# 6 3 Piecewise Functions Warm Up Lesson Presentation

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6 -3 Piecewise Functions Warm Up Lesson Presentation Lesson Quiz Holt. Mc. Dougal Algebra 2 Holt

6 -3 Piecewise Functions Warm Up Write the equation of each line in slopeintercept form. 1. slope of 3 and passes through the point (50, 200) y = 3 x + 50 2. slope of – 1 2 and passes through the point (6, 40) y = – 1 x + 43 2 Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Objectives Write and graph piecewise functions. Use piecewise functions to describe real -world situations. Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Vocabulary piecewise function step function Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions A piecewise function is a function that is a combination of one or more functions. The rule for a piecewise function is different for different parts, or pieces, of the domain. For instance, movie ticket prices are often different for different age groups. So the function for movie ticket prices would assign a different value (ticket price) for each domain interval (age group). Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Remember! When using interval notation, square brackets [ ] indicate an included endpoint, and parentheses ( ) indicate an excluded endpoint. (Lesson 1 -1) Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Example 1: Consumer Application Create a table and a verbal description to represent the graph. Step 1 Create a table Because the endpoints of each segment of the graph identify the intervals of the domain, use the endpoints and points close to them as the domain values in the table. Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Example 1 Continued The domain of the function is divided into three intervals: Holt Mc. Dougal Algebra 2 Weights under 2 [0, 2) Weights 2 and under 5 [2, 5) Weights 5 and over [5, ∞)

6 -3 Piecewise Functions Example 1 Continued Step 2 Write a verbal description. Mixed nuts cost $8. 00 per pound for less than 2 lb, $6. 00 per pound for 2 lb or more and less than 5 lb, and $5. 00 per pound for 5 or more pounds. Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Check It Out! Example 1 Create a table and a verbal description to represent the graph. Step 1 Create a table Because the endpoints of each segment of the graph identify the intervals of the domain, use the endpoints and points close to them as the domain values in the table. Holt Mc. Dougal Algebra 2

Piecewise Functions 6 -3 Check It Out! Example 1 Continued The domain of the function is divided into three intervals: Green Fee ($) Time Range (h) 28 8 A. M. – noon $28 [8, 12) 24 noon – 4 P. M. $24 [12, 4) 12 4 P. M. – 9 P. M. $12 [4, 9) Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Check It Out! Example 1 Continued Step 2 Write a verbal description. The green fee is $28 from 8 A. M. up to noon, $24 from noon up to 4 P. M. , and $12 from 4 up to 9 P. M. Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions A piecewise function that is constant for each interval of its domain, such as the ticket price function, is called a step function. You can describe piecewise functions with a function rule. The rule for the movie ticket prices from Example 1 on page 662 is shown. Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Read this as “f of x is 5 if x is greater than 0 and less than 13, 9 if x is greater than or equal to 13 and less than 55, and 6. 5 if x is greater than or equal to 55. ” Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions To evaluate any piecewise function for a specific input, find the interval of the domain that contains that input and then use the rule for that interval. Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Example 2 A: Evaluating a Piecewise Function Evaluate each piecewise function for x = – 1 and x = 4. 2 x + 1 if x ≤ 2 x 2 – 4 if x > 2 h(x) = h(– 1) = 2(– 1) + 1 = – 1 h(4) = 42 – 4 = 12 Holt Mc. Dougal Algebra 2 Because – 1 ≤ 2, use the rule for x ≤ 2. Because 4 > 2, use the rule for x > 2.

6 -3 Piecewise Functions Example 2 B: Evaluating a Piecewise Function 2 x if x ≤ – 1 5 x if x > – 1 g(x) = g(– 1) = 2(– 1) = 1 2 g(4) = 5(4) = 20 Holt Mc. Dougal Algebra 2 Because – 1 ≤ – 1, use the rule for x ≤ – 1. Because 4 > – 1, use the rule for x > – 1.

6 -3 Piecewise Functions Check It Out! Example 2 a Evaluate each piecewise function for x = – 1 and x = 3. 12 if x < – 3 f(x) = 15 if – 3 ≤ x < 6 20 if x ≥ 6 f(– 1) = 15 f(3) = 15 Holt Mc. Dougal Algebra 2 Because – 3 ≤ – 1 < 6, use the rule for – 3 ≤ x < 6. Because – 3 ≤ 3 < 6, use the rule for – 3 ≤ x < 6.

6 -3 Piecewise Functions Check It Out! Example 2 b Evaluate each piecewise function for x = – 1 and x = 3. 3 x 2 + 1 if x < 0 g(x) = 5 x – 2 if x ≥ 0 g(– 1) = 3(– 1)2 + 1 = 4 g(3) = 5(3) – 2 = 13 Holt Mc. Dougal Algebra 2 Because – 1 < 0, use the rule for x < 0. Because 3 ≥ 0, use the rule for x ≥ 0.

6 -3 Piecewise Functions You can graph a piecewise function by graphing each piece of the function. Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Example 3 A: Graphing Piecewise Functions Graph each function. g(x) = 1 4 x+3 – 2 x + 3 if x < 0 if x ≥ 0 The function is composed of two linear pieces that will be represented by two rays. Because the domain is divided by x = 0, evaluate both branches of the function at x = 0. Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Example 3 A Continued For the first branch, the function is 3 when x = 0, so plot the point (0, 3) with an open circle and draw a ray with the slope 0. 25 to the left. For the second branch, the function is 3 when x = 0, so plot the point (0, 3) with a solid dot and draw a ray with the slope of – 2 to the right. Holt Mc. Dougal Algebra 2 ● O

6 -3 Piecewise Functions Example 3 B: Graphing Piecewise Functions Graph each function. x 2 – 3 g(x) = 1 2 x– 3 (x – 4)2 – 1 if x < 0 if 0 ≤ x < 4 if x ≥ 4 The function is composed of one linear piece and two quadratic pieces. The domain is divided at x = 0 and at x = 4. Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Example 3 B Continued No circle is required at (0, – 3) and (4, – 1) because the function is connected at those points. Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Check It Out! Example 3 a Graph the function. f(x) = 4 if x ≤ – 1 – 2 if x > – 1 The function is composed of two constant pieces that will be represented by two rays. Because the domain is divided by x = – 1, evaluate both branches of the function at x = – 1. Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Check It Out! Example 3 a Continued The function is 4 when x ≤ – 1, so plot the point (– 1, 4) with a closed circle and draw a horizontal ray to the left. The function is – 2 when x > – 1, so plot the point (– 1, – 2) with an open circle and draw a horizontal ray to the right. Holt Mc. Dougal Algebra 2 ● O

6 -3 Piecewise Functions Check It Out! Example 3 b Graph the function. g(x) = – 3 x if x < 2 x+3 if x ≥ 2 The function is composed of two linear pieces. The domain is divided at x = 2. Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Check It Out! Example 3 b Continued x – 3 x – 4 12 – 2 6 0 0 2 – 6 4 x+3 ● 5 7 O Add an open circle at (2, – 6) and a closed circle at (2, 5) and so that the graph clearly shows the function value when x = 2. Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Notice that piecewise functions are not necessarily continuous, meaning that the graph of the function may have breaks or gaps. To write the rule for a piecewise function, determine where the domain is divided and write a separate rule for each piece. Combine the pieces by using the correct notation. Remember! The distance formula d = rt can be arranged to find rates: r = Holt Mc. Dougal Algebra 2 d t .

6 -3 Piecewise Functions Example 4: Sports Application Jennifer is completing a 15. 5 -mile triathlon. She swims 0. 5 mile in 30 minutes, bicycles 12 miles in 1 hour, and runs 3 miles in 30 minutes. Sketch a graph of Jennifer’s distance versus time. Then write a piecewise function for the graph. Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Example 4 Continued Step 1 Make a table to organize the data. Use the distance formula to find Jennifer’s rate for each leg of the race. Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Example 4 Continued Step 2 Because time is the independent variable, determine the intervals for the function. Swimming: 0 ≤ t ≤ 0. 5 She swims for half an hour. Biking: 0. 5 < t ≤ 1. 5 She bikes for the next hour. Running: 1. 5 < t ≤ 2 She runs the final half hour. Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Example 4 Continued Step 3 Graph the function. After 30 minutes, Jennifer has covered 0. 5 miles. On the next leg, she reaches a distance of 12 miles after a total of 1. 5 hours. Finally she completes the 15. 5 miles after 2 hours. Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Example 4 Continued Step 4 Write a linear function for each leg. Use point-slope form: y – y 1 = m(x – x 1). Swimming: d = t Use m = 0. 5 and (0, 0). Biking: d = 12 t – 5. 5 Use m = 12 and (0. 5, 0. 5). Use m = 6 and (1. 5, 12. 5). Running: d = 6 t + 3. 5 The function rule is d(t) = t if 0 ≤ t ≤ 0. 5 12 t – 5. 5 if 0. 5 < t ≤ 1. 5 6 t + 3. 5 Holt Mc. Dougal Algebra 2 if 1. 5 < t ≤ 2

6 -3 Piecewise Functions Check It Out! Example 4 Shelly earns $8 an hour. She earns $12 an hour for each hour over 40 that she works. Sketch a graph of Shelly’s earnings versus the number of hours that she works up to 60 hours. Then write a piecewise function for the graph. Step 1 Make a table to organize the data. Holt Mc. Dougal Algebra 2 Shelly’s Earnings Hours worked Pay ($/hr) 0– 40 8 >40 12

6 -3 Piecewise Functions Check It Out! Example 4 Continued Step 2 Because the number of hours worked is the independent variable, determine the intervals for the function. 0 ≤ h ≤ 40 She works equal to or less than 40 hours. h > 40 She works more than 40 hours. Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Check It Out! Example 4 Continued Step 3 Graph the function. Shelly earns $8 per hour for 0– 40. After 40 hours, she earns $12 per hour. Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Check It Out! Example 4 Continued Step 4 Write a linear function for each leg. Use a point-slope form: y – y 1 = m(x – x 1). 0– 40 hours: 8 h Use m = 8. Hours > 40: 12(h – 4) + 320 Use m = 12 and (40, 320). The function rule is f(h) = 8 h if 0 ≤ h ≤ 40 12(h – 40) + 320 if h > 40 Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Lesson Quiz: Part I 1. Graph the function, and evaluate at x = 1 and x = 3. p(x) = 1 2 x 2 + 2 if x ≤ 2 1 2 x+3 if x > 2 Holt Mc. Dougal Algebra 2

6 -3 Piecewise Functions Lesson Quiz: Part II 2. Write and graph a piecewise function for the following situation. A house painter charges $12 per hour for the first 40 hours he works, time and a half for the 10 hours after that, and double time for all hours after that. How much does he earn for a 70 -hour week? Holt Mc. Dougal Algebra 2

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