Contents 4. Functions and Graphs 2 4.5 Calculator Graphs and Solving Equations Graphically........... 2 Example 4.53................................ 2 Example 4.54................................ 3 Example 4.55................................ 3 Example 4.56................................ 4 Example 4.57................................ 5 Example 4.58................................ 6 Example 4.61................................ 6 4.6 Introduction to Modeling.......................... 8 Example 4.62................................ 8 Example 4.63................................ 9 Example 4.64................................ 11 Example 4.65................................ 12 1
Peterson, Technical Mathematics, 3rd edition 2 4.5 Calculator Graphs and Solving Equations Graphically A graphing calculator or computer can remove much of the drudgery of plotting equations by hand. In this section we will use a graphing calculator to demonstrate some mathematical concepts and to show how it can relieve you of some work. Moreover, we will show you how to interpret the information displayed on the machine. Examples in this section were run on a Texas Instruments TI-83 graphing calculator. However, you could have used a different graphing calculator or a computer program. Rather than teach you how to write a graphing program for a computer, we will assume you have access to a program that can be used to graph functions. Some, such as Excel, are computer spreadsheets that have graphing capabilities. Other more specialized programs include Mathematica, Derive, and Maple, all of which are classified as computer algebra systems. A purpose of this book is not to explain how to use the graphing capabilities of these programs, but to help you use these computer programs to graph curves and to interpret their graphs. Using a Graphing Calculator We begin with a discussion on using graphing calculators. The examples and the instructions in this book are meant to supplement the user s guide for your calculator, not to replace it. You should always consult the user s guide to get answers to how to questions. All examples described in the text used a Texas Instruments TI-86 graphing calculator. This special print out works those same examples using a TI-83 graphing calculator. The first equation to graph will be of a straight line, y = 2x + 7. Before we begin, we must determine the domain of this function and what portion we want to see. The viewing range of this function is partially determined by the allowable values of y that can appear on the calculator s screen. Example 4.53 Use a calculator to sketch the graph of f(x) = 2x + 7. FIGURE 4.26a Solution We begin by entering the function. First, press Y= and then type 2x+7. For the x key you can press the key labeled X,T,θ,n. What you see should look like Figure 4.26a The domain of this function is all real numbers. We can only view a portion of the domain in the viewing window. We select 5 x 4. To set the size of the viewing window press the WINDOW key. The part of the x-coordinate system that will be shown on the screen is indicated by Xmin (the smallest value of x displayed on the screen), Xmax (the corresponding largest x-value), and Xscl (the x-axis scale or, more precisely, the distance between tick marks on the x-axis). The y-coordinate is similarly restricted by Ymin, Ymax, and Yscl. For this function we use the values in Figure 4.26b. To display the graph press GRAPH. The result is shown in Figure 4.26c.
Peterson, Technical Mathematics, 3rd edition 3 FIGURE 4.26b FIGURE 4.26c Example 4.54 Use a graphing calculator to sketch the graph of g(x) = 0.2x 2 + 15. FIGURE 4.27a Solution Again, press Y=. If you want to erase the previous function, press CLEAR. If you want to graph this new function and the previous function, press ENTER and then type 0.2x 2 +15. To get x 2 you can type either X,T,θ,n x 2 or X,T,θ,n 2. When you press graph you get the result shown in Figure 4.27a. This is not very satisfactory because we don t see any of the graph. This is because we used the same window settings as in the previous example. There are several ways to correct this. One way is to decide what x-values you want to see in the viewing windows. Suppose for this example, you wanted to see the graph when 6 x 5. Press WINDOW and set xmin=-6 and xmax=5. If you then use ZoomFit by pressing ZOOM and holding down the key until you see 0: ZoomFit, as shown in Figure 4.27b. Pressing ENTER graphs the function and gives Figure 4.27c. FIGURE 4.27b FIGURE 4.27c Whenever we are not using the default window settings on a calculator, we will adopt the notation [Xmin, Xmax, Xscl] [Ymin, Ymax, Yscl]. Thus, the viewing window used in Figure 4.27c is [ 6, 5, 1] [15, 22.2, 1]. If both Xscl and Yscl are 1 we may shorten this to [ 6, 5] [15, 22.2]. The default window settings, written ZStandard for Zoom Standard, are [ 10, 10, 1] [ 10, 10, 1]. Example 4.55 Use a graphing calculator to sketch the graph of f(x) = x2 + 1 x + 2. Solution In graphing this function you must be careful to insert enough parentheses and to insert them in the right spots. Remember that everything over the fraction bar is one group and everything under the fraction bar is another group. Parentheses are used to group terms. So, we end up writing the equation as y = (x 2 + 1)/(x + 2)
Peterson, Technical Mathematics, 3rd edition 4 FIGURE 4.28a When this is graphed on a TI-83 in the standard viewing window. you get the graph in Figure 4.28a. (You get the standard viewing window by pressing ZOOM 6 [6:ZStandard].) Sometimes, as in Figure 4.28a, the calculator draws in some points (usually shown as vertical lines) that are really not part of the graph. This may also happen with some computer graphing software. It may be possible to adjust the viewing window, as in Figure 4.28b, to eliminate these extraneous points. Another possibility is to set the graph format to draw individual dots (Dot on a TI-83) rather than connecting the dots. When this is done you get a graph like the one in Figure 4.28c. ON a TI-83, you get the calculator to draw individual dots by pressing MODE ENTER. FIGURE 4.28b FIGURE 4.28c Solving Equations Graphically To solve an equation graphically means to locate the x-intercepts of the graph. For a function these points are called the zeros of the function because these are the values of x where f(x) = 0. Other places these are called the solutions or roots of the equation. Steps to Solving an Equation Graphically 1. Write the equation in the form [expression in x] = 0 or f(x) = [expression in x]. 2. Graph y = [expression in x]. 3. Find the x-intercepts of the graph. These are the solutions of the equation. Example 4.56 Graphically approximate the roots of x 2 3x 1 = 0. Solution First, set f(x) = x 2 3x 1. Then set up a partial table of values for the function f. x 2 1 0 1 2 3 4 5 y = f(x) 9 3 1 3 3 1 3 9 Graphing the curve determined by these points, we get the curve in Figure 4.29 in the text. From the graph we can see that there appear to be roots at x = 0.25 and x = 3.25. If we evaluate the function at these two values, we get f( 0.25) = 0.1875 and f(3.25) = 0.1875. This shows two things. First, it shows that 0.25 and 3.25 are not roots of this function, since f(x) 0 at either of these two points. Second, it does show that the roots of this function are close to 0.25 and 3.25.
Peterson, Technical Mathematics, 3rd edition 5 If we wanted more accurate approximations of the roots of the function in Example 4.56, we could substitute different values for x until we got an approximation that was as accurate as we wanted. In a later chapter, we will find out how to determine the exact roots to this function. In the next example, we will show how to use a graphing calculator to approximate this function s roots. Example 4.57 Use a graphing calculator to find the approximate roots of f(x) = x 2 3x 1. FIGURE 4.30a Solution This is the same function we graphed in Example 4.56. We already know that the roots are near x = 0.25 and x = 3.25. When you graph the function using the calculator s default window settings, you obtain the graph in Figure 4.30a on a TI-83. As you can see from this figure, the graph crosses the x-axis at two points; these are the x-intercepts or roots. On a TI-83 press 2nd TRACE [CALC] 2 [2:zero]. [Note: Some calculators, here the TI-83, use the term zero rather than root.] This function has two roots, and we will find the one on the left first. At the bottom left of the calculator screen you should see the question Left Bound? as in Figure 4.30b. Use the or to move the cursor to the left of the root you are trying to find. When the cursor is left of this root press ENTER and you see something like Figure 4.30c. FIGURE 4.30b FIGURE 4.30c Now, you should see the question Right Bound? at the bottom left of the screen. Use the key to move the cursor to the right of the root you are trying to find as in Figure 4.30c. Again, press ENTER. You should now see the question Guess?. Move the cursor near the x-intercept and press ENTER a third time. Shortly, you should see Figure 4.30d on your calculator screen. This indicates that the root is x 0.3024456 (A TI-83 gives an answer of x 0.3027756377.) The screen also displays y=0. Now, approximate the other root of this function. Press 2nd TRACE [CALC] 2 [2:zero] and repeat the process as shown in Figures Figure 4.30e Figure 4.30g. The second root is x 3.3027756. (Notice in the text that the TI-83 found that the root is x 3.3027756377.)
Peterson, Technical Mathematics, 3rd edition 6 FIGURE 4.30d FIGURE 4.30e FIGURE 4.30f FIGURE 4.30g FIGURE 4.31 Example 4.58 Use a graphing calculator or computer graphing software with zoom capability to determine any x-intercepts of y = x 2 + 1.5 Solution The graph of this function is shown in Figure 4.31. As you can see, the graph does not cross the x-axis. Thus, this function does not have any real roots. * * * Go to the textbook until you finish Example 4.60. * * * Example 4.61 Use a graphing calculator to rework parts of Example 4.60. FIGURE 4.36 (a) Sketch the graph of C(m) = 50 m 2 + 4, m 0. (b) Use the graph and your calculator to estimate how far you are from the factory if the concentration of the pollutant is 3.08 ppm. Solutions: (a) You must make two changes before we begin graphing. First, you must replace C(m) with y. Second, your calculator will only allow you to use certain variables when you graph a function. At present the only variable you can use is x. So, rewrite the given function as y = 50 x 2, x 0. Don t forget the correct + 4 50 use of parentheses. You will want to graph this as y = (x 2. Also, to get + 4) the graph to look like the one in Figure 4.35b, change the viewing window to [0, 14, 1] [0, 14, 1]. The result is the graph shown in Figure 4.36.
Peterson, Technical Mathematics, 3rd edition 7 (b) We want to know when the concentration of the pollutant is 3.08 ppm. This means we want to know the value of x when y = 3.08. Press TRACE and move the cursor until you have a y-value near 3.08. On this particular calculator, the y-values change from around 2.98 to 3.17. Zooming in around one of these points and using the trace function, we see that when y 3.08, then x 3.4976. Once again, we conclude that the pollutant was collected about 3.5 mi from the factory.
Peterson, Technical Mathematics, 3rd edition 8 4.6 Introduction to Modeling We are not always given a function or an equation and asked to use it. Instead, we collect some data and have to interpret it. The first step is to organize the data in a list or table. Next, graph the points and use a calculator or computer program to draw a curve through the points. The points seldom lie on a nice curve. You can connect the points using a series of line segments in what is called a broken line graph. A broken line graph makes the graph easier to see but is not very useful. If at all possible we want to develop a model for the data. A model is a function that comes close to going through the data points. It may not go through them all in fact it may not go through any of them but it comes the closest of any given type of function. Determining these models uses a technique known as regression and until it was built into calculators it was very difficult to do. A look to the Future The next example will show a plot of some data, a broken line graph for that data, and one model that seems to fit the data. We will use that model to predict an outcome that is not in the table. This example is intended as a demonstration of what we will be doing later in this section and later in the book. Example 4.62 Table 4.3 gives the annual domestic fuel consumption, in billion gallons, of all motor vehicles in the United States for selected years from 1970 through 1999. (a) Plot the points. (b) Connect the points to form a broken line graph. (c) Determine a regression curve that will fit the data. (d) Use the regression curve to determine the fuel consumption in 1998 and 2000. Table 4.3: Annual Motor Vehicle Fuel Consumption in the US, 1970 1999 Year 1970 1975 1980 1985 1990 1995 1997 1999 Fuel consumption 92.3 109.0 115.0 121.3 130.8 143.8 150.4 160.7 (billion gallons) Solutions: Note that the entry for 1970 is 92.3. The fuel data is given in billion gallons so 92.3 represents 92,300,000,000 gallons. (a) Using the statistical plotting feature of the calculator, we are given several plot types. The first is the scatter plot. When it is selected and the data in Table 4.3 is plotted using the zoom setting 9:ZoomStat you obtain the result in Figure 4.37a. (b) Changing the plot type to xyline you obtain the broken line graph in Figure 4.37b.
Peterson, Technical Mathematics, 3rd edition 9 FIGURE 4.37a FIGURE 4.37b FIGURE 4.37c FIGURE 4.37d (c) A cubic regression that fits this data is C(t) 0.005t 3 1.306t 2 + 109.465t 2972.151 billion gallons t years after 1900. When the actual regression equation produced by the calculator is graphed on the scatter plot we obtain the graph in Figure 4.37c. (d) We can use the regression formula stored in the calculator to approximate the fuel consumption in 1998 and 2000. According to the calculator (see Figure 4.37d) C(98) = 155.6. Note here that in the regression formula t represented years after 1900 and so 1998 would use the value t = 98. Similarly, fuel consumption in 2000 would be C(100) 166.2 billion gallons. The actual fuel consumption for 1998 was 155.4 billion gallons; very close to the predicted value. The actual value for 2000 was 162.3 billion gallons. Linear Curve Fitting In Example 4.60 we looked at the graph of some data, had the calculator determine the formula for a curve that would come close to passing through the data, and made some predictions based on the formula. Now lets look at this more closely. We will begin with some data that seems to fit a line. This process is called linear curve fitting or linear regression. Graphing calculators and many computer software programs with graphing ability contain built-in programs that find the best fit equation for a collection of points in a scatter diagram. Different software has different names for this ability. For example, in Excel it is referred to as Add Trendline and a TI-83 calls it LinReg. Example 4.63 The total vehicle highway miles of travel in the United States from 1980 1999, in billions of miles, is shown in Table 4.4. (a) Plot the points. (b) Determine a linear regression curve that will fit the data. Solutions: (a) On a calculator you begin by entering the data in two lists. On a TI-83 calculator you begin by pressing the STAT key and then selecting 1:EDIT by either
Peterson, Technical Mathematics, 3rd edition 10 Table 4.4: Motor Vehicle Miles of Travel Year 1980 1985 1990 1992 1993 1995 1997 1999 Billion Miles 1527 1775 2143 2247 2296 2423 2562 2691 FIGURE 4.38a pressing ENTER or 1. The first two columns are called L1 and L2. Place the independent variable, in this case the years, in the L1 list and place the dependent variable, the number of miles, in the L2 list. In Figure 4.38a you will see that only the last two digits of the year were entered in L1. If you do this type of transformation of the data you must be sure to indicate that fact when you write the linear equation and also when you use the equation to make predictions. To graph these data first press Y= and delete any functions that are listed. Next, press 2nd Y= [STAT PLOT]. The result, shown in Figure 4.38b, shows three possible stat plot screens. Press ENTER or 1 to select the first plot screen. The result is shown in Figure 4.38c. Set your calculator so it looks like the screen in Figure 4.38c. FIGURE 4.38b FIGURE 4.38c To plot the data press ZOOM 9 [9:ZoomStat]. The result should look like Figure 4.38d. To get the regression equation, press STAT 4 [Lin- Reg (ax+b)] ENTER with the result in Figure 4.38e. The second, third, and fourth lines of Figure 4.38e indicate that the linear regression line is y = ax + b and that a = 61.95951417 and b = 3453.550607. We would probably write our answer as the function M(t) = 61.95951417t 3453.550607 billion miles t years after 1900. FIGURE 4.38d FIGURE 4.38e
Peterson, Technical Mathematics, 3rd edition 11 FIGURE 4.38f To graph the regression line with the original data first press Y= and then press VARS 5 [5:Statistics] 1 [1:RegEq] ENTER. This pastes the regression equation on the Y1 line of the Y= screen. Notice that the regression equation shows more digits for a and b than were given in Figure 4.38e. Now press GRAPH to get the result in Figure 4.38f. As we will see later when we study linear equations, 61.96 is the slope of this line and 3453.6 is the y-intercept. * * * Go to the textbook and read the subsection Correlation Coefficients. * * * Interpolation and Extrapolation When you predict values of the dependent variable for independent values that are within the interval of the given data in your table or scatter plot, you are using a process called interpolation. On the other hand, when you are predicting values of the dependent variable for values of the independent variable that are outside the interval of the given data you are using a process call extrapolation. Because you do not know what happens outside the given data, estimates obtained through extrapolation should be viewed with more caution than estimates found with interpolation. Example 4.64 FIGURE 4.40a FIGURE 4.40b Use the regression curve from Example 4.63 to determine the total vehicle highway miles of travel in the United States for 1996 and 2000. Solution In Example 4.63, we entered the regression equation from Example 4.63 as Y1 in the Y=. We can use this to evaluate M(1996) and M(2000). 1. Use the function feature of the calculator. Key VARS 1 [1:Function] ENTER ( 96 ) ENTER and VARS 1 [1:Function] ENTER ( 100 ) ENTER on the home screen of the calculator. You should obtain something like Figure 4.40a. 2. Use the table feature of the calculator. Since you want the data for 1996 and 2000 press 2nd WINDOW [TBLSET] and set TblStart=96. If you leave Tbl=1 and press 2nd GRAPH [TABLE] you should obtain Figure 4.40b. Both procedures yield the same results. According to our regression formula and interpolation all the motor vehicles in the United States traveled around 2495 billion miles. (The actual value was 2486 billion miles.) In 2000, extrapolation predicts that the vehicles traveled about 1742 billion miles.
Peterson, Technical Mathematics, 3rd edition 12 Example 4.65 The following table gives the height, in inches, and the weight, in pounds, for 12 students. Student A B C D E F G H I J K L Height 66 67 68 69 70 71 72 70 72 71 72 73 Weight 130 140 180 160 185 190 200 195 195 210 210 180 (a) Plot the points. (b) Determine a linear regression curve that will fit the data. (c) Determine the correlation coefficient for this curve. (d) Use the regression curve to predict the weight of a student who is 6 6 tall. Solutions: (a) A graph of these 12 points is shown in Figure 4.41a. A quick glance at this graph tells you that these points do not all lie on the same straight line. But, in general, the taller a person is, the more he or she weighs. Thus, we think that it is possible that there might be some straight line that would come close to these points and from which none of the points differ by very much. FIGURE 4.41a (b) The calculator gives a regression curve of y 9.614173228x 492.5433071. We will write this as the function w(h) 9.614173228h 492.5433071 where h is the height in inches and w is the weight in pounds. A graph of the original 12 points and the regression line is in Figure 4.41b. FIGURE 4.41b (c) The correlation coefficient is about r = 0.82. (d) A student who is 6 6 tall is 78 inches tall. Evaluating, we obtain w(78) 257. So, a student who is 6 6 tall would weigh about 257 pounds.