Section 1.1 Graphs Graphs

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2 Section 1.1 Graphs Graphs Much of algebra is concerned with solving equations. Many algebraic techniques have been developed to provide insights into various sorts of equations, and those techniques are essential for understanding the basic ideas of calculus and more advanced mathematical topics. In recent decades people have also used calculators and computers for calculating approximate solutions to certain types of equations. Such numerical solution methods are often essential for solving real problems in engineering and science. In this course we investigate problems both algebraically and graphically. It is often useful to consider a situation from several viewpoints: a different direction can provide insights that are not obvious from the original position. The geometrical approach to solving equations has been made more productive by the invention of the graphing calculator. These calculators quickly construct graphs that are fairly accurate and reliable. In pre-calculator years, the graphical approach was much more tedious. In the next two sections (1. and 1.3) we will explore various methods to solve equations graphically. (It is assumed that we already know the algebraic methods. Those methods are outlined in sections.1 and. but we will not spend time to review them now.) In this section, we will discuss the rectangular coordinate system, properties of graphs, and how to use the calculator to display a graph. The Rectangular Coordinate System: (x, y)-plane The rectangular coordinate system is a system for labeling points in the plane. The coordinate plane is constructed from two number lines, the x-axis and the y-axis. The x-axis and the y-axis are perpendicular to each other and intersect at the point 0 on both lines. (Perpendicular means that the two lines meet at 90 angles.) The point of intersection is called the origin. See Figure 1 below. II y-axis I origin x-axis III IV Figure 1

3 56 Chapter 1 Graphs and Calculators The plane is divided into four quadrants, labeled I, II, III, and IV as in Figure 1. These intersecting number lines allow us to represent each point in the plane by an ordered-pair written as (x, y). Each point of the plane is either on an axis or in one of the quadrants. This coordinate system provides a major benefit: it allows us to solve geometrical problems using algebra. Two useful features are the formulas to compute the distance between two points and for finding midpoints of line segments. Distance Formula and The Distance Formula states that the distance d between the points ( x, y 1 1 ) ( x, y ) is given by the formula: d = ( x x ) + ( y ) 1 1 y The derivation of the distance formula uses the Pythagorean Theorem and is included as an exercise at the end of the section. Example 1 (Distance Formula) Find the distance between the points (8, 76) and (59, 3). To solve this problem we use the distance formula and let (59, 3) be ( x, y ) 1 1 and (8, 76) be ( x, y ). d = ( 59 8) + ( 3 76) Substitute the values into the formula. d = ( 31) + ( 44 ) d = 897 d = (an infinite decimal) Note that 897 is the exact answer but it is often more useful to have an approximate (rounded off) answer. So, after rounding to two decimal places, the distance between the two points is approximately Midpoint Formula The midpoint of a line segment is the point that is half-way between the endpoints. That is, it is the only point on the line segment that is the same distance from each end. On a number line, the number that is halfway between two numbers is the average of the two numbers. For instance, to find the number halfway a + b between a and b, we compute the value. For points in the plane, the x- coordinate of the midpoint is halfway between the x-coordinates of the endpoints, x and similarly for the y-coordinate of the midpoint. So if we let the points ( ) 1, y 1

4 Section 1.1 Graphs 57 ( ) and x, y be the endpoints of the line segment, then the Midpoint Formula states that the midpoint is: x1 + x y1 + y, Example (Midpoint Formula) Find the midpoint of the line segment between the two points (4, 9) and (7, 6) Here we let ( x, y 1 1 ) be the point (4, 9) and ( x, y ) be the point (7, 6). Then the midpoint equals: ( 6), 31 3, or (15.5, 1.5) 31 3 To be sure that, is the midpoint, we can check that the distance 31 3 between (4, 9) and, is the same as the distance between (7, 6) 31 3 and,. This is left as an exercise in using the distance formula. Computing distances and midpoints is useful but the real utility of the rectangular coordinate system can be seen when we look at an expression like 3x 4. This expression involves a single variable x, which represents an unknown number. It can be viewed as a command, change the number x by multiplying by 3 and then subtracting 4. Thinking geometrically on a single number line, we may view this as the command move x to a spot 3 times as far from the origin and then shift that point to the left 4 units. The behavior of such an expression is represented more clearly when we introduce a new variable y and consider the equation y = 3x 4 as a relationship between two variables. The relationship can be described using a table as in Table 1. Note that the table does not contain all of the possible values for x since there are infinitely many.

5 58 Chapter 1 Graphs and Calculators X Y Table 1 Since we have both an x-value and y-value for each entry in the table, we can plot each row of the table as a point in the plane. In this example we would put dots at the points (0, 4), (1, 1), (, ), (3, 5), and (4, 8). By representing all possible entries from the table in such a fashion, we obtain the graph of y = 3x 4. We can not draw every dot separately since there are infinitely many, but we should plot enough of them to see the pattern. Each of the solutions (or possible entries in the table) is included on the graph and the graph represents the set of all solutions to the equation. In this case, the graph is a straight line. A graph gives a picture of the behavior of an equation and represents the set of all solutions to the equation. Remember that the graph sketch is a finite representation of an infinite relationship. Just as the table is limited in the entries that it can display, the graph is limited too. At any time, we see only a small portion of the graph of an equation in two variables. Since this is the case, we need to be sure that we are looking at a picture which best describes the behavior and displays the important features of the graph. This is particularly true when using a graphing calculator. On CD: Basics of Graphing Getting Your Function Into the Proper Format for Graphing Graphs and the Graphing Calculator Entering equations into the TI 83 calculator is done by using the Equation Window. This feature allows you to store various equations in the memory of the calculator. Graphs of those equations are displayed in the viewing window. The standard viewing window of the TI 83 displays the values of x from 10 to 10, and the values of y from 10 to 10, each with a scale of 1. (The scale refers to the distance between the tic marks on the axes.) To view the dimensions of the current viewing window, press the [WINDOW] key. To get a good picture of the graph, we will develop some strategies for selecting a window that shows all the important features of the graph. The trace, zoom, and table features will be used to help determine good viewing windows. To use the calculator effectively, we need to be able to access and understand all of these features. A summary of the calculator features is provided in Table. Please note: Additional help on using the calculator can be found on the included CD-ROM.

6 Section 1.1 Graphs 59 Feature Calculator Key Uses Equation Window [Y=] Used for entering equations Graphs Graphing Equations [GRAPH] Allows the user to view the graphs of any or all of the equations entered in the calculator. Viewing Window [WINDOW] Displays the dimensions of the current viewing window. These dimensions can be adjusted by the user. Standard Window [ZOOM]: select Returns the calculator to the standard Zoom In Zoom Out ZStandard [6] [ZOOM]: select Zoom In [] [ZOOM]: select Zoom Out [3] viewing window Adjusts the viewing window in ways: 1. re-centers the window at a different point in the plane;. decreases the window values by a factor of 4. (This makes the picture of the graph look bigger.) Does the same as Zoom In except that it increases the window values by a factor of 4. (This makes the picture of the graph look smaller.) Tracing the graph [TRACE] Allows the user to move along the graph using the and arrows. It displays the coordinates of the highlighted point of the graph at the bottom of the viewing window. Tables Table for equations [ nd ] [GRAPH] This displays a table with entries representing solutions to the selected equation (or equations). These are points on the graph. Scroll through the values using the and keys. Table Setup [ nd ] [WINDOW] Displays the starting value for the table as well as the increment for determining the next x value. These values can be adjusted by the user. Table

7 60 Chapter 1 Graphs and Calculators One of the most difficult aspects of using a graphing calculator is finding an appropriate window in which to view the graph. Remember that we can see only a finite portion of an infinite relationship so it is important to develop an understanding of what the graph of the given equation should look like. The following examples explore how to use the features from Table to adjust the window and to find solutions to equations in two variables. In some cases we need to see only a portion of the graph, such as an intercept, but other times we need to see a complete graph. A complete graph is a graph that displays all of the important features of the given equation such as peaks, valleys, and intercepts. Finding good viewing windows is not an exact science and there are many different windows that show a complete graph for a given equation. Example 3 (Graphing Equations) 3 a) Graph the equation y = x + 0x + 5x in the standard window and in the window with x-values in the interval [ 15, 10] and y-values in the interval [ 10, 300]. Adjust the y-scale to 30. a) To view the graphs we must first enter the equation into the calculator using the equation window. Once the equation is entered, we view the graph in the standard viewing window. We now must adjust the window so that we are using the given values. We then display the graph in the new window. (What happens when the y-scale is left at 1?) On CD: Window Menu Overview Notice that this window gives us a much better picture of the graph and it also shows a complete graph of the equation. It can be hard to believe that both of the pictures show the same graph since they appear to be very different. This new window does provide a complete graph because it turns

8 Section 1.1 Graphs 61 out that there are no more wiggles that occur outside the window. This knowledge comes with experience with graphing equations and recognizing that certain families of equations have some features in common. You should already be familiar with the characteristics of the linear family and the quadratic family. b) Use the graph to find two solutions to the equation. Use the table feature to find two more solutions to the equation. Also, find y when x is 5 and when x is 6.3. b) We will use the trace feature of the calculator to find two solutions to the equation. Once we enter the trace mode, the calculator displays the solutions at the bottom of the calculator. On CD: Tracing the Function: The Trace Key There are many choices for solutions. Using the arrows we find one at approximately (1.755, 81.16) and another one at approximately ( 4.096, ). Remember that, in most cases, the calculator will give only approximate values. Using the table feature we can display more solutions at one time. Here we see a list of seven solutions. This also illustrates the table setup procedure. To find the y-values for the given values of x, we can use the trace or table features. To use the trace feature, enter the trace mode and type the value for x. In this case we enter 5. Once we press [Enter] the machine displays the coordinates for the point with x-coordinate 5.

9 6 Chapter 1 Graphs and Calculators So y = 775 when x = 5. To find the value for y when x = 6.3, we can use the table feature. (If you try to use the trace method you will get an error unless you adjust the window.) Enter the table setup and set the value for TblStart = 6 and Tbl= 0.1. This produces the following table. From this we can see that when x = 6.3, y is approximately Note that the table can display only 6 characters in the y column. Example 4 (Complete Graphs) Find the complete graph of the equation y =.008x.3x x.004x We first enter the equation in the equation window and look at the graph in the standard window. The standard window is a good place to start when trying to find a complete graph. On CD: Graphing Polynomials Notice that we seem to be missing a piece of the graph when x is between 0 and 5. So this picture is not a complete graph. We need to find a window that shows all of the important features of the graph and still gives the graph

10 Section 1.1 Graphs 63 a nice shape. Since some of the graph is outside the current window, we should zoom out. After we zoom out we have the following picture. On CD: Zoom Menu Overview We can now see more of the graph but we also discover that we are missing another piece. We should adjust the minimum values for y to see the rest of the graph. To do this, trace the graph and watch the values for y. We then reach the smallest value for y when x is about 5.5. Now we use this information from the first two pictures of the graph to adjust our window and get a complete graph. To check that this is a complete graph we can use the trace feature to see that the values for y that are outside the window get larger as we move to the right and smaller as we move to the left. This indicates that we have, more than likely, found a complete graph. However, the top portion of the graph here is indistinguishable from the x-axis since there is a big difference between the Ymin and Ymax values and the Ymax is close to the x-axis. Sometimes a better representation of a complete graph can be made by a hand-drawn sketch than from a calculator picture. The sketch may not be drawn to scale but it might provide a clearer display of the graphs important features.

11 64 Chapter 1 Graphs and Calculators Example 4 shows that finding a good window is an art form more than it is a science. For each problem that you encounter, you will need to first have an idea of the general shape of the graph. This will come with practice and experience with algebra. Everyone should know that quadratic equations all have a similar type of graph. The same is true for linear equations. More complicated polynomial graphs have a general shape which is determined by the degree (or highest power) of the polynomial. Rational, exponential, and logarithmic equations all have distinctive shapes. Developing the properties of these families of functions is left for future courses. There are a few issues about graphing calculators and graphs that were mentioned earlier and should be restated. First: The calculator provides approximate solutions to equations. It provides the answers as decimals and has a limited accuracy for the numbers it can display. Second: You should always try to find a complete graph of the equation unless the context of the problem suggests otherwise. We will use calculators to help us solve many problems and we need to know when it is appropriate to focus on a particular portion of the graph rather than the entire graph. Exercises 1.1 For problems 1 6, find the distance between the two points and the midpoint of the line segment joining them. 1. ( 5, 6) and (6, 5). ( 8, ) and (8, ) 3. (7, 8) and (7, 14) 4. (3, 9) and ( 4, 9) 5. (13, 1) and (7, 59) 6. (x, y) and (1, ) 7. Two points ( x, 1 0) and ( x, 0) on the x-axis., 0, y on the y-axis. 8. Two points ( 0 y ) and ( ) 1 For problems 9 11, solve. 9. Find the perimeter of the triangle with vertices at the points (1, 1), (5, 4) and ( 1, ). 10. Find the point that is one quarter the distance from point (9, 1) to ( 5, 4).

12 Section 1.1 Graphs A 4 foot by 6 foot map is divided into squares that are 1-inch by 1-inch in size. The legend of the map states that one inch on the map is equivalent to miles. a. If town A is located at point (19, 3) and town B is located at point (45, 60), what are the coordinates of the town that is half-way between the two towns? b. If town C is located at the point (7, 49), find the distance in miles from town C to A. c. Which of the towns A or B is closer to C? For problems 1 15, draw the graph of the equation by hand (making a table of at least 6 entries) and then graph the equation in the standard viewing window. 1. y = x 13. y = x x y = x 15. y = x For problems 16 17, determine if the given ordered-pairs are points on the graph of the given equation. Explain your reasoning. 16. y = 4x a. (0, 7) c. (3, 108) b. (1, 6935) d. ( 6, 886) 17. y = 5 x a., 6 c., b. (0, 6) d. (7.8, 1.5) For problems 18 0, use the calculator to graph the following equations and then use the trace or table feature of the calculator to estimate the values for x or y. 18. y = x 1 Find y when x =, x = 4, and x = y = 3x x + 6 Find x when y = 7, y = 4.5, and y = y = 5x + 7x 4x Find y when x = 5 and x = 11. For problems 1 3, graph the equations in the given windows. Which window best displays a complete graph of the equation? Justify your answer Equation: y = 8x + 7x a. Window: Standard Window b. Window: y [.5,.5], x [.5,.5] y 8, 500, x.5,.5 (Note: is read as is in.) c. Window: [ ] [ ]

13 66 Chapter 1 Graphs and Calculators. Equation: y = 100x 5 a. Window: Standard Window y.5,.5, x.5,.5 y 8, 500, x.5,.5 b. Window: [ ] [ ] c. Window: [ ] [ ] 3. Equation: y = ( x.) ( x + 4.5) a. Window: Standard Window b. Window: y [ 0, 00], x [ 10, 10] y 0, 1000, x, 1 c. Window: [ ] [ ] For problems 4 6, find a viewing window that produces a complete graph of the equation. There are many correct answers. Remember that the graph needs to display all of the intercepts (both x and y) as well as all of the peak and valleys. (Hint: You may want to use the table feature to help you.) x 4. y = x y =.1x + x + x + x A cardiac test measures the concentration y of a dye x seconds after a known amount is injected into a vein near the heart. In a normal heart 4 3 y =.006x +.14x.053x + 179x. 7. Below is a graph in the standard viewing window. After adjusting the viewing window, which are possible graphs of the same equation? Give your reasoning as to why it is or isn t the same graph. a. b.

14 Section 1.1 Graphs 67 c. d. e. f. For problems 4 8, graph the two equations and state whether the following equations have the same graph. 8. y = 4 x + 3 and y = 4 x y = x and y = x. (Note: The means absolute value function and is found on the calculator under the [Math] Num menu as abs.) x y = log 10 and y = x 30. ( ) 31. log( x ) 3. y = and y = log( x) 3 x 1 y = and y = x + x + 1 x 1 For problem 9, if possible, determine whether the two equations are equal. In order for the equations to be equal, each x-value must produce the same y- value for both equations. Justify your answers y = ( 1 x) and y = 1 6x + 15x 0x + 15x 6x + x 34. Given two points, ( x, y 1 1 ) and ( x, y ), draw an associated right triangle and use the Pythagorean Theorem to prove the Distance Formula: ( ) ( ) d = x 1 x + y1 y.

15 68 Chapter 1 Graphs and Calculators 1. Solving Equations Graphically Part 1: The Zero Method When we solve an equation involving an unknown quantity x, we are trying to find all of the values for x that make the statement true. Take for instance the equation 3 x 4 = 0. This equation states that a certain number x, when changed by multiplying by 3 and then subtracting by 4 ends up at the point 0. To determine all the values x with that property we can use algebra: add 4 to both sides to get 4 3x = 4; divide both sides by 3 to find x =. Using geometry on the number line 3 that solution can be stated as follows: Suppose a certain number is moved 3 times as far from the origin and then shifted left 4 units, and ends up at 0. Then the original number was 3 4. The geometric description in English seems more difficult and complicated than taking the algebraic steps. The equation 3x 4 = 0 can also be understood using -dimensional pictures. In the previous section, we introduced the extra variable y and graphed the equation y = 3x 4. To solve the equation 3x 4 = 0 we can think of this as asking which points on that graph have y = 0. Since the set of points where y = 0 is the x-axis, the solutions to the equation are exactly the x-values of the points on the graph that lie on the x-axis. These are called the x-intercepts of the graph of the equation y = 3x 4. This idea motivates the graphical method introduced in this section. The x-intercepts of a graph are sometimes called roots or zeros. (These should not be confused with square roots.) In other words, for the graph of an equation a root or zero is a value of x that makes y = 0. For example, the roots of the graph in Figure 1 are x = 6, x =, and x = 5. y 6 5 x Figure 1

16 Section 1. The Zero Method 69 To find the zeros algebraically, we set y = 0 and solve for x. To do this graphically, we use an accurate graph of the equation and find where it crosses the x-axis. Example 1 (Finding Roots) Find the roots of x + 10x + 1 algebraically. Since we are asked to find the roots algebraically set x + 10x + 1 equal to 0. 0 = x + 10x + 1 This is just solving a quadratic equation. 0 = ( x + 3)( x + 7) We factor the right-side. 0 = x + 3 or 0 = x + 7 One of the factors must equal zero. x = 3 or x = 7 Solve for x. Here the roots occur when x = 3 and x = 7. Unfortunately, algebra techniques are not always powerful enough to find the roots. This is where the graphing calculator can help. We can use the calculator to graph the equation and to find approximate values for the roots. This can be done by using the built-in root finder of the calculator. (For some of the older calculator models, this is not an option. Instead, use the zoom and trace functions to approximate the solution.) Remember, since we are using the calculator, it is likely that there will be some loss of accuracy when finding a root. Example (Finding Roots) Find the roots of the given quantities graphically. I. 3 x 5 To do this problem, we enter the polynomial equation y = 3x 5 into the calculator, and then press the [GRAPH] key to get a picture of the graph in the standard viewing window. Note: This is a complete graph and it has two roots.

17 70 Chapter 1 Graphs and Calculators Finding the roots: To find the roots we use the built-in root finder. On the TI-83, we must enter the CALC menu which is accessed by pressing [nd] and then [Trace]. In the CALC menu, the TI-83 uses the word zero instead of root so we select option. On CD: Finding the Zeros (Roots) of a Polynomial With that selection, the calculator asks which root we would like to find. It does this by asking three questions: the first two establish an interval and the third approximates the location. Answers to these questions can be entered by using the arrow keys, or by typing numerical values before pressing [ENTER]. In this example, let s find the positive root first. These three pictures show the process that the calculator goes through when finding roots. The first two steps select the left and right bounds for the interval. This limits the values for x that the calculator must check when finding the answer. Using the arrow keys we told the calculator that the root is in the interval [ , ]. This seems correct in this case since the values for y change from negative to positive. The Guess? tells the calculator where to start when computing the root.

18 Section 1. The Zero Method 71 After entering the information above, the calculator displays the root or zero value x = Using the same process, we can find that the other root occurs when x = Since the equation in this example was a quadratic, we could have used algebraic methods, finding the exact solutions to 0 = 3x 5. In the next example the algebra option is much harder but the graphical method provides fairly accurate solutions. 5 4 II. x + 3x x First enter the equation y = x + 3x x 6 into the calculator and find a suitable viewing window that will display the complete graph. The standard window displays 3 x-intercepts. This graph continues very steeply to the left and right (beyond the viewing window). This is seen by using the trace feature and checking the y-values. Therefore it seems likely that this is a complete graph and that the graph does not have any more x-intercepts. Once we are confident that we have a complete graph, we use the root finder once again to find that the three roots occur when x = , x = and x = On CD: Scientific Notation and Your Calculatror s Use of E Note that for the last root, the calculator found the value for x when 4 y = = which is a value very close to zero. It is not exactly zero since calculators usually make small rounding errors.

19 7 Chapter 1 Graphs and Calculators Calculators seldom give truly exact answers but they are accurate enough for many purposes. The root or zero method to solve equations begins in the same way we would use when algebraically solving a quadratic equation: we must make one side of the equation 0. In the next examples, which involve solving an equation with a single variable, we might need to set one side of the equation equal to zero before when can use the graph to find the zeros. Example 3 (Solving Equations) Solve the equation. Round your answer to two decimal places. I. 4x 4 3 = 0 We are asked to find the values for x that make the expression 4x 4 3 equal to 0. (Note that there is no need to use algebra first in this situation since the expression is already equal to zero.) This is the same as finding the roots of the equation y = 4x 4 3. To do this we enter the equation into the calculator, graph, and approximate the roots by using the root finder. On CD: Finding Solutions of Equations Calculator work provides roots near to the values when x = and x = Rounding to two decimal places we have the solutions x = 0.93 and x = (Solve the equation algebraically to see that the exact answers are x = and 4 answers?) 3 x = Do these agree with the calculator

20 Section 1. The Zero Method 73 II. 7 x + 5 = 3x 3 7 In order to use the root method for this problem, we must first make one side of the equation 0. To do this, we use some algebra. 3 0 = 3x 7x 1 We subtract 7x and 5 from both sides. (This new equation is equivalent to the first so it has the same solutions.) Finding solutions to this equation is the same as finding where the graph of y = 3x 3 7x 1 crosses the x-axis. Enter the equation into the calculator, find a complete graph, and use the zero feature to find the x-intercepts. (Note that the graph shown is missing a piece but the viewing window is adequate to find a root for the equation. Could there be other roots outside of this window? Experimentation with the trace feature or table feature suggest that the graph, in both directions, does not bend back to towards the x-axis. However, a true skeptic would not be convinced by this evidence!) This tells us that the root occurs near to x = After rounding, the solution to 7x + 5 = 3x 3 7 is approximately x =.07. If we were to check this solution by plugging it into the original equation, we would find that the right-side and left-side are not exactly equal. This is due to the rounding of the answer. In this case the equation had only one solution.

21 74 Chapter 1 Graphs and Calculators III. 4x = 5 We rewrite this equation using algebra. 4x = 0 Subtract 5 from both sides. This problem is now reduced to analyzing the graph of the equation y = 4x To do this we enter that equation into the calculator, find the complete graph, and use the root finder. This shows that the root occurs near x = After rounding, the approximate solution to the original equation is x = Remarks: 1. Does the graph actually stop there in the third Quadrant? Where is the endpoint? Hint: When is 4x Can you find the exact solution algebraically and check that it matches the estimated solution x = 1.65? Exercises 1. For problems 1 11, find the zeros, if they exist, of the equation both graphically and algebraically. 1. y = 5 x + 4. y = x + x 3 3. y = 5x + 3

22 Section 1. The Zero Method y = x + x 3 5. y = x 3 + 8x 33x 6. y = 4x y = x y = x 4 x 8 9. y = x 4 13x y = 4x + x y = ( x )( x + 11x 60) 1. y = ( x + 5)( x + 7x 10) When using the calculator to find the zeros, Joe Student entered the following information. For problems 13 and 14, explain what Joe did wrong and why it did not work. x Joe wanted to find the zeros of, so he entered the following 3 equation in his calculator and found the answers given below.

23 76 Chapter 1 Graphs and Calculators To find the zeros of x 3 + 4x +, Joe entered the corresponding equation 0 in the calculator and looked at the graph. a) He noticed that the graph appears to have two zeros. When trying to find the zero on the right, he got this message: b) After figuring out what he did wrong in part a), he then tried to find the zero to the left. When trying to find this root, he entered the following information into the calculator and got an error message:

24 Section 1. The Zero Method 77 For problems 15 17, determine graphically, by using the zero method, the number of solutions to the equation. You do not need to find the roots x + 5 = 3x + x 16. x 7 10x x + 10 = x + 500x 8,000x = 16x 3, 000 For problems 18 1, use the zero finder to find the approximate solution(s) to the equation. (Round your solutions to decimal place accuracy.) x + x 3 = x 6x + 6 = x 5 3x + x 6 = x = x + x 1

25 78 Chapter 1 Graphs and Calculators 1.3 Solving Equations Graphically Part : The Intersection Method In this section we discuss another graphical method to solve equations. This is the intersection method, used for a system of two equations. Before we discuss this graphical approach, we review the algebraic methods used to solve a system of equations. Two common algebraic methods to solve a system of equations are elimination and substitution. You may be familiar with using these methods for linear equations. x + y = 7 3x y = 0 Linear System x + 3xy = 1 y = 4x Non-linear System Figure 1 A non-linear system is a system in which at least one of the equations is not a linear equation. (See figure 1.) We want to be able to move beyond solving linear systems to those that are non-linear. Since the substitution method is more useful than the elimination method for solving non-linear systems, we will focus here on the substitution method. Substitution To use the substitution method, solve one of the two equations for one of the unknowns and substitute this expression into the other equation. This produces an equation involving only one unknown. Example 1 (Substitution) Solve the following systems of equations using the substitution method. x y = 1 I. Linear System 3x + y = 5 First solve one of the equations for one of the unknowns. The first equation seems simplest to handle. y = x 1 Add y and subtract 1 from both sides. Next, substitute that formula for y in the other equation. 3 x + (x 1) = 5 Substitute x 1 into the second equation. 3 x + 4x = 5 Distribute the.

26 Section 1.3 The Intersection Method 79 7 x = 5 Combine like terms. 7 x = 7 Add to both sides. x = 1 Divide by 7. We have found the value for x. Remember that there were two unknowns. We have found the x-value but still need to compute y. To do this substitute the value that we found in x back into one of the original equations. (1) y = 1 Substitute x = 1 into the first equation. y = 1 Subtract from both sides. y = 1 Multiply both sides by 1. This gives us the solution (x, y) = (1, 1). Remember to check that the ordered-pair is a solution to both equations. II. x + 4 = y x + y = 7 Non-linear System Even though this is a non-linear system, we can still use the substitution method. Solve one of the equations for one of the unknowns and substitute this into the other equation. Since the first equation is already solved for y, we will just plug this value for y into the other equation. x + x + 4 = Substitute y into the second equation. ( ) 7 x x + 4 = 7 Rearrange the terms. x x 3 = 0 Subtract 7 from both sides. ( x + 1)( x 3) = 0 Factor. 0 = x 3 or 0 = x + 1 One of the factors must equal zero. x = 3 or x = 1 Solve for x. For each of these possible values for x, we need to find the corresponding values for y. To do this, substitute each value of x back into one of the original equations and solve for y. x = 3: y = = 13 Substitute x into the first equation. x = 1: y = ( 1) + 4 = 5 Substitute x into the first equation. This gives us two ordered-pair solutions (3, 13) and ( 1, 5). When solving a system of equations we are finding the ordered-pairs (a, b) that are solutions to both equations. Since a graph of an equation represents the set of all ordered-pair solutions to the equation, finding the solutions to both equations would mean that we are finding points that are on both of the graphs. This is the same as finding where the two graphs have a point in common; the places where they intersect. The graph intersection feature in the calculator provides us with a graphical approach to solving systems of equations. Here are some examples using this graphical approach.

27 80 Chapter 1 Graphs and Calculators Example (Graphical Approach) Solve the following systems of equations using graphs. x y = 1 I. 3x y = 4 First input the two equations into the calculator. This requires that we use 4 3x algebra to solve both of the equations for y, y = x + 1 and y =. We know that there is only one point of intersection since this is a linear system (two intersecting lines). We then need to find a suitable viewing window that displays the point of intersection. On CD: Finding Solutions of Equations The picture above displays the two graphs in the standard window. After adjusting the window, we have a better picture of where the two lines intersect. To solve the system of equations, we use the intersection finder of the calculator to locate the point of intersection. Enter the calc menu and select option 5. Then select the two graphs and offer a guess as to where the two equations intersect. On CD: Finding the Intersection of Two Polynomials

28 Section 1.3 The Intersection Method 81 After completing this process, the calculator returns an answer of ( 6, 11). This system can also be solved algebraically. II. x + 4 = y x + y = 7 (Previously done as Example 1, II) Solve both equations for y and enter the formulas in the calculator. The second equation becomes y = x + 7. Since this is a non-linear system, there may be more than one solution. The picture indicates that we will have two points of intersection. So we adjust the viewing window to see both points of intersection. Now use the intersection finder twice to find those points of intersection. The intersection finder will locate only one point of intersection at a time.

29 8 Chapter 1 Graphs and Calculators This gives us that the two solutions to the system of equations are ( 1, 5) and (3, 13). These answers match the solutions found algebraically in Example 1, II. As we have just seen, the calculator can compute a point of intersection of two graphs provided we feed it an estimated answer. To solve a single equation we can use the intersection method by rewriting that equation as a system of two equations. The following examples demonstrate this technique. Example 3 (Intersection Method) Solve the following equations using the intersection method. 3 I. x 5x = x 4 To solve this single equation we make it into a system by introducing a new variable y and setting each side of the equation equal to y: y = x 3 5x and y = x 4. The x-coordinates of the ordered-pair solutions to this system correspond to solutions of the original equation. To solve a system on the calculator we graph the two equations and find the intersections. We enter these equations into the calculator as Y = X 3 5X and Y = X 4. 1 Looking at the standard window, we see three points of intersection. Experience with such graphs tells us that there are no more intersection points. It is not necessary to adjust the viewing window, but in order to get a more usable picture we will adjust the y-axis values down by 5 units.

30 Section 1.3 The Intersection Method 83 Unlike solving a system of equations, we need to give only the values for x as the solutions instead of an ordered-pair. Because of this, we find that the solutions are approximately x =.73, x = 0.73 and x =. Note that we could have solved this problem using the root method by rewriting the 3 original equation as x 6x + 4 = 0. (This equation can also be solved algebraically. Can you find the exact solutions using algebra?) II. t 3 + 5t 7 = 9 Like problem I, to solve this we introduce another variable y and solve the system. However, when entering the values into the calculator, we will let t be represented by x. Also note that this problem is not easy to solve using the algebra tools that we have available. (Can you find the exact solutions?) Again, looking at the standard window, we see three points of intersection. However, this time to get a nicer picture, we will adjust the y-axis values up by 5 units.

31 84 Chapter 1 Graphs and Calculators Remember that we let x represent t and we are trying to find the values of t that satisfy t 3 + 5t 7 = 9. These values are approximately t = 4, t =.5616 and t = We do not need to give an ordered-pair solution since our original problem was a single equation. III. x + 7 = x x 6 This problem will also involve setting up and solving a system of equations. This problem has no simple solution by using algebra alone. Here we can see two points of intersection on the screen and the graphs don t seem to intersect elsewhere. We use the intersection finder to compute the points of intersection. This shows that the two solutions to our original equation are approximately x = and x = Since we discussed algebraic methods and graphical methods to solve a system of equations, now is a good time to mention that it is possible to use a mixture of algebra and graphing to solve a system of equations. Example 4 uses an algebraic/graphing hybrid approach to solve the problem originally done in example 1, II and re-done in example, II. Example 4 (Hybrid Method) Solve the following system of equations. x + 4 = y x + y = 7

32 Section 1.3 The Intersection Method 85 To use the root method, first we must use substitution and algebra to rewrite the equation x + x + 4 = Substitute y into the second equation. ( ) 7 x x + 4 = 7 Rearrange the terms. x x 3 = 0 Subtract 7 from both sides. Enter this equation into the calculator and view the graph. Note that the graph is a quadratic and all of the roots are visible. Use the root finder to find the values. Remember that this is giving us only the x-coordinates of our ordered-pair solution so we need to substitute each value into one of the original equations to find y. x = 1: y = ( 1) + 4 = 5 x = 3: y = (3) + 4 = 13 This gives us the ordered-pairs of ( 1, 5) and (3, 13). Note that the hybrid approach involves using less algebra than the algebraic approach (Example 1, II) and doesn t require us to adjust the viewing window to find the second point of intersection (Example, II). As we have just seen in the first three sections of the book, graphing calculators allow us to solve more types of equations then we could do just by using algebra. However, to quote a recent movie, with great power comes great responsibility. Now that you can use the calculator as a tool to solve equations, you need to be aware that it is not always the best option. The calculator sometimes causes you to spend more time than necessary on a problem, or it might not give reliable answers. This is where you will need to exercise responsibility. Here are

33 86 Chapter 1 Graphs and Calculators some examples showing that caution is sometimes necessary when working with a calculator. Example 5 (Potential Pitfalls) I. Solve using the intersection method. x 3 x + x = y = y As we will see, this innocent looking non-linear system is particularly troublesome. As usual, we input the two equations into the calculator and view the graph in the standard window. As we can see, the standard window does not clearly show what is happening for the positive values of x. So we adjust the window and view the graph again. From this graph, it appears that there are two places where the graphs meet. So we use the intersection finder to find them. (This is where it gets interesting.)

34 Section 1.3 The Intersection Method 87 This is not a misprint. When you try to find the second intersection point, which occurs when x = 1, the calculator returns to you the point (0, 0) as the result. Why is this? Well, at this point the two curves meet but do not cross. They are tangent to each other at this intersection point. The point (1, ) is a solution to the system of equations, but the calculator cannot find it. This is one of those times when you need to choose your method of solving the problem wisely. This problem is not too difficult to solve using algebra (or even the hybrid method). II. Solve the given problem using the root method. x x x 3 = 0 In this problem, we enter the equation into the calculator and look at the graph. All the features of the graph can be seen in the standard viewing window so all we try to use the root finder to determine the zeros. When trying to do this, we get an error. The calculator has a problem here since it is looking for the place in the interval where the y-values change sign. This doesn t happen with this graph since the values for y are always nonnegative (greater than or equal to zero). There is a way to solve this problem using the root method, though. We recall the following fact: a = 0 only when a = So, we solve 0 = x + x x 3 by graphing y = x + x x 3 and finding the roots.

35 88 Chapter 1 Graphs and Calculators This shows that the two solutions to the original equation are approximately x = and x = III. Solve the following equation. 1 0 = + 3 x x + π To do this problem we need to enter the equation into the calculator and view the graph. We find the two visible roots by using the zero or root finder. This gives us two solutions at approximately x = 1.98 and

36 Section 1.3 The Intersection Method 89 x = This graph looks very similar to a graph of a quadratic equation except for the fact that it appears to stop. Does this graph exhibit the same behavior as the graph from Example 3, III in section 1.? In this case the graph that we see doesn t tell the whole story. The algebraic 1 formula helps explain the difficulty. From y = + 3 x we see that x + π a value of y is defined whenever x + π > 0, that is, whenever x > π. We get an idea of the behavior by looking at the table starting at x = 3.15 and going by steps of The sign change of y-values between x = 3.1 and x = 3.11 indicates that there is a zero in that interval. The [CALC] feature of the calculator computes that zero more accurately. (Can you find a window making the calculator display this graph more accurately?) This example shows that calculator graphs can sometimes be misleading and incomplete. Exercises 1.3 For problems 1 6, solve the systems of equations algebraically by substitution and also by graphing. Check your answers to be sure that they are the same for both methods. For each problem decide which method provides a more efficient method of solution. Explain the reasons for your choices.

37 90 Chapter 1 Graphs and Calculators 3x y = 1 1. x + y = y = 3x + 11 x y = 6 r + s = 0 r s = 5 x y = 0 3x + y = y = x y = x y = x y = x + 3x + 5 4x 7 y = 85x y = 15x 35x x y = 3x 3y = x 5.5x For problems 9 10, find approximate solutions to the equations using the intersection method. (Round your answers to decimal places.) 3 9. x + 3 = x + 5x x + 5 = 3x + 4x For problems 11 18, solve. You may use any method that works. You may want to think about the problem before you choose your strategy. When appropriate, round your answers to decimal places x 3 + 3x + = x + 3 = x x + 13x + 36 = x x 1 = x x x + x x 3 = 0

38 Section 1.3 The Intersection Method x 3 4x + 4x = = y y x x = x x 18. = 1 x + 5 For problems 19 and 0, find an exact (No decimals allowed!) solution for the equation in the given interval. You can use a calculator to find an approximate solution. This decimal value should then be converted to the exact value, e.g x 3 x + 3x = 0, x ( 0, 1) 0. 1 x 4 x 3 1x + 5x = 0, x ( 0, 1) Applications 1. According to data from the National Center for Educational Statistics and the College Board, the average cost y of tuition and fees (in thousands of dollars) at public four-year institutions in year x is approximated by the equation 3 y =.00044x.00039x +.114x +.195, where x = 0 corresponds to If this model continues to be accurate, in what year will tuition and fees reach $4,000?. According to data from the US Department of Health and Human Services, the cumulative number y of AIDS cases (in thousands) diagnosed in the United States during is approximated by y = 3.3x 14.81x , where x = 0 corresponds to In what year did the cumulative number of cases reach 50,000?

39 9 Chapter 1 Graphs and Calculators Chapter 1 Review Midpoint and Distance For problems 1 4, find the distance between the two points, and find the midpoint of the line segment joining them. 1. (9, 7) and (7, 9). ( 3.5, 45.1) and ( 1.75, 79) 3. (7, 0) and ( 50, 0) 4. (0, 4.75) and (0, 9) For problems 5 & 6, find the midpoint of the two x-intercepts of the graph of the equation. 5. y = x 7x y = x 1x + 7 Complete Graphs For problems 7 & 8, draw a complete graph of the equation. Find and label all of the zeros y = 5x + 64x 745x y = 7x + 5x 361x + x Systems of Equations For problems 9 & 10, solve the system of equations. 3 x + 5x = y 9. 5x + 18 = y 10. x x + 3x + 5 = y 10x + 45 = y For problems 11 15, solve if possible x + 3 = x x 1 = x x 5x + 10x = 17x x + 85x + 35 = 105x The profit earned in a month by the law firm Dewey, Cheatum, and Howe is modeled by the equation y = 9.756x x, where x is the number of cases handled. Knowing that the law firm can handle at most 40 cases a month, how many cases do they need to handle to earn $41,356. a month?

40 A.4 Answers to Odd-Numbered Exercises Chapter 1 Section 1.1, page d = ,, 3. d = 6, ( 7, 11) 5. d = 81.16, ( 0, 19) x1 + x 7. d = x1 x,, 0 Possible graphs: b, c, d *Other interpretations possible based on good reasoning. 9. Yes 31. No 33. Possibly True. Can t prove by looking at the graph. Section 1., page x = a. (3, 41.5) b miles 3. No zeros c. B 5. x = 3, 0, or x = 3 or 3 9. x = 3, 3,, or x = 15, 4, or Joe forgot parentheses in the numerator. 17. a. Yes b. No c. Yes d. No y = 7, x = 3 y = 4.5, x x y = 8, No value of x exists 1. b 1. x = 0 or approx b Section 1.3, page Must show the following 6 13 points: 1., 5 5 (-3.518, 0), (11.308, 0) (0, 50), (8.15, 4.7) 5 3. r =, s 7. Not possible graphs: a, e, f = 5

41 Answers to Odd-Numbered Exercises A ,, ( 1, 3) 13. x cases 7. (, 140), (3, 40) 9. x = x = 13.x = 3.9 or x = 1.4 or x = 0 or x = Chapter 1 Review, page (1, 0) & ( 5, 0): (3.5, 0) 7. Roots at 13, 1, and (1.34, 9.07) 11. 3