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  1. 5.1 Linear Functions Section 5.1 p1

  2. Linear functions describe quantities that have a constant rate of increase (or decrease). For instance, suppose a video store charges $2.50 for overnight rentals plus a late fee of $2.99 per day. If C = f(t) is the total cost as a function of t, the number of days late, then f(0) = Rental plus 0 days late fee = 2.50 = 2.50 + 0(2.99) = 2.50 f(1) = Rental plus 1 day late fee = 2.50 + 2.99 = 2.50 + 1(2.99) = 5.49 ... So C = f(t) = 2.50 + 2.99t. See Table 5.1 and Figure 5.1. (See next slide.) Section 5.1 p2

  3. Notice that the values of C go up by 2.99 each time t increases by 1. This has the effect that the points in Figure 5.1 lie on a line. Table 5.1 Figure 5.1: Cost, C, of renting a video as a function of t, the number of days late Section 5.1 p3

  4. The Family of Linear Functions The function f(t) = 2.50 + 2.99t belongs to the family of linear functions: A linear function is a function that can be written f(t) = b + mt, for constants b and m. We call the constants b and m the parameters for the family. For instance, the function f(t) = 2.50 + 2.99t has b = 2.50 and m = 2.99. Section 5.1 p4

  5. The Family of Linear Functions(continued) What is the Meaning of the Parameters b and m? The population of a town t years after it is founded is given by the linear function P(t) = 30,000 + 2000t. (a) What is the town’s population when it is founded? Example 1a Solution (a) The town is founded when t = 0, so Initial population = P(0) = 30,000 + 2000(0) = 30,000. Thus, the 30,000 in the expression for P(t) represents the starting population of the town. In the family of functions representing proportionality, y = kx, we found that the constant k has an interpretation as a rate of change. The next example shows that m has the same interpretation for linear functions. Section 5.1 p5

  6. The Family of Linear Functions(continued) What is the Meaning of the Parameters b and m? (continued) The population of a town t years after it is founded is given by the linear function P(t) = 30,000 + 2000t. (b) What is the population of the town one year after it is founded? By how much does the population increase every year? Example 1b Solution (b) After one year, we have t = 1, and P(1) = 30,000 + 2000(1) = 32,000, an increase of 2000 over the starting population. After two years we have t = 2, and P(2) = 30,000 + 2000(2) = 34,000, an increase of 2000 over the year 1 population. In fact, the 2000 in the expression for P(t) represents the amount by which the population increases every year. Section 5.1 p6

  7. The Family of Linear Functions(continued) What is the Meaning of the Parameters b and m? (continued) The population of a town t years after it is founded is given by the linear function P(t) = 30,000 + 2000t. (c) Sketch a graph of the population. Example 1c Solution The population is 30,000 when t = 0, so the graph passes through the point (0, 30,000). It slopes upward since the population is increasing at a rate of 2000 people per year. See Figure 5.2. Figure 5.2: Population of a town Section 5.1 p7

  8. The Family of Linear Functions(continued) What is the Meaning of the Parameters b and m? (continued) The value of a car, in dollars, t years after it is purchased is given by the linear function V(t) = 18,000 − 1700t. (a) What is the value of the car when it is new? Example 2a Solution (a) As in Example 1, the 18,000 represents the starting value of the car, when it is new: Initial value = V(0) = 18,000 − 1700(0) = $18,000. The next example gives a decreasing linear function—one whose output value goes down when the input value goes up. Section 5.1 p8

  9. The Family of Linear Functions(continued) What is the Meaning of the Parameters b and m? (continued) The value of a car, in dollars, t years after it is purchased is given by the linear function V(t) = 18,000 − 1700t. (b) What is the value of the car after one year? How much does the value decrease each year? Example 2b Solution (b) The value after one year is V(1) = 18,000 − 1700(1) = $16,300, which is $1700 less than the value when new. The car decreases in value by $1700 every year until it is worthless. Another way of saying this is that the rate of change in the value of the car is −$1700 per year. Section 5.1 p9

  10. The Family of Linear Functions(continued) What is the Meaning of the Parameters b and m? (continued) The value of a car, in dollars, t years after it is purchased is given by the linear function V(t) = 18,000 − 1700t. (c) Sketch a graph of the value. Example 2c Solution (c) Since the value of the car is 18,000 dollars when t = 0, the graph passes through the point (0, 18,000). It slopes downward, since the value is decreasing at the rate of 1700 dollars per year. See Figure 5.3. Figure 5.3: Value of a car Section 5.1 p10

  11. The Family of Linear Functions(continued) What is the Meaning of the Parameters b and m? (continued) Notice the pattern: Example 1 involves a population given by P(t) = 30,000 + 2000t, so In Example 2, the value is given by V(t) = 18,000 + (−1700)t, so In each case the coefficient of t is a rate of change, and the resulting function is linear, Q = b + mt. Section 5.1 p11

  12. Example 3a Solution • See • Table 5.2 • and • Figure 5.4. (a) Make a table of values for the function f(x) = 5 + 2x, and sketch its graph. Graphical and Numerical Interpretation of Linear Functions We can also interpret the constants b and m in terms of the graph of f(x) = b + mx. Section 5.1 p12

  13. Solution Example 3b (b) The constant 5 gives the y-value where the graph crosses the y-axis. This is the y-intercept or vertical intercept of the graph. From the table and the graph we see that 2 represents the amount by which y increases when x is increased by 1 unit. This is called the slope. (b) Interpret the constants 5 and 2 in terms of the table and the graph. Graphical and Numerical Interpretation of Linear Functions (continued) In Example 3, the constant m is positive, so the y-value increases when x is increased by 1unit, and consequently the graph rises from left to right. If the value of m is negative, the y-value decreases when x is increased by 1 unit, resulting in a graph that falls from left to right. Section 5.1 p13

  14. Example 4a For the function g(x) = 16 − 3x, (a) What does the coefficient of x tell you about the graph? Solution (a) Since the coefficient of x is negative, the graph falls from left to right. The value of the coefficient, −3, tells us that each time the value of x is increased by 1 unit, the value of y goes down by 3 units. The slope of the graph is −3. Graphical and Numerical Interpretation of Linear Functions (continued) Section 5.1 p14

  15. Example 4b For the function g(x) = 16 − 3x, (b) What is the y-intercept of the graph? Solution (b) Although we cannot tell from Figure 5.5 where the graph crosses the y-axis, we know that x = 0 on the y-axis, so the y-intercept is y = 16 − 3 · 0 = 16. Table 5.3 Figure 5.5: Graph of y = 16 − 3x Graphical and Numerical Interpretation of Linear Functions (continued) Section 5.1 p15

  16. Graphical and Numerical Interpretation of Linear Functions (continued) In general, we have For the linear function Q = f(t) = b + mt, • b = f(0) is the initial value, and gives the vertical intercept of the graph. • m is the rate of change and gives the slope of the graph. • If m > 0, the graph rises from left to right. • If m < 0, the graph falls from left to right. Section 5.1 p16

  17. Example 5a Find the vertical intercept and slope, and use this information to graph the functions: (a) f(x) = 100 + 25x Solution • The vertical intercept is 100 and the slope is 25, so f(x) starts at 100 when x = 0 and increases by 25 each time x increases by 1. See Figure 5.6. • Figure 5.6: Graph of y = 100 + 25x Graphical and Numerical Interpretation of Linear Functions (continued) Section 5.1 p17

  18. Example 5b Find the vertical intercept and slope, and use this information to graph the functions: (b) g(t) = 6 − 0.5t Solution (b) The vertical intercept is 6 and the slope is −0.5, so g(t) starts at 6 when t = 0 and decreases by 0.5 each time t increases by 1. See Figure 5.7. Figure 5.7: Graph of y = 6 − 0.5t Graphical and Numerical Interpretation of Linear Functions (continued) Section 5.1 p18

  19. Interpreting Linear Functions Using Units It is often useful to consider units of measurement when interpreting a linear function Q = b + mt. Since b is the initial value of Q, the units of b are the same as the units of Q. Since m is the rate of change of Q with respect to t, the units of m are the units of Q divided by the units of t. Section 5.1 p19

  20. Example 6a Worldwide, soda is the third most popular commercial beverage, after tea and milk. The global consumption of soda rose at an approximately constant rate from 150 billion liters in 1995 to 179 billion liters in 2000. (a) Find a linear function for the quantity of soda consumed, S, in billions of liters, t years after 1995. Solution • To find the linear function, we first find the rate of change, or slope. Since consumption increased from S = 150 to S = 179 over a period of 5 years, we have • When t = 0 (the year 1995), we have S = 150, so the vertical intercept is 150. Therefore • S = 150 + 5.8t. Interpreting Linear Functions Using Units (continued) Section 5.1 p20

  21. Example 6b Worldwide, soda is the third most popular commercial beverage, after tea and milk. The global consumption of soda rose at an approximately constant rate from 150 billion liters in 1995 to 179 billion liters in 2000. (b) Give the units and practical interpretation of the slope and the vertical intercept. Solution (b) Since the rate of change is equal to S/t, its units are S-units over t-units, or billion liters per year. The slope tells us that world soda production has been increasing at a constant rate of 5.8 billion liters per year. The vertical intercept 150 is the value of S when t is zero. Since it is a value of S, the units are S-units, or billion liters of soda. The vertical intercept tells us that the global consumption of soda in 1995 was 150 billion liters. Interpreting Linear Functions Using Units (continued) Section 5.1 p21

  22. Interpreting Linear Functions Using Units(continued) A borehole is a hole dug deep in the earth, for oil or mineral exploration. Often temperature gets warmer at greater depths. Suppose that the temperature in a borehole at the surface is 4°C and rises by 0.02°C with each additional meter of depth. Express the temperature T in °C as a function of depth d in meters. Example 7 Solution The temperature starts at 4°C, and increases at the rate of 0.02 degrees per meter as you go down. Thus T = 4 + 0.02d. Functions Where the Independent Variable Is Not Time Units are particularly useful in interpreting functions where the independent variable is not time. Section 5.1 p22

  23. 5.1 LINEAR FUNCTIONS Key Points • Definition of a linear function • Interpreting linear functions • Initial value and rate of change • Slope and vertical intercept • Units Section 5.1 p23

  24. 5.2Working With Linear Expressions Section 5.2 p24

  25. An expression, such as 3 + 2t, that defines a linear function is called a linear expression. When we are talking about the expression, rather than the function it defines, we call b the constant term and m the coefficient. Section 5.2 p25

  26. Identify the constant term and the coefficient in the expression for the following linear functions. (a) u(t) = 20 + 4t (b) v(t) = 8 − 0.3t (c) w(t) = t/7 + 5 Example 1 Solution (a) We have constant term 20 and coefficient 4. (b) Writing v(t) = 8 + (−0.3)t, we see that the constant term is 8 and the coefficient is −0.3. (c) Writing w(t) = 5 + (1/7)t, we see that the constant term is 5 and the coefficient is 1/7. Different forms for linear expressions reveal different aspects of the functions they define. Section 5.2 p26

  27. The Slope-Intercept Form The form that we have been using for linear functions is called the Slope-Intercept Form The form f(x) = b + mx or y = b + mx for expressing a linear function is called slope-intercept form, because it shows the slope, m, and the vertical intercept, b, of the graph. Expressing a function in slope intercept form is helpful in reading its initial value and rate of change. Section 5.2 p27

  28. The cost C of a vacation lasting d days consists of the air fare, $350, plus accommodation expenses of $55 times the number of days, plus food expenses of $40 times the number of days. (a) Give an expression for C as a function of d that shows air fare, accommodation, and food expenses separately. Example 2a Solution (a) The cost is obtained by adding together the air fare of $350, the accommodation, and the food. The accommodation for d days costs 55d and the food costs 40d. Thus C = 350 + 55d + 40d. The Slope-Intercept Form (continued) Section 5.2 p28

  29. The cost C of a vacation lasting d days consists of the air fare, $350, plus accommodation expenses of $55 times the number of days, plus food expenses of $40 times the number of days. (b) Express the function in slope-intercept form. What is the significance of the vertical intercept and the slope? Example 2b Solution (b) Collecting like terms, we get C = 350 + 95d dollars, which is linear in d with slope 95 and vertical intercept 350. The vertical intercept represents the initial cost (the air fare) and the slope represents the total daily cost of the vacation, $95 per day. The Slope-Intercept Form (continued) Section 5.2 p29

  30. The Point-Slope Form Although slope-intercept form is the simplest form, sometimes another form reveals a different aspect of a function. Point-Slope Form The form f(x) = y0 + m(x − x0) or y = y0 + m(x − x0) for expressing a linear function is called point-slope form, because • The graph passes through the point (x0, y0). • The slope, or rate of change, is m. Section 5.2 p30

  31. The population of a town t years after it is founded is given by P(t) = 16,000 + 400(t − 5). (a) What is the practical interpretation of the constants 5 and 16,000 in the expression for P? Example 3a The Point-Slope Form (continued) Solution (a) The difference between point-slope form and slope-intercept form is that the coefficient 400 multiplies the expression t − 5, rather than just t. Whereas the slope-intercept form tells us the value of the function when t = 0, this form tells us the value when t = 5: P(5) = 16,000 + 400(5 − 5) = 16,000 + 0 = 16,000. Thus, the population is 16,000 after 5 years. Section 5.2 p31

  32. The population of a town t years after it is founded is given by P(t) = 16,000 + 400(t − 5). (b) Express P(t) in slope-intercept form and interpret the slope and intercept. Example 3b Solution (b) We have P(t) = 16,000 + 400(t − 5) Population in year 5 is 16,000. = 16,000 + 400t − 400 · 5 Five years ago, it was 2000 fewer. = 14,000 + 400tThus, the starting value is 14,000. From the slope-intercept form we see that 400 represents the growth rate of the population per year. When we subtract 400 · 5 from 16,000 in the above calculation, we are deducting five years of growth in order to obtain the initial population, 14,000. The Point-Slope Form (continued) Section 5.2 p32

  33. The Point-Slope Form (continued) Example 4a Find formulas for the linear functions graphed. Solution • We use any two points • to find the slope. If we use • the points (0,−10) and (5, 10), we have • The slope is m = 4. We see in the graph that the y-intercept is −10, so a formula for the function is y = −10 + 4x. How Do We Find the Slope? In Section 4.4 we saw how to find the average rate of change of a function between two points. For a linear function this rate of change is equal to the slope and is the same no matter which two points we choose. Section 5.2 p33

  34. The Point-Slope Form (continued) Example 4b Find formulas for the linear functions graphed. How Do We Find the Slope? (continued) Solution (b) We use the two points (0, 5) and (4, 0) to find the slope: The slope is m = −5/4. The slope is negative since the graph falls from left to right. We see in the graph that the y-intercept is 5, so a formula for the function is Section 5.2 p34

  35. The Point-Slope Form (continued) Find linear functions satisfying: (a) f(5) = −8 and the rate of change is −3 Solution Example 5a (a) The graph passes through (x0, y0) = (5,−8) and has slope m = −3, so we use point-slope form f(x) = y0 + m(x − x0) = (−8) + (−3)(x − 5) = −8 − 3x + 15 = 7 − 3x. How Do We Find the Slope? (continued) Section 5.2 p35

  36. The Point-Slope Form (continued) Example 5b Solution Find linear functions satisfying: (b) g(5) = 20 and g(8) = 32. (b) The graph passes through (5, 20) and (8, 32), so the slope is Using point-slope form with the point (x0, y0) = (5, 20), we have g(x) = y0 + m(x − x0) = 20 + 4(x − 5) = 20 + 4x − 20 = 4x. Notice that in either case, we could have left the equation of the line in point-slope form instead of putting it in slope-intercept form. How Do We Find the Slope? (continued) Section 5.2 p36

  37. How Do We Recognize a Linear Expression? Is the expression linear in x? If it is, give the constant term and the coefficient. (a) 5 + 0.2x2 (d) 2(x + 1) + 3x − 5 (f) ax + x + b Example 6adf Sometimes a simple rearrangement of the terms is enough to recognize an expression as linear. Solution (a) This expression is not linear because of the x2 term. (d) We distribute the 2 and combine like terms: 2(x + 1) + 3x − 5 = 2x + 2 + 3x − 5 = −3 + 5x, so we see that the expression is linear. The constant term is −3, and the coefficient is 5. (f) We collect the x terms to get ax + x + b = b + (a + 1)x, so we see that the expression is linear. The constant term is b, and the coefficient is a + 1. Section 5.2 p37

  38. Example 7a Is each function linear? (a) The share of a community garden plot with area A square feet divided between 5 families is f(A) = A/5 square feet per family. Solution • This function is linear, since we can rewrite it in the form • f(A) = 0 + (1/5)A. The initial value is 0 and the rate of change • is 1/5. How Do We Recognize a Linear Expression? (continued) How Do We Decider If a Function Is Linear? To decide if a function f(x) given by an expression is linear, we focus on the form of the expression and see if it can be put in the form f(x) = b + mx. Section 5.2 p38

  39. Example 7bcd How Do We Recognize a Linear Expression? (continued) How Do We Decider If a Function Is Linear? Is each function linear? (b) The gasoline remaining in an electric generator running for h hours is G = 0.75 − 0.3h. (c) The circumference of a circle of radius r is C = 2πr. (d) The time it takes to drive 300 miles at v mph is T = 300/v hours. Solution (b) Since G = 0.75 + (−0.3)h, we see this is linear with initial value 0.75 and rate of change −0.3. (c) This function is linear, with initial value 0 and rate of change 2π. (d) This function is not linear, since it involves dividing by v, not multiplying v by a constant. Section 5.2 p39

  40. How Do We Recognize a Linear Expression? (continued) Expressions Involving More Than One Letter In Example 6(f) there were constants a and b in the expression ax + x + b, in addition to the variable x. We could change perspective on this expression, and regard a as the variable and x and b as the constants. Writing it in the form xa + (x + b) = Constant · a + Constant, we see that it gives a linear function of a. In this case we say that the expression is linear in a. Section 5.2 p40

  41. How Do We Recognize a Linear Expression? (continued) Example 8 Expressions Involving More Than One Letter (a) Is the expression xy2 + 5xy + 2y − 8 linear in x? In y? (b) Is the expression r2h linear in r? In h? Solution (a) To see if the expression is linear in x, we try to match it with the form b + mx. We have xy2 + 5xy + 2y − 8 = (2y − 8) + (y2 + 5y)x which is linear in x with b = 2y − 8 and m = y2 + 5y. Thus the expression is linear in x. It is not linear in y because of the y2 term. (b) The expression is not linear in r because of the r2 term. It is linear in h, with constant term 0 and coefficient r2. Section 5.2 p41

  42. 5.2 WORKING WITH LINEAR EXPRESSIONS Key Points • Identifying linear expressions • Different purposes of different forms • Slope-intercept and point-slope form Section 5.2 p42

  43. 5.3Solving Linear Equations Section 5.3 p43

  44. Example 1 For the town in Example 1 on page 115 (section 5.1), the population t years after incorporation is given by P(t) = 30,000 + 2000t. How many years does it take for the population to reach 50,000? Solution We want to know the value of t that makes P(t) equal to 50,000, so we solve the equation 30,000 + 2000t = 50,000 2000t = 20,000 subtract 30,000 from both sides t = 20,000/2000 = 10. divide both sides by 2000 Thus, it takes 10 years for the population to reach 50,000. We are often interested in knowing which input values to a linear function give a particular output value. This gives rise to a linear equation. Section 5.3 p44

  45. Finding Where Two Linear Functions Are Equal Example 2 Incandescent light bulbs are cheaper to buy but more expensive to operate than fluorescent bulbs. The total cost in dollars to purchase a bulb and operate it for t hours is given by f(t) = 0.50 + 0.004t (for incandescent bulbs) g(t) = 5.00 + 0.001t (for fluorescent bulbs). How many hours of operation gives the same cost with either choice? Solution (See next slide for solution.) Linear equations also arise when we want to know what input makes two linear functions f(x) and g(x) have the same output. Section 5.3 p45

  46. Example 2 (Continued from previous slide.) f(t) = 0.50 + 0.004t (for incandescent bulbs) g(t) = 5.00 + 0.001t (for fluorescent bulbs). How many hours of operation gives the same cost with either choice? Solution We need to find a value of t that makes f(t) = g(t), so 0.50 + 0.004t = 5.00 + 0.001tset expressions for f(t) and g(t) equal 0.50 + 0.003t = 5.00 subtract 0.001t 0.003t = 4.50 subtract 0.50 t = 4.50/0.003 = 1500 divide by 0.003. After 1500 hours of use, the cost to buy and operate an incandescent bulb equals the cost to buy and operate a fluorescent bulb. Let’s verify our solution: Cost for incandescent bulb: f(1500) = 0.50 + 0.004(1500) = 6.50 Cost for fluorescent bulb: g(1500) = 5.00 + 0.001(1500) = 6.50. We see that the cost of buying and operating either type of bulb for 1500 hours is the same: $6.50. Finding Where Two Linear Functions Are Equal (continued) Section 5.3 p46

  47. Using a Graph to Visualize Solutions Example 3 Solution In Example 2 we saw that the cost to buy and operate an incandescent bulb for 1500 hours is the same as the cost for a fluorescent bulb. (a) Graph the cost for each bulb and indicate the solution to the equation in Example 2. (b) Which bulb is cheaper if you use it for less than 1500 hours? More than 1500 hours? See Figure 5.10. The t-coordinate of the point where the two graphs intersect is 1500. At this point, both functions have the same value. In other words, t = 1500 is a solution to the equation f(t) = g(t). (See next slide for Figure 5.10.) If we graph two functions f and g on the same axes, then the values of t where f(t) = g(t) correspond to points where the two graphs intersect. Section 5.3 p47

  48. Solution Example 3 (continued) Figure 5.10 shows that the incandescent bulb is cheaper if you use it for less than 1500 hours. For example, it costs $4.50 to operate the incandescent bulb for 1000 hours, whereas it costs $6.00 to operate the fluorescent bulb for the same time. On the other hand, if you operate the bulb for more than 1500 hours, the fluorescent bulb is cheaper. (Figure 5.10.) (a) Graph the cost for each bulb and indicate the solution to the equation in Example 2. (b) Which bulb is cheaper if you use it for less than 1500 hours? More than 1500 hours? Using a Graph to Visualize Solutions (continued) Section 5.3 p48

  49. Example 4a Solution (a) We have 3x + 5 = 3(x + 5) 3x + 5 = 3x + 15 5 = 15. What is going on here? The last equation is false no matter what the value of x, so this equation has no solution. Figure 5.11 shows why the equation has no solution. The graphs of y = 3x + 5 and y = 3(x + 5) have the same slope, so never meet. Figure 5.11: Graphs of y = 3x + 5 and y = 3(x + 5) do not intersect. Solve (a) 3x + 5 = 3(x + 5) How Many Solutions Does a Linear Equation Have? In general a linear equation has one solution. If we visualize the equation graphically as in Example 3, we see that the two lines intersect at one point. Section 5.3 p49

  50. Example 4b Solve (b) 3x + 5 = 3(x + 1) + 2. Solution (b) We have 3x + 5 = 3(x + 1) + 2 3x + 5 = 3x + 5 0 = 0 You might find this result surprising as well. The last equation is true no matter what the value of x. So this equation has infinitely many solutions; every value of x is a solution. In this case, the graphs of y = 3x + 5 and y = 3(x + 1) + 2 are the same line, so they intersect everywhere. How Many Solutions Does a Linear Equation Have? Section 5.3 p50