Linear functions equations

examples of nonlinear functions and linear functions algebra 1 and linear functions formula
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Linear Functions Figure 1 A bamboo forest in China (credit: “j Fxie”/Flickr) Introduction Imagine placing a plant in the ground one day and finding that it has doubled its height just a few days later. Although it may seem incredible, this can happen with certain types of bamboo species. These members of the grass family are 6 the fastest-growing plants in the world. One species of bamboo has been observed to grow nearly 1.5 inches every hour. In a twenty-four hour period, this bamboo plant grows about 36 inches, or an incredible 3 feet A constant rate of change, such as the growth cycle of this bamboo plant, is a linear function. Recall from Functions and Function Notation that a function is a relation that assigns to every element in the domain exactly one element in the range. Linear functions are a specific type of function that can be used to model many real-world applications, such as plant growth over time. In this chapter, we will explore linear functions, their graphs, and how to relate them to data. 6 http://www.guinnessworldrecords.com/records-3000/fastest-growing-plant/ 279280 CHAPTER 4 l i Near f u Ncti o Ns l eARnIng Obje CTIveS In this section, you will: • Represent a linear function. • Determine whether a linear function is increasing, decreasing, or constant. • Interpret slope as a rate of change. • Write and interpret an equation for a linear function. • Graph linear functions. • Determine whether lines are parallel or perpendicular. • Write the equation of a line parallel or perpendicular to a given line. 4.1 lI ne AR FUnCTIOnS Figure 1 Shanghai magl ev Train (credit: “kanegen”/Flickr) Just as with the growth of a bamboo plant, there are many situations that involve constant change over time. Consider, for example, the first commercial maglev train in the world, the Shanghai MagLev Train ( Figure 1). It carries 7 passengers comfortably for a 30-kilometer trip from the airport to the subway station in only eight minutes. Suppose a maglev train travels a long distance, and that the train maintains a constant speed of 83 meters per second for a period of time once it is 250 meters from the station. How can we analyze the train’s distance from the station as a function of time? In this section, we will investigate a kind of function that is useful for this purpose, and use it to investigate real-world situations such as the train’s distance from the station at a given point in time. Representing l inear Functions e f Th unction describing the train’s motion is a linear function, which is defined as a function with a constant rate of change, that is, a polynomial of degree 1. There are several ways to represent a linear function, including word form, function notation, tabular form, and graphical form. We will describe the train’s motion as a function using each method. Representing a Linear Function in Word Form Let’s begin by describing the linear function in words. For the train problem we just considered, the following word sentence may be used to describe the function relationship. • e t Th rain’s distance from the station is a function of the time during which the train moves at a constant speed plus its original distance from the station when it began moving at constant speed. e s Th peed is the rate of change. Recall that a rate of change is a measure of how quickly the dependent variable changes with respect to the independent variable. The rate of change for this example is constant, which means that it is the same for each input value. As the time (input) increases by 1 second, the corresponding distance (output) increases by 83 meters. The train began moving at this constant speed at a distance of 250 meters from the station. 7 http://www.chinahighlights.com/shanghai/transportation/maglev-train.htmSECTION 4.1 l i Near f u Nctio Ns 281 Representing a Linear Function in Function Notation Another approach to representing linear functions is by using function notation. One example of function notation is an equation written in the slope-intercept form of a line, where x is the input value, m is the rate of change, and b is the initial value of the dependent variable. Equation form y = mx + b Function notation f (x) = mx + b In the example of the train, we might use the notation D(t) in which the total distance D is a function of the time t. The rate, m, is 83 meters per second. The initial value of the dependent variable b is the original distance from the station, 250 meters. We can write a generalized equation to represent the motion of the train. D(t) = 83t + 250 Representing a Linear Function in Tabular Form A third method of representing a linear function is through the use of a table. e  Th relationship between the distance from the station and the time is represented in Figure 2. From the table, we can see that the distance changes by 83 meters for every 1 second increase in time. 1 second 1 second 1 second t 0 1 2 3 D(t) 250 333 416 499 83 meters 83 meters 83 meters Figure 2 Tabular representation of the function D showing selected input and output values Q & A… Can the input in the previous example be any real number? No. The input represents time, so while nonnegative rational and irrational numbers are possible, negative real numbers are not possible for this example. The input consists of non-negative real numbers. Representing a Linear Function in Graphical Form Another way to represent linear functions is visually, using a graph. We can use the function relationship from above, D(t) = 83t + 250, to draw a graph, represented in Figure 3. Notice the graph is a line. When we plot a linear function, the graph is always a line. e r Th ate of change, which is constant, determines the slant, or slope of the line. The point at which the input value is zero is the vertical intercept, or y-intercept, of the line. We can see from the graph that the y-intercept in the train example we just saw is (0, 250) and represents the distance of the train from the station when it began moving at a constant speed. 500 400 300 200 100 0 1 2 3 45 Time (s) Figure 3 The graph of D(t) = 83t + 250. graphs of linear functions are lines because the rate of change is constant. Notice that the graph of the train example is restricted, but this is not always the case. Consider the graph of the line f (x) = 2x + 1. Ask yourself what numbers can be input to the function. In other words, what is the domain of the function? The domain is comprised of all real numbers because any number may be doubled, and then have one added to the product. Distance (m)282 CHAPTER 4 l i Near f u Ncti o Ns linear function A linear function is a function whose graph is a line. Linear functions can be written in the slope-intercept form of a line f (x) = mx + b where b is the initial or starting value of the function (when input, x = 0), and m is the constant rate of change, or slope of the function. The y-intercept is at (0, b). Example 1 Using a Linear Function to Find the Pressure on a Diver The pressure, P, in pounds per square inch (PSI) on the diver in Figure 4 depends upon her depth below the water surface, d, in feet. This relationship may be modeled by the equation, P(d) = 0.434d + 14.696. Restate this function in words. Figure 4 (credit: Ilse Reijs and j an-noud hutten) Solution To restate the function in words, we need to describe each part of the equation. The pressure as a function of depth equals four hundred thirty-four thousandths times depth plus fourteen and six hundred ninety-six thousandths. Analysis The initial value, 14.696, is the pressure in PSI on the diver at a depth of 0 feet, which is the surface of the water. The rate of change, or slope, is 0.434 PSI per foot. This tells us that the pressure on the diver increases 0.434 PSI for each foot her depth increases. determining Whether a l inear Function Is Increasing, decreasing, or Constant e l Th inear functions we used in the two previous examples increased over time, but not every linear function does. A linear function may be increasing, decreasing, or constant. For an increasing function, as with the train example, the output values increase as the input values increase. The graph of an increasing function has a positive slope. A line with a positive slope slants upward from left to right as in Figure 5(a). For a decreasing function, the slope is negative. e o Th utput values decrease as the input values increase. A line with a negative slope slants downward from left to right as in Figure 5(b). If the function is constant, the output values are the same for all input values so the slope is zero. A line with a slope of zero is horizontal as in Figure 5(c). Increasing function Decreasing function Constant function f (x) f (x) f (x) f f f xx x (a) (b) (c) Figure 5SECTION 4.1 l i Near f u Ncti o Ns 283 increasing and decreasing functions e s Th lope determines if the function is an increasing linear function, a decreasing linear function, or a constant function. • f (x) = mx + b is an increasing function if m 0. • f (x) = mx + b is an decreasing function if m 0. • f (x) = mx + b is a constant function if m = 0. Example 2 Deciding Whether a Function Is Incr easing, Decreasing, or Constant 8 Some recent studies suggest that a teenager sends an average of 60 texts per day. For each of the following scenarios, find the linear function that describes the relationship between the input value and the output value. Then, determine whether the graph of the function is increasing, decreasing, or constant. a. e t Th o tal number of texts a teen sends is considered a function of time in days. The input is the number of days, and output is the total number of texts sent. b. A t een has a limit of 500 texts per month in his or her data plan. The input is the number of days, and output is the total number of texts remaining for the month. c. A teen has an unlimited number of texts in his or her data plan for a cost of 50 per month. The input is the number of days, and output is the total cost of texting each month. Solution Analyze each function. a. e f Th un ction can be represented as f (x) = 60x where x is the number of days. The slope, 60, is positive so the function is increasing. This makes sense because the total number of texts increases with each day. b. The f un ction can be represented as f (x) = 500 − 60x where x is the number of days. In this case, the slope is negative so the function is decreasing. This makes sense because the number of texts remaining decreases each day and this function represents the number of texts remaining in the data plan aer ft x days. c. e cos Th t f unction can be represented as f (x) = 50 because the number of days does not ae ff ct the total cost. The slope is 0 so the function is constant. Interpreting Slope as a Rate of Change In the examples we have seen so far, we have had the slope provided for us. However, we often need to calculate the slope given input and output values. Recall that given two values for the input, x and x , and two corresponding values 1 2 for the output, y and y —which can be represented by a set of points, (x , y ) and (x , y )—we can calculate the slope m. 1 2 1 1 2 2 y − y change in output (rise) Δy 2 1 __ _ _ m = = = x − x change in input (run) Δx 2 1 Note in function notation two corresponding values for the output y and y for the function f, y = f (x ) and y = f (x ), 1 2 1 1 2 2 so we could equivalently write f (x ) − f (x ) 2 1 _ m = x − x 2 1 Figure 6 indicates how the slope of the line between the points, (x , y ) and (x , y ), is calculated. Recall that the slope 1 1 2 2 measures steepness, or slant. The greater the absolute value of the slope, the steeper the line is. 8 http://www.cbsnews.com/8301-501465_162-57400228-501465/teens-are-sending-60-texts-a-day-study-says/ 284 CHAPTER 4 l i Near f u Ncti o Ns y 10 9 x – x 2 1 8 (x , y ) 7 2 2 y – y 2 1 6 5 (x , y ) 1 1 4 y – y Δy 3 2 1 m = = 2 x – x Δx 2 1 1 x 0 –1 14 2 3 Figure 6 The slope of a function is calculated by the change in y divided by the change in x. It does not matter which coordinate is used as the (x , y ) and which is the (x , y ), as long as each calculation is started with the elements from the same coordinate pair. 2 2 1 1 Q & A… units for the output __ Are the units for slope always ? units for the input Yes. Think of the units as the change of output value for each unit of change in input value. An example of slope could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for the input. calculate slope The slope, or rate of change, of a function m can be calculated according to the following: y − y change in output (rise) Δy 2 1 __ _ _ m = = = x − x change in input (run) Δx 2 1 where x and x are input values, y and y are output values. 1 2 1 2 How To… Given two points from a linear function, calculate and interpret the slope. 1. Determine the units for output and input values. 2. Calculate the change of output values and change of input values. 3. Interpret the slope as the change in output values per unit of the input value. Example 3 Finding the Slope of a Linear Function If f (x) is a linear function, and (3, −2) and (8, 1) are points on the line, find the slope. Is this function increasing or decreasing? Solution e c Th oordinate pairs are (3, −2) and (8, 1). To find the rate of change, we divide the change in output by the change in input. change in output 1 − (−2) 3 __ ________ __ m = = = 8 − 3 5 change in input We could also write the slope as m = 0.6. The function is increasing because m 0. Analysis As noted earlier, the order in which we write the points does not matter when we compute the slope of the line as long as the first output value, or y-coordinate, used corresponds with the first input value, or x-coordinate, used. Note that if we had reversed them, we would have obtained the same slope. (−2) − (1) −3 3 _________ ___ _ m = = = 3 − 8 −5 5 SECTION 4.1 l i Near f u Ncti o Ns 285 Try It 1 If f (x) is a linear function, and (2, 3) and (0, 4) are points on the line, find the slope. Is this function increasing or decreasing? Example 4 Finding the Population Change from a Linear Function e p Th opulation of a city increased from 23,400 to 27,800 between 2008 and 2012. Find the change of population per year if we assume the change was constant from 2008 to 2012. Solution The rate of change relates the change in population to the change in time. The population increased by 27,800 − 23,400 = 4,400 people over the four-year time interval. To find the rate of change, divide the change in the number of people by the number of years. 4,400 people people __ __ = 1,100 year 4 years So the population increased by 1,100 people per year. Analysis Because we are told that the population increased, we would expect the slope to be positive. This positive slope we calculated is therefore reasonable. Try It 2 e p Th opulation of a small town increased from 1,442 to 1,868 between 2009 and 2012. Find the change of population per year if we assume the change was constant from 2009 to 2012. Writing and Interpreting an equation for a l inear Function Recall from Equations and Inequalities that we wrote equations in both the slope-intercept form and the point-slope form. Now we can choose which method to use to write equations for linear functions based on the information we are given. That information may be provided in the form of a graph, a point and a slope, two points, and so on. Look at the graph of the function f in Figure 7. y f 10 8 (0, 7) 6 (4, 4) 4 2 x –10–8 –6–4 –2 28 4 6 10 –2 –4 Figure 7 We are not given the slope of the line, but we can choose any two points on the line to find the slope. Let’s choose (0, 7) and (4, 4). We can use these points to calculate the slope. y − y 2 1 _ m = x − x 2 1 4 − 7 _____ = 4 − 0 3 _ = −   4 Now we can substitute the slope and the coordinates of one of the points into the point-slope form. y − y = m(x − x ) 1 1 3 _ y − 4 = −   ( x − 4) 4286 CHAPTER 4 l i Near f u Ncti o Ns If we want to rewrite the equation in the slope-intercept form, we would find 3 _ y − 4 = −  (x − 4) 4 3 _ y − 4 = −  x + 3 4 3 _ y = −   x + 7 4 If we wanted to find the slope-intercept form without first writing the point-slope form, we could have recognized that the line crosses the y-axis when the output value is 7. e Th refore, b = 7. We now have the initial value b and the slope m so we can substitute m and b into the slope-intercept form of a line. f (x) = mx + b ↑ ↑ 3 _ −  7 4 3 _ f (x) = −  x + 7 4 3 3 _ _ So the function is f (x) = −  x + 7, and the linear equation would be y = −  x + 7. 4 4 How To… Given the graph of a linear function, write an equation to represent the function. 1. Identify two points on the line. 2. Use the two points to calculate the slope. 3. Determine where the line crosses the y-axis to identify the y-intercept by visual inspection. 4. Substitute the slope and y-intercept into the slope-intercept form of a line equation. Example 5 W riting an Equation for a Linear Function Write an equation for a linear function given a graph of f shown in Figure 8. y 10 8 f 6 4 2 x –10–8 –6–4 –2 28 4 6 10 –2 –4 –6 –8 –10 Figure 8 Solution Identify two points on the line, such as (0, 2) and (−2, −4). Use the points to calculate the slope. y − y 2 1 _ m = x − x 2 1 −4 − 2 _______ = −2 − 0 −6 ___ = −2 = 3 Substitute the slope and the coordinates of one of the points into the point-slope form. y − y = m(x − x ) 1 1 y − (−4) = 3(x − (−2)) y + 4 = 3(x + 2)SECTION 4.1 l i Near f u Nctio Ns 287 We can use algebra to rewrite the equation in the slope-intercept form. y + 4 = 3(x + 2) y + 4 = 3x + 6 y = 3x + 2 Analysis This makes sense because we can see from Figure 9 that the line crosses the y-axis at the point (0, 2), which is the y-intercept, so b = 2. y 10 8 6 4 (0, 2) 2 x –10–8 –6–4 –2 28 4 6 10 –2 (−2, −4) –4 –6 –8 –10 Figure 9 Example 6 W riting an Equation for a Linear Cost Function Suppose Ben starts a company in which he incurs a fixed cost of 1,250 per month for the overhead, which includes his oc ffi e rent. His production costs are 37.50 per item. Write a linear function C where C(x) is the cost for x items produced in a given month. Solution e fi Th xed cost is present every month, 1,250. The costs that can vary include the cost to produce each item, which is 37.50 for Ben. The variable cost, called the marginal cost, is represented by 37.5. The cost Ben incurs is the sum of these two costs, represented by C(x) = 1250 + 37.5x. Analysis If Ben produces 100 items in a month, his monthly cost is found by substitution 100 for x. C(100) = 1,250 + 37.5(100) = 5,000 So his monthly cost would be 5,000. Example 7 Writing an Equation for a Linear Function Given Two Points If f is a linear function, with f (3) = −2 , and f (8) = 1 , find a n equation for the function in slope-intercept form. Solution We can write the given points using coordinates. f (3) = −2 → (3, −2) f (8) = 1 → (8, 1) We can then use the points to calculate the slope. y − y 2 1 _ m = x − x 2 1 1 − (−2) ________ = 8 − 3 3 _ =   5 Substitute the slope and the coordinates of one of the points into the point-slope form. y − y = m(x − x ) 1 1 3 _ y − (−2) =    (x − 3) 5288 CHAPTER 4 l i Near f u Ncti o Ns We can use algebra to rewrite the equation in the slope-intercept form. 3 _ y + 2 =    (x − 3) 5 3 9 _ _ y + 2 =   x −    5 5 3 19 _ _ y =   x − 5 5 Try It 3 If f (x) is a linear function, with f (2) = −11, and f (4) = −25, fin d an equation for the function in slope-intercept form. modeling Real-World Problems with l inear Functions In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know, or can figure out, the initial value and the rate of change of a linear function, we can solve many die ff rent kinds of real-world problems. How To… Given a linear function f and the initial value and rate of change, evaluate f (c). 1. Determine the initial value and the rate of change (slope). 2. Substitute the values into f (x) = mx + b. 3. Evaluate the function at x = c. Example 8 Using a Linear Function to Determine the Number of Songs in a Music Collection Marcus currently has 200 songs in his music collection. Every month, he adds 15 new songs. Write a formula for the number of songs, N, in his collection as a function of time, t, the number of months. How many songs will he own in a year? Solution e i Th nitial value for this function is 200 because he currently owns 200 songs, so N(0) = 200, which means that b = 200. e n Th umber of songs increases by 15 songs per month, so the rate of change is 15 songs per month. Therefore we know that m = 15. We can substitute the initial value and the rate of change into the slope-intercept form of a line. f (x) = mx + b ↑ ↑ 15 200 N(t) = 15t + 200 Figure 10 We can write the formula N(t) = 15t + 200. With this formula, we can then predict how many songs Marcus will have in 1 year (12 months). In other words, we can evaluate the function at t = 12. N(12) = 15(12) + 200 = 180 + 200 = 380 Marcus will have 380 songs in 12 months. Analysis Notice that N is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well.SECTION 4.1 l i Near f u Ncti o Ns 289 Example 9 Using a Linear Function to Calculate Salary Based on Commission Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya’s weekly income, I, depends on the number of new policies, n, he sells during the week. Last week he sold 3 new policies, and earned 760 for the week. The week before, he sold 5 new policies and earned 920. Find an equation for I(n), and interpret the meaning of the components of the equation. Solution e g Th iven information gives us two input-output pairs: (3,760) and (5,920). We start by finding the rate of change. 920 − 760 ________ m = 5 − 3 160 _______ = 2 policies = 80 per policy Keeping track of units can help us interpret this quantity. Income increased by 160 when the number of policies increased by 2, so the rate of change is 80 per policy. Therefore, Ilya earns a commission of 80 for each policy sold during the week. We can then solve for the initial value. I(n) = 80n + b 760 = 80(3) + b when n = 3, I(3) = 760 760 − 80(3) = b 520 = b e v Th alue of b is the starting value for the function and represents Ilya’s income when n = 0, or when no new policies are sold. We can interpret this as Ilya’s base salary for the week, which does not depend upon the number of policies sold. We can now write the final equation. I(n) = 80n + 520 Our final interpretation is that Ilya’s base salary is 520 per week and he earns an additional 80 commission for each policy sold. Example 10 Using Tabular Form to Write an Equation for a Linear Function Table 1 relates the number of rats in a population to time, in weeks. Use the table to write a linear equation. Number of weeks, w 0 2 4 6 Number of rats, P(w) 1,000 1,080 1,160 1,240 Table 1 Solution We can see from the table that the initial value for the number of rats is 1,000, so b = 1,000. Rather than solving for m, we can tell from looking at the table that the population increases by 80 for every 2 weeks that pass. This means that the rate of change is 80 rats per 2 weeks, which can be simplie fi d to 40 rats per week. P(w) = 40w + 1000 If we did not notice the rate of change from the table we could still solve for the slope using any two points from the table. For example, using (2, 1080) and (6, 1240) 1240 − 1080 __________ m = 6 − 2 160 ___ = 4 = 40 Q & A… Is the initial value always provided in a table of values like Table 1? No. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of 0, then the initial value would be the corresponding output. If the initial value is not provided because there is no value of input on the table equal to 0, find the slope, substitute one coordinate pair and the slope into f (x) = mx + b, and solve for b.290 CHAPTER 4 l i Near f u Ncti o Ns Try It 4 A new plant food was introduced to a young tree to test its ee ff ct on the height of the tree. Table 2 shows the height of the tree, in feet, x months since the measurements began. Write a linear function, H(x), where x is the number of months since the start of the experiment. x 0 2 4 8 12 H(x) 12.5 13.5 14.5 16.5 18.5 Table 2 graphing l inear Functions Now that we’ve seen and interpreted graphs of linear functions, let’s take a look at how to create the graphs. There are three basic methods of graphing linear functions. The first is by plotting points and then drawing a line through the points. The second is by using the y-intercept and slope. And the third method is by using transformations of the identity function f (x) = x. Graphing a Function by Plotting Points To find points of a function, we can choose input values, evaluate the function at these input values, and calculate output values. The input values and corresponding output values form coordinate pairs. We then plot the coordinate pairs on a grid. In general, we should evaluate the function at a minimum of two inputs in order to find at least two points on the graph. For example, given the function, f (x) = 2x, we might use the input values 1 and 2. Evaluating the function for an input value of 1 yields an output value of 2, which is represented by the point (1, 2). Evaluating the function for an input value of 2 yields an output value of 4, which is represented by the point (2, 4). Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error. How To… Given a linear function, graph by plotting points. 1. Choose a minimum of two input values. 2. Evaluate the function at each input value. 3. Use the resulting output values to identify coordinate pairs. 4. Plot the coordinate pairs on a grid. 5. Draw a line through the points. Example 11 Graphing by Plotting Points 2 __ Graph f (x) = −  x + 5 by plotting points. 3 Solution Begin by choosing input values. This function includes a fraction with a denominator of 3, so let’s choose multiples of 3 as input values. We will choose 0, 3, and 6. Evaluate the function at each input value, and use the output value to identify coordinate pairs. 2 __ x = 0 f (0) = −  (0) + 5 = 5 ⇒ (0, 5) 3 2 __ x = 3 f (3) = −  (3) + 5 = 3 ⇒ (3, 3) 3 2 __ x = 6 f (6) = −   (6) + 5 = 1 ⇒ (6, 1) 3 Plot the coordinate pairs and draw a line through the points. Figure 11 represents the graph of the function 2 __ f (x) = − x + 5. 3SECTION 4.1 l i Near f u Ncti o Ns 291 f(x) 6 (0, 5) f 5 4 (3, 3) 3 2 (6, 1) 1 – x –1 16 2 3 4 5 7 2 __ Figure 11 The graph of the linear function f (x) = −  x + 5. 3 Analysis e g Th raph of the function is a line as expected for a linear function. In addition, the graph has a downward slant, which indicates a negative slope. This is also expected from the negative, constant rate of change in the equation for the function. Try It 5 3 _ Graph f (x) = −   x + 6 by plotting points. 4 Graphing a Function Using y-intercept and Slope Another way to graph linear functions is by using specic c fi haracteristics of the function rather than plotting points. e fi Th rst characteristic is its y-intercept, which is the point at which the input value is zero. To find the y-intercept, we can set x = 0 in the equation. e o Th t her characteristic of the linear function is its slope. Let’s consider the following function. 1 _ f (x) =   x + 1 2 1 _ e s Th lope is . B ecause the slope is positive, we know the graph will slant upward from left to right. e Th y-intercept is 2 the point on the graph when x = 0. The g raph crosses the y-axis at (0, 1). Now we know the slope and the y-intercept. rise _ We can begin graphing by plotting the point (0, 1). We know that the slope is rise over run, m = . From our run 1 _ example, we have m =   , w hich means that the rise is 1 and the run is 2. So starting from our y-intercept (0, 1), we can 2 rise 1 and then run 2, or run 2 and then rise 1. We repeat until we have a few points, and then we draw a line through the points as shown in Figure 12. y 5 f 4 y-intercept 3 2 ←Rise = 1 (0, 1) ↑Run = 2 x –2 –1 16 2 3 4 5 7 Figure 12 graphical interpretation of a linear function In the equation f (x) = mx + b • b is the y-intercept of the graph and indicates the point (0, b) at which the graph crosses the y-axis. • m is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope: y − y change in output (rise) Δy 2 1 ___ _ _ m = = = x − x change in input (run) Δx 2 1 292 CHAPTER 4 l i Near f u Ncti o Ns Q & A… Do all linear functions have y-intercepts? Yes. All linear functions cross the y-axis and therefore have y-intercepts. (Note: A vertical line parallel to the y-axis does not have a y-intercept, but it is not a function.) How To… Given the equation for a linear function, graph the function using the y-intercept and slope. 1. Evaluate the function at an input value of zero to find the y-intercept. 2. Identify the slope as the rate of change of the input value. 3. Plot the point represented by the y-intercept. rise _ 4. Use to determine at least two more points on the line. run 5. Sketch the line that passes through the points. Example 12 Graphing by Using the y-intercept and Slope 2 _ Graph f (x) = −  x + 5 using the y-intercept and slope. 3 Solution Evaluate the function at x = 0 to fi nd the y-intercept. e o Th utput value when x = 0 is 5, so the graph will cross the y-axis at (0, 5). 2 _ According to the equation for the function, the slope of the line is −  . This tel ls us that for each vertical decrease in 3 the “rise” of −2 units, the “run” increases by 3 units in the horizontal direction. We can now graph the function by first plotting the y-intercept on the graph in Figure 13. From the initial value (0, 5) we move down 2 units and to the right 3 units. We can extend the line to the left and right by repeating, and then draw a line through the points. f(x) 6 5 f 4 3 2 1 x – –1 16 2 3 4 5 7 2 __ Figure 13 graph of f (x ) = − x + 5 and shows how to calculate the rise over run for the slope. 3 Analysis The graph slants downward from left to right, which means it has a negative slope as expected. Try It 6 Find a point on the graph we drew in Example 12 that has a negative x-value. Graphing a Function Using Transformations Another option for graphing is to use a transformation of the identity function f (x) = x. A function may be transformed by a shift up, down, left, or right. A function may also be transformed using a ree fl ction, stretch, or compression. Vertical Stretch or Compression In the equation f (x) = mx, the m is acting as the vertical stretch or compression of the identity function. When m is negative, there is also a vertical ree fl ction of the graph. Notice in Figure 14 that multiplying the equation of f (x) = x by m stretches the graph of f by a factor of m units if m 1 and compresses the graph of f by a factor of m units if 0 m 1. This means the larger the absolute value of m, the steeper the slope.SECTION 4.1 l i Near f u Ncti o Ns 293 y 6 5 4 3 2 1 x –6 –5 –4 –3 –2 –1 1 2 3 4 56 –1 –2 –3 –4 –5 –6 Figure 14 vertical stretches and compressions and reflections on the function f (x) = x. Vertical Shift In f (x) = mx + b, the b acts as the vertical shift, moving the graph up and down without ae ff cting the slope of the line. Notice in Figure 15 that adding a value of b to the equation of f (x) = x shifts the graph of f a total of b units up if b is positive and ∣b∣ units down if b is negative. y f(x) = x + 4 f(x) = x + 2 10 8 f(x) = x 6 f(x) = x – 2 4 f(x) = x – 4 2 x –10 –8 –6 –4 –2 28 4 6 10 –2 –4 –6 –8 –10 Figure 15 This graph illustrates vertical shifts of the function f (x) = x. Using vertical stretches or compressions along with vertical shifts is another way to look at identifying die ff rent types of linear functions. Although this may not be the easiest way to graph this type of function, it is still important to practice each method. How To… Given the equation of a linear function, use transformations to graph the linear function in the form f (x) = mx + b. 1. Graph f (x) = x. 2. Vertically stretch or compress the graph by a factor m. 3. Shift the graph up or down b units. Example 13 Graphing by Using Transformations 1 __ Graph f (x) =   x − 3 using transformations. 2 1 1 __ __ Solution e e Th quation for the function shows that m =   s o the identity function is vertically compressed by . eTh 2 2 equation for the function also shows that b = −3 so the identity function is vertically shifted down 3 units. First, graph the identity function, and show the vertical compression as in Figure 16. e Th n show the vertical shift as in Figure 17.294 CHAPTER 4 l i Near f u Ncti o Ns y y 5 5 1 y = x y = x 4 4 2 3 3 2 1 2 1 y x = 2 y = x − 3 1 1 2 x x –7–6 –5–4 –3–2 –1 14 2 3 56 7 –7 –6 –5 –4 –3 –2 –1 14 2 3 56 7 –1 –1 –2 –2 –3 –3 –4 –4 –5 –5 1 __ 1 Figure 16 The function, y = x, compressed by a factor of . __ Figure 17 The function y =    x, shifted down 3 units. 2 2 Try It 7 Graph f (x) = 4 + 2x, using transformations. Q & A… In Example 13, could we have sketched the graph by reversing the order of the transformations? No. The order of the transformations follows the order of operations. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. This is why we performed the compression first. For example, following the order: Let the input be 2. 1 __ f (2) =    (2) − 3 2 = 1 − 3 = −2 Writing the equation for a Function from the graph of a l ine Earlier, we wrote the equation for a linear function from a graph. Now we can extend what we know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at Figure 18. We can see right away that the graph crosses the y-axis at the point (0, 4) so this is the y-intercept. y f 10 8 6 4 2 x –10–8 –6–4 –2 28 4 6 10 –2 –4 –6 –8 –10 Figure 18 e Th n we can calculate the slope by finding the rise and run. We can choose any two points, but let’s look at the point (−2, 0). To get from this point to the y-intercept, we must move up 4 units (rise) and to the right 2 units (run). So the slope must be rise 4 _ _ m = =    = 2 run 2 Substituting the slope and y-intercept into the slope-intercept form of a line gives y = 2x + 4SECTION 4.1 l i Near f u Nctio Ns 295 How To… Given a graph of linear function, find the equation to describe the function. 1. Identify the y-intercept of an equation. 2. Choose two points to determine the slope. 3. Substitute the y-intercept and slope into the slope-intercept form of a line. Example 14 Matching Linear Functions to Their Graphs Match each equation of the linear functions with one of the lines in Figure 19. 1 __ a. f (x) = 2x + 3 b. g(x) = 2x − 3 c. h(x) = −2x + 3 d. j(x) =    x + 3 2 y II I 5 III 4 3 2 1 x –7–6 –5–4 –3–2 –1 14 2 3 56 7 –1 –2 –3 –4 –5 IV Figure 19 Solution Analyze the information for each function. a. Thi s func tion has a slope of 2 and a y-intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. We can use two points to find the slope, or we can compare it with the other functions listed. Function g has the same slope, but a different y-intercept. Lines I and III have the same slant because they have the same slope. Line III does not pass through (0, 3) so f must be represented by line I. b. This func tion also has a slope of 2, but a y-intercept of −3. It must pass through the point (0, −3) and slant upward from left to right. It must be represented by line III. c. This func tion has a slope of −2 and a y-intercept of 3. This is the only function listed with a negative slope, so it must be represented by line IV because it slants downward from left to right. 1 _ d. Thi s fun ction has a slope of and a y-intercept of 3. It must pass through the point (0, 3) and slant upward 2 from left to right. Lines I and II pass through (0, 3), but the slope of j is less than the slope of f so the line for j must be flatter. This function is represented by Line II. Now we can re-label the lines as in Figure 20. y g(x) = 2x − 3 5 4 3 2 1 j(x) = x + 3 2 1 x –7 –6 –5–4 –3 –2 –1 14 2 3 56 7 –1 –2 –3 –4 –5 f (x) = 2x + 3 h(x) = −2x + 3 Figure 20296 CHAPTER 4 l i Near f u Ncti o Ns Finding the x-intercept of a Line So far, we have been finding the y-intercept of a function: the point at which the graph of the function crosses the y-axis. Recall that a function may also have an x-intercept, which is the x-coordinate of the point where the graph of the function crosses the x-axis. In other words, it is the input value when the output value is zero. To find the x-intercept, set a function f (x) equal to zero and solve for the value of x. For example, consider the function shown. f (x) = 3x − 6 Set the function equal to 0 and solve for x. 0 = 3x − 6 6 = 3x 2 = x x = 2 e g Th raph of the function crosses the x-axis at the point (2, 0). Q & A… Do all linear functions have x-intercepts? No. However, linear functions of the form y = c, where c is a nonzero real number, are the only examples of linear functions with no x-intercept. For example, y = 5 is a horizontal line 5 units above the x-axis. This function has no x-intercepts, as shown in Figure 21. y y = 5 5 4 3 2 1 x –5 –4 –3 –2 –1 14 2 3 5 –1 –2 –3 –4 –5 Figure 21 x-intercept The x-intercept of the function is value of x when f (x) = 0. It can be solved by the equation 0 = mx + b. Example 15 Finding an x-intercept 1 _ Find the x-intercept of f (x) =    x − 3. 2 Solution Set the function equal to zero to solve for x. 1 _ 0 =   x − 3 2 1 __ 3 =    x 2 6 = x x = 6 e g Th raph crosses the x-axis at the point (6, 0). Analysis A graph of the function is shown in Figure 22. We can see that the x-intercept is (6, 0) as we expected.SECTION 4.1 l i Near f u Ncti o Ns 297 y 4 2 x –10–8 –6–4 –2 28 4 6 10 –2 –4 –6 –8 –10 Figure 22 Try It 8 1 _ Find the x-intercept of f (x) =   x − 4. 4 Describing Horizontal and Vertical Lines e Th re are two special cases of lines on a graph—horizontal and vertical lines. A horizontal line indicates a constant output, or y-value. In Figure 23, we see that the output has a value of 2 for every input value. The change in outputs between any two points, therefore, is 0. In the slope formula, the numerator is 0, so the slope is 0. If we use m = 0 in the equation f (x) = mx + b, the equation simplie fi s to f (x) = b. In other words, the value of the function is a constant. This graph represents the function f (x) = 2. y 5 x −4 −2 0 2 4 4 3 f y 2 2 2 2 2 2 1 x –5 –4 –3 –2 –1 14 2 3 5 Figure 23 A horizontal line representing the function f (x) = 2. A vertical line indicates a constant input, or x-value. We can see that the input value for every point on the line is 2, but the output value varies. Because this input value is mapped to more than one output value, a vertical line does not represent a function. Notice that between any two points, the change in the input values is zero. In the slope formula, the denominator will be zero, so the slope of a vertical line is undefined. change of output ← Non-zero real number __ m = ← 0 change of input Figure 24 example of how a line has a vertical slope. 0 in the denominator of the slope. Notice that a vertical line, such as the one in Figure 25, has an x-intercept, but no y-intercept unless it’s the line x = 0. This graph represents the line x = 2. y 5 4 3 2 x 2 2 2 2 2 1 x y −4 −2 0 2 4 –5 –4 –3 –2 –1 14 2 3 5 –1 –2 –3 –4 –5 Figure 25 The vertical line, x = 2, which does not represent a function. 298 CHAPTER 4 l i Near f u Ncti o Ns horizontal and vertical lines Lines can be horizontal or vertical. A horizontal line is a line defined by an equation in the form f (x) = b. A vertical line is a line defined by an equation in the form x = a. Example 16 Writing the Equation of a Horizontal Line Write the equation of the line graphed in Figure 26. y x –10–8 –6–4 –2 28 4 6 10 –2 –4 f –6 –8 –10 Figure 26 Solution For any x-value, the y-value is −4, so the equation is y = −4. Example 17 Writing the Equation of a Vertical Line Write the equation of the line graphed in Figure 27. y 10 8 f 6 4 2 x –10–8 –6–4 –2 28 4 6 10 –2 –4 –6 –8 –10 Figure 27 Solution e Th constant x-value is 7, so the equation is x = 7. determining Whether l ines are Parallel or Perpendicular e t Th wo lines in Figure 28 are parallel lines: they will never intersect. They have exactly the same steepness, which means their slopes are identical. The only difference between the two lines is the y-intercept. If we shifted one line vertically toward the y-intercept of the other, they would become coincidentt. y 8 7 6 5 4 3 2 1 x –6 –5 –4–3 –2 –1 14 2 3 5 6 –1 –2 –3 Figure 28 Parallel lines We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the y-intercepts are die ff rent, the lines are parallel. If the slopes are die ff rent, the lines are not parallel.