: {x|x ≠ –3} Horizontal asymptote: y = –2 Check Graph the function on a graphing calculator. The graph suggests that the function has asymptotes at x = –3 and y = –2. 1 x + 3 1 x – (–3) g(x) = – 2 Range: {y|y ≠ –2} 8-4

Identify the asymptotes, domain, and range of the function g(x) = – 5. Vertical asymptote: x = 3 Domain: {x|x ≠ 3} Horizontal asymptote: y = –5 Check Graph the function on a graphing calculator. The graph suggests that the function has asymptotes at x = 3 and y = –5. 1 x – 3 1 x – (3) g(x) = – 5 Range: {y|y ≠ –5} 8-4

The graphs of some rational functions are not hyperbolas. Consider the rational function f(x) = and its graph. (x – 3)(x + 2) x + 1 The numerator of this function is 0 when x = 3 or x = –2. Therefore, the function has x-intercepts at –2 and 3 The denominator of this function is 0 when x = –1. As a result, the graph of the function has a vertical asymptote at the line x = –1. 8-4

Identify the zeros and vertical asymptotes of f(x) =. (x 2 + 3x – 4) x + 3 (x + 4)(x – 1) x + 3 f(x) = Step 1 Find the zeros and vertical asymptotes. Zeros: –4 and 1 Vertical asymptote: x = –3 8-4

Identify the zeros and vertical asymptotes of f(x) =. (x 2 + 3x – 4) x + 3 Step 2 Graph the function. Plot the zeros and draw the asymptote. Then make a table of values to fill in missing points. Vertical asymptote: x = –3 x –8–4–3.5–2.5 014 y –7.2 04.5 –10.5–1.3 03.4 8-4

Identify the zeros and vertical asymptotes of f(x) =. (x 2 + 7x + 6) x + 3 (x + 6)(x + 1) x + 3 f(x) = Step 1 Find the zeros and vertical asymptotes. Zeros: –6 and –1 Vertical asymptote: x = –3 8-4

Step 2 Graph the function. Plot the zeros and draw the asymptote. Then make a table of values to fill in missing points. x –7–5–2–1237 y –1.52–404.8610.4 Identify the zeros and vertical asymptotes of f(x) =. (x 2 + 7x + 6) x + 3 Vertical asymptote: x = –3 8-4

Algebra II (Bell Work) Day 2 1)Summarize the Know it note Below 2)Summarize the Know it Note pg. 596 8-4

A discontinuous function is a function whose graph has one or more gaps or breaks. (Hyperbola) A continuous function is a function whose graph has no gaps or breaks. The functions you have studied before this, including linear, quadratic, polynomial, exponential, and logarithmic functions, are continuous functions. 8-4

Identify the zeros and asymptotes of the function. Then graph. x 2 – 3x – 4 x f(x) = (x-4)(x+1) x f(x) = Degree of p > degree of q. Zeros: 4 and –1 Vertical asymptote: x = 0 Horizontal asymptote: none 8-4

Identify the zeros and asymptotes of the function. Then graph. Graph with a graphing calculator or by using a table of values. Vertical asymptote: x = 0 Notice how they use parenthesis to write the equation into the calculator Always graph zeros/asympototes first 8-4

Identify the zeros and asymptotes of the function. Then graph. x – 2 x2 – 1x2 – 1 f(x) = x – 2 (x – 1)(x + 1) f(x) = Degree of p < degree of q. Zero: 2 Vertical asymptote: x = 1, x = –1 Horizontal asymptote: y = 0 8-4

Identify the zeros and asymptotes of the function. Then graph. Graph with a graphing calculator or by using a table of values. 8-4

Identify the zeros and asymptotes of the function. Then graph. x 2 + 2x – 15 x – 1 f(x) = (x – 3)(x + 5) x – 1 f(x) = Degree of p > degree of q. Zeros: 3 and –5 Vertical asymptote: x = 1 Horizontal asymptote: none 8-4

Identify the zeros and asymptotes of the function. Then graph. Graph with a graphing calculator or by using a table of values. 8-4

In some cases, both the numerator and the denominator of a rational function will equal 0 for a particular value of x. As a result, the function will be undefined at this x-value. If this is the case, the graph of the function may have a hole. A hole is an omitted point in a graph. 8-4

Math Joke Q: What’s the difference between Tiger Woods and f(x) = A: One gets a hole in 1, one has a hole in it X 2 -1 X - 1 8-4

(x – 3)(x + 3) x – 3x – 3 f(x) =Identify holes in the graph of f(x) = Then graph. x 2 – 9 x – 3x – 3 There is a hole in the graph at x = 3. (x – 3)(x + 3) (x – 3) For x ≠ 3, f(x) = = x + 3 8-4

The graph of f is the same as the graph of y = x + 3, except for the hole at x = 3. On the graph, indicate the hole with an open circle. The domain of f is {x|x ≠ 3}. Hole at x = 3 8-4

(x – 2)(x + 3) x – 2x – 2 f(x) =Identify holes in the graph of f(x) = Then graph. x 2 + x – 6 x – 2x – 2 There is a hole in the graph at x = 2. For x ≠ 2, f(x) = = x + 3 (x – 2)(x + 3) (x – 2) 8-4

The graph of f is the same as the graph of y = x + 3, except for the hole at x = 2. On the graph, indicate the hole with an open circle. The domain of f is {x|x ≠ 2}. Hole at x = 2 8-4

HW Section 8.4 # 2- 16, 33-36, 46, Find everything algebraically, use calc to sketch Challenge: #32, 42, 44,

Algebra II (Bell work) 1.Turn in all of 8.4

8. 5 Solving Rational Equations and Inequalities Algebra II

A rational equation is an equation that contains one or more rational expressions. Step 1 In solving rational equations To solve a rational equation, start by multiplying each term of the equation by the least common denominator (LCD) of all of the expressions in the equation. 8-5

Solve the equation x – = 3. 18 x x(x) – (x) = 3(x) 18 x Multiply each term by the LCD, x. x 2 – 18 = 3x Simplify. Note that x ≠ 0. x 2 – 3x – 18 = 0 Write in standard form. (x – 6)(x + 3) = 0 Factor. x – 6 = 0 or x + 3 = 0 Apply the Zero Product Property. x = 6 or x = –3 Solve for x. 8-5

Multiply each term by the LCD, 3x. 10x = 12 + 6x Simplify. Note that x ≠ 0. 4x = 12 Combine like terms. x = 3 Solve for x. Solve the equation = + 2. 4 x 10 3 (3x) = (3x) + 2(3x) 10 3 4 x 8-5

Multiply each term by the LCD, 4x. 24 + 5x = –7x Simplify. Note that x ≠ 0. 24 = –12x Combine like terms. x = –2 Solve for x. Solve the equation + = –. 5 4 6 x 7 4 (4x) + (4x) = – (4x) 6 x 5 4 7 4 Optional 8-5

An extraneous solution is a solution of an equation derived from an original equation that is not a solution of the original equation. When you solve a rational equation, it is possible to get extraneous solutions. These values should be eliminated from the solution set. Always check your solutions by substituting them into the original equation. 8-5

Solve each equation. The solution x = 2 is extraneous because it makes the denominators of the original equation equal to 0. Therefore, the equation has no solution. Divide out common factors. Multiply each term by the LCD, x – 2. Simplify. Note that x ≠ 2. 5x x – 2 3x + 4 x – 2 = 5x x – 2 3x + 4 x – 2 (x – 2) = (x – 2) 5x x – 2 3x + 4 x – 2 (x – 2) = (x – 2) 5x = 3x + 4 x = 2 Solve for x. 8-5

Solve each equation. Multiply each term by the LCD, 2(x – 8). Simplify. Note that x ≠ 8. 2(2x – 5) + x(x – 8) = 11(2) 4x – 10 + x 2 – 8x = 22 Use the Distributive Property. 2x – 5 x – 8 11 x – 8 + = x 2 2x – 5 x – 8 2(x – 8) + 2(x – 8) = 2(x – 8) 11 x – 8 x 2 2x – 5 x – 8 2(x – 8) + 2(x – 8) = 2(x – 8) 11 x – 8 x 2 8-5

x 2 – 4x – 32 = 0 Write in standard form. (x – 8)(x + 4) = 0 Factor. x – 8 = 0 or x + 4 = 0 Apply the Zero Product Property. x = 8 or x = –4 Solve for x. The solution x = 8 us extraneous because it makes the denominator of the original equation equal to 0. The only solution is x = –4. 8-5

Math Joke Q: Why did the rational equation get a poor grade in chemistry? A: It kept producing extraneous solutions

16 (x – 4)(x + 4) 2 x – 4 (x – 4)(x + 4) = (x – 4 )(x + 4) Solve the equation. The solution x = 4 is extraneous because it makes the denominators of the original equation equal to 0. Therefore, the equation has no solution. Multiply each term by the LCD, (x – 4)(x +4). Simplify. Note that x ≠ ±4. 16 x 2 – 16 2 x – 4 = 16 (x – 4)(x + 4) 2 x – 4 (x – 4)(x + 4) = (x – 4 )(x + 4) 16 = 2x + 8 x = 4 Solve for x. 8-5

Natalie can finish a 500-piece puzzle in about 8 hours. When Natalie and Renzo work together, they can finish a 500-piece puzzle in about 4.5 hours. About how long will it take Renzo to finish a 500-piece puzzle if he works by himself? 1 8 Natalie’s rate: of the puzzle per hour 1 h Renzo’s rate: of the puzzle per hour, where h is the number of hours needed to finish the puzzle by himself. Algebra II (Bell work) 1.Summarize the example below 8-5

Natalie’s rate  hours worked Renzo’s rate  hours worked 1 complete puzzle + = 1 8 (4.5) 1 h + = 1 1 8 (4.5)(8h)+ 1 h (4.5)(8h) = 1(8h) Multiply by the LCD,8h. Simplify. 4.5h + 36 = 8h Solve for h. 36 = 3.5h 10.3 = h It will take Renzo about 10.3 hours, or 10 hours 17 minutes to complete a 500-piece puzzle working by himself. 8-5

Julien can mulch a garden in 20 minutes. Together Julien and Remy can mulch the same garden in 11 minutes. How long will it take Remy to mulch the garden when working alone? Julien’s rate: of the garden per minute 1 20 Remy’s rate: of the garden per minute, where m is the number of minutes needed to mulch the garden by himself. 1 m 8-5

Julien’s rate  min worked Remy’s rate  min worked 1 complete job + = 1 20 (11) 1 m + = 1 1 20 (11)(20m)+ 1 m (11)(20m) = 1(20m) Multiply by the LCD, 20m. Simplify. 11m + 220 = 20m Solve for m. 220 = 9m 24 ≈ m It will take Remy about 24 minutes to mulch the garden working by himself. 8-5

You can also solve rational inequalities algebraically. You start by multiplying each term by the least common denominator (LCD) of all the expressions in the inequality. However, you must consider two cases: the LCD is positive or the LCD is negative. 8-5

Solve ≤ 3 algebraically. 6 x – 8x – 8 Case 1 LCD is positive. Step 1 Solve for x. 6 x – 8 (x – 8) ≤ 3(x – 8) Multiply by the LCD. Simplify. Note that x ≠ 8. Solve for x. 6 ≤ 3x – 24 30 ≤ 3x 10 ≤ x x ≥ 10 Rewrite with the variable on the left. 8-5

Step 2 Consider the sign of the LCD. LCD is positive. Solve for x. x – 8 > 0 x > 8 For Case 1, the solution must satisfy x ≥ 10 and x > 8, which simplifies to x ≥ 10. Solve ≤ 3 algebraically. 6 x – 8x – 8 8-5

Solve ≤ 3 algebraically. 6 x – 8x – 8 Case 2 LCD is negative. Step 1 Solve for x. 6 x – 8 (x – 8) ≥ 3(x – 8) Multiply by the LCD. Reverse the inequality. Simplify. Note that x ≠ 8. Solve for x. 6 ≥ 3x – 24 30 ≥ 3x 10 ≥ x x ≤ 10 Rewrite with the variable on the left. 8-5

Step 2 Consider the sign of the LCD. LCD is positive. Solve for x. x – 8 > 0 x > 8 For Case 2, the solution must satisfy x ≤ 10 and x < 8, which simplifies to x < 8. Solve ≤ 3 algebraically. 6 x – 8x – 8 The solution set of the original inequality is the union of the solutions to both Case 1 and Case 2. The solution to the inequality ≤ 3 is x < 8 or x ≥ 10, or {x|x < 8  x ≥ 10}. 6 x – 8x – 8 8-5

Case 1 LCD is positive. Step 1 Solve for x. Solve ≥ –4 algebraically. 6 x – 2x – 2 6 x – 2 (x – 2) ≥ –4(x – 2) Multiply by the LCD. Simplify. Note that x ≠ 2. Solve for x. 6 ≥ –4x + 8 –2 ≥ –4x Rewrite with the variable on the left. ≤ x 1 2 x ≥x ≥ 1 2 8-5

Step 2 Consider the sign of the LCD. LCD is positive. Solve for x. x – 2 > 0 x > 2 Solve ≥ –4 algebraically. 6 x – 2x – 2 For Case 1, the solution must satisfy and x > 2, which simplifies to x > 2. x ≥x ≥ 1 2 8-5

Step 2 Consider the sign of the LCD. LCD is negative. Solve for x. x – 2 < 0 x < 2 Solve ≥ –4 algebraically. 6 x – 2x – 2 For Case 2, the solution must satisfy and x < 2, which simplifies to x ≤x ≤ 1 2 x ≤x ≤ 1 2 The solution set of the original inequality is the union of the solutions to both Case 1 and Case 2. The solution to the inequality ≥ –4 is x > 2 is x > 2 or x ≤, or {x|  x > 2}. 6 x – 2x – 2 x ≤x ≤ 1 2 1 2 8-5

Solve ≥ –4 algebraically. 6 x – 2x – 2 Case 2 LCD is negative. Step 1 Solve for x. 6 x – 2 (x – 2) ≤ –4(x – 2) Simplify. Note that x ≠ 2. Solve for x. 6 ≤ –4x + 8 –2 ≤ –4x Rewrite with the variable on the left. ≥ x 1 2 x ≤x ≤ 1 2 Multiply by the LCD. Reverse the inequality. 8-5

HW Section 8.5 # 2-10, 12, 16-18, 19-25 (Odd), 53, 64 Challenge: 36, 51,

The nth root of a real number a can be written as the radical expression where n is the index (plural: indices) of the radical and a is the radicand. When a number has more than one root, the radical sign indicates only the principal, or positive, root. When a radical sign shows no index, it represents a square root. Reading Math Algebra II (Bell work) 8-6

8.6 Radical Expressions and Rational Exponents Algebra II

Find all real roots. Example 1: Finding Real Roots A. sixth roots of 64 A positive number has two real sixth roots. Because 2 6 = 64 and (–2) 6 = 64, the roots are 2 and –2. B. cube roots of –216 A negative number has one real cube root. Because (–6) 3 = –216, the root is –6. C. fourth roots of –1024 A negative number has no real fourth roots. 8-6

Find all real roots. a. fourth roots of –256 A negative number has no real fourth roots. Check It Out! Example 1 b. sixth roots of 1 A positive number has two real sixth roots. Because 1 6 = 1 and (–1) 6 = 1, the roots are 1 and –1. c. cube roots of 125 A positive number has one real cube root. Because (5) 3 = 125, the root is 5. 8-6

When an expression contains a radical in the denominator, you must rationalize the denominator. To do so, rewrite the expression so that the denominator contains no radicals. Remember! 8-6

Math Joke Q: How is an artificial tree like the fourth root of -68? A: Neither has real roots.

Simplify each expression. Assume that all variables are positive. Example 2A: Simplifying Radical Expressions Factor into perfect fourths. Product Property. Simplify. 3  x  x  x 3x33x3 8-6

Example 2B: Simplifying Radical Expressions Quotient Property. Product Property. Simplify the numerator. Rationalize the numerator. Simplify. 8-6

Simplify the expression. Assume that all variables are positive. Check It Out! Example 2a Product Property. Simplify. Factor into perfect fourths. 2  x 2x2x 44 16x 4 2 4 4 x 4 4 2 4x 4 8-6

Check It Out! Example 2b Quotient Property. Product Property. Rationalize the numerator. Simplify. Simplify the expression. Assume that all variables are positive. 8 4 4 3 x 4 2 27 3 x 8-6

A rational exponent is an exponent that can be expressed as where m and n are integers and n ≠ 0. Radical expressions can be written by using rational exponents. m n 8-6

Example 3: Writing Expressions in Radical Form Method 1 Evaluate the root first. (–2) 3 Write with a radical. Write the expression (–32) in radical form and simplify. 3 5 –8 Evaluate the root. Evaluate the power. Method 2 Evaluate the power first. Write with a radical. –8 Evaluate the power. Evaluate the root.   3 5 32  5 32,768 8-6

Method 1 Evaluate the root first. (4) 1 Write with a radical. 64 1 3 4 Evaluate the root. Evaluate the power. Check It Out! Example 3a Write the expression in radical form, and simplify. Method 2 Evaluate the power first. Write will a radical. 4 Evaluate the power. Evaluate the root.  1 3 64  1 3 3 8-6

Write each expression by using rational exponents. Simplify. a.b. 3 4 81 10 3 Check It Out! Example 4 9 3 10 1000 Simplify. 5 1 2 c. 2 4 5 8-6

Example 5A: Simplifying Expressions with Rational Exponents Product of Powers. Simplify each expression. Simplify. Evaluate the Power. 7272 49 Check Enter the expression in a graphing calculator. 8-6

Example 5B: Simplifying Expressions with Rational Exponents Quotient of Powers. Simplify each expression. Simplify. Negative Exponent Property. 1 4 Evaluate the power. 8-6

Product of Powers. Simplify each expression. Simplify. Evaluate the Power. 6 Check It Out! Example 5a Check Enter the expression in a graphing calculator. 8-6

HW 8.6 Pg. 614 1-27 Odd, (Show All Work) 30, 33, 35, 38, 39

8.7 Radical Functions Algebra II

Algebra II (Bell work) A radical function is a function whose rule is a radical expression. A square-root function is a radical function involving. The square-root parent function is. The cube-root parent function is. 1.Read the information below 2.Turn In Quiz Corrections 8-7

Math Joke Q: How can you predict how many protestors will show up at a rally? A: By using a Radical Function? 8-7

Graph each function and identify its domain and range. 8-7

Example 1A Continued x (x, f(x)) 3(3, 0) 4(4, 1) 7(7, 2) 12(12, 3) The domain is {x|x ≥3}, and the range is {y|y ≥0}. ● ● ● ● 8-7

Graph each function and identify its domain and range. 8-7

x(x, f(x)) –6 (–6, –4) 1(1,–2) 2(2, 0) 3(3, 2) 10(10, 4) Example 1B Continued ● ● ● ● ● The domain is the set of all real numbers. The range is also the set of all real numbers 8-7

Using the graph of as a guide, describe the transformation and graph the function. Example 2: Transforming Square-Root Functions Translate f 5 units up. g(x) = x + 5 f(x) = x 8-7

Using the graph of as a guide, describe the transformation and graph the function. Check It Out! Example 2b g is f vertically compressed by a factor of. 1 2 f(x) = x 8-7

Transformations of square-root functions are summarized below. 8-7

Example 3: Applying Multiple Transformations Reflect f across the x-axis, and translate it 4 units to the right. Using the graph of as a guide, describe the transformation and graph the function. f(x)= x 8-7

g is f reflected across the y-axis and translated 3 units up. ● ● Check It Out! Example 3a Using the graph of as a guide, describe the transformation and graph the function. f(x)= x 8-7

g is f vertically stretched by a factor of 3, reflected across the x-axis, and translated 1 unit down. ● ● Check It Out! Example 3b Using the graph of as a guide, describe the transformation and graph the function. f(x)= x g(x) = –3 x – 1 8-7

Example 4: Writing Transformed Square-Root Functions Use the description to write the square-root function g. The parent function is reflected across the x-axis, compressed vertically by a factor of, and translated down 5 units. 1 5 f(x)=x 1 5 g(x) =  x + ( 5)       8-7

Use the description to write the square-root function g. Check It Out! Example 4 The parent function is reflected across the x-axis, stretched vertically by a factor of 2 and translated 1 unit up. f(x)= x 8-7

8.7 HW Pg. 624 3-15 Odd, 24-38 Even, 39, 41 Sketch All graphs Bonus 56, 59, 84

For a square root, the index of the radical is 2. Remember! A radical equation contains a variable within a radical. Recall that you can solve quadratic equations by taking the square root of both sides. Similarly, radical equations can be solved by raising both sides to a power. 8-8

Algebra II Schedule 1.Monday: 8.8 (D1) 1 st Hour, 8.8 (Notes) 3 rd Hour 2.Tuesday: 8.8 (D2) 3.Wednesday: Vocabulary Quiz --  Finals Review 4.Thursday: Quiz 8.4-8.6 -  Finals Review 5.Friday: Finals Review

8.8 Solving Radical Equations and Inequalities Algebra II

Solve each equation. Example 1A: Solving Equations Containing One Radical Subtract 5. Simplify. Square both sides. Solve for x. Simplify. 8-8 Always double check Answers by putting them Back into the original equation

Example 1B: Solving Equations Containing One Radical Divide by 7. Simplify. Cube both sides. Solve for x. Simplify. Solve each equation. 3 7 7 5x  7 84 7  8-8

Divide by 6. Square both sides. Solve for x. Simplify. Check It Out! Example 1c Solve the equation. 8-8

Example 2: Solving Equations Containing Two Radicals Solve 7x + 2 = 9(3x – 2) 7x + 2 = 27x – 18 20 = 20x 1 = x 8-8

Solve each equation. x + 6 = 8(x – 1) Check It Out! Example 2b x + 6 = 8x – 8 14 = 7x 2 = x 8-8

Solve. Example 3: Solving Equations with Extraneous Solutions Method 1 Use a graphing calculator. The solution is x = – 1 The graphs intersect in only one point, so there is exactly one solution. Let Y1 = and Y2 = 5 – x. You can use the intersect feature on a graphing calculator to find the point where the two curves intersect. Helpful Hint 8-8

Example 3 Continued Method 2 Use algebra to solve the equation. Step 1 Solve for x. –3x + 33 = 25 – 10x + x 2 0 = x 2 – 7x – 8 0 = (x – 8)(x + 1) x – 8 = 0 or x + 1 = 0 x = 8 or x = –1 8-8

Example 4A: Solving Equations with Rational Exponents Solve each equation. (5x + 7) = 3 5x + 7 = 27 5x = 20 x = 4 1 3 Algebra II (Bell work) 8-8

Example 4B: Solving Equations with Rational Exponents 2x = (4x + 8) 1 2 (2x) 2 = [(4x + 8) ] 2 1 2 4x 2 = 4x + 8 4x 2 – 4x – 8 = 0 4(x 2 – x – 2) = 0 4(x – 2)(x + 1 ) = 0 4 ≠ 0, x – 2 = 0 or x + 1 = 0 x = 2 or x = – 1 Step 1 Solve for x.

(2x + 15) = x 1 2 [(2x + 15) ] 2 = (x) 2 1 2 2x + 15 = x 2 x 2 – 2x – 15 = 0 (x – 5)(x + 3) = 0 x – 5 = 0 or x + 3 = 0 x = 5 or x = – 3 Step 1 Solve for x. Check It Out! Example 4b 8-8

A radical inequality is an inequality that contains a variable within a radical. You can solve radical inequalities by graphing or using algebra. A radical expression with an even index and a negative radicand has no real roots. Remember! 8-8 Algebra II (Bell work)

Example 5 Continued Method 2 Use algebra to solve the inequality. Step 1 Solve for x. 2x ≤ 42 x ≤ 21 8-8

Example 5 Continued Method 2 Use algebra to solve the inequality. Step 2 Consider the radicand. The radicand cannot be negative. Solve for x. 2x – 6 ≥ 0 2x ≥ 6 x ≥ 3 The solution of is x ≥ 3 and x ≤ 21, or 3 ≤ x ≤ 21. 8-8

Method 2 Use algebra to solve the inequality. Step 1 Solve for x. Subtract 2. Solve for x. Simplify. Square both sides. x – 3 ≤ 9 x ≤ 12 Check It Out! Example 5a Continued 8-8

Method 2 Use algebra to solve the inequality. Step 2 Consider the radicand. The radicand cannot be negative. Solve for x. x – 3 ≥ 0 x ≥ 3 Check It Out! Example 5a Continued The solution of is x ≥ 3 and x ≤ 12, or 3 ≤ x ≤ 12. 8-8

Method 2 Use algebra to solve the inequality. Step 1 Solve for x. Solve for x. Cube both sides. x + 2 ≥ 1 x ≥ – 1 Check It Out! Example 5b Continued 8-8

Method 2 Use algebra to solve the inequality. Step 2 Consider the radicand. The radicand cannot be negative. Solve for x. x + 2 ≥ 1 x ≥ –1 Check It Out! Example 5b Continued The solution of is x ≥ – 1. 8-8

HW 8.8 pg. 632 1: 2-16, 27-32 2: 17-25, 33-38, 42-44, Bonus: 52 Combined: 2-25, 27-38, 42-44, Bonus: 52 (TEC) pg. 636 Multi Step Test Prep (Due Friday May 10 th )

Chapter 8 - Rational and Radical Functions Algebra 2.

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Chapter 8 - Rational and Radical Functions Algebra 2.

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