science. In this course we investigate problems both algebraically and graphically.

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1 Section. Graphs. Graphs Much of algebra is concerned with solving equations. Man 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 approimate solutions to certain tpes of equations. Such numerical solution methods are often essential for solving real problems in engineering and science. In this course we investigate problems both algebraicall and graphicall. 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 b the invention of the graphing calculator. These calculators quickl construct graphs that are fairl accurate and reliable. In pre-calculator ears the graphical approach was much more tedious. In the net two sections (. and.) we will eplore various methods to solve equations graphicall. (It is assumed that we alread know the algebraic methods. Those methods are outlined in sections. and. but we will not spend time to review them now.) In this section we will discuss the rectangular coordinate sstem properties of graphs and how to use the calculator to displa a graph. The Rectangular Coordinate Sstem: ( )-plane The rectangular coordinate sstem is a sstem for labeling points in the plane. The coordinate plane is constructed from two number lines the -ais and the -ais. The -ais and the -ais 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 below. -ais II I origin -ais III IV Figure

2 6 Chapter Graphs and Calculators The plane is divided into four quadrants labeled I II III and IV as in Figure. These intersecting number lines allow us to represent each point in the plane b an ordered-pair written as ( ). Each point of the plane is either on an ais or in one of the quadrants. This coordinate sstem 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 The Distance Formula states that the distance d between the points and is given b the formula: d The derivation of the distance formula uses the Pthagorean Theorem and is included as an eercise at the end of the section. Eample (Distance Formula) Find the distance between the points (8 76) and (9 ). To solve this problem we use the distance formula and let (9 ) be and (8 76) be. d Substitute the values into the formula. d d 897 d =.8786 (an infinite decimal) Note that 897 is the eact answer but it is often more useful to have an approimate (rounded off) answer. So after rounding to two decimal places the distance between the two points is approimatel.8. Midpoint Formula The midpoint of a line segment is the point that is half-wa between the endpoints. That is it is the onl point on the line segment that is the same distance from each end. On a number line the number that is halfwa between two numbers is the average of the two numbers. For instance to find the number halfwa between a and b we compute the value a b. For points in the plane the - coordinate of the midpoint is halfwa between the -coordinates of the endpoints and similarl for the -coordinate of the midpoint. So if we let the points Section. Graphs 7 and be the endpoints of the line segment then the Midpoint Formula states that the midpoint is: Eample (Midpoint Formula) Find the midpoint of the line segment between the two points ( 9) and (7 6) Here we let Then the midpoint equals: 7 9 ( 6) be the point ( 9) and or (..) be the point (7 6). To be sure that is the midpoint we can check that the distance between ( 9) and is the same as the distance between (7 6) and. This is left as an eercise in using the distance formula. Computing distances and midpoints is useful but the real utilit of the rectangular coordinate sstem can be seen when we look at an epression like. This epression involves a single variable which represents an unknown number. It can be viewed as a command change the number b multipling b and then subtracting. Thinking geometricall on a single number line we ma view this as the command move to a spot times as far from the origin and then shift that point to the left units. The behavior of such an epression is represented more clearl when we introduce a new variable and consider the equation as a relationship between two variables. The relationship can be described using a table as in Table. Note that the table does not contain all of the possible values for since there are infinitel man.

3 On CD: Basics of Graphing Getting Your Function Into the Proper Format for Graphing 8 Chapter Graphs and Calculators X Y 0 8 Table Since we have both an -value and -value for each entr in the table we can plot each row of the table as a point in the plane. In this eample we would put dots at the points (0 ) ( ) ( ) ( ) and ( 8). B representing all possible entries from the table in such a fashion we obtain the graph of. We can not draw ever dot separatel since there are infinitel man 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 displa the graph is limited too. At an time we see onl 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 displas the important features of the graph. This is particularl true when using a graphing calculator. Graphs and the Graphing Calculator Entering equations into the TI 8 calculator is done b using the Equation Window. This feature allows ou to store various equations in the memor of the calculator. Graphs of those equations are displaed in the viewing window. The standard viewing window of the TI 8 displas the values of from 0 to 0 and the values of from 0 to 0 each with a scale of. (The scale refers to the distance between the tic marks on the aes.) To view the dimensions of the current viewing window press the [WINDOW] ke. 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 effectivel we need to be able to access and understand all of these features. A summar of the calculator features is provided in Table. Please note: Additional help on using the calculator can be found on the included CD-ROM. Section. Graphs 9 Feature Calculator Ke Uses Equation Window [Y=] Used for entering equations Graphs Graphing Equations [GRAPH] Allows the user to view the graphs of an or all of the equations entered in the calculator. Viewing Window [WINDOW] Displas the dimensions of the current viewing window. These dimensions can be adjusted b the user. Standard Window [ZOOM]: select ZStandard [6] Zoom In [ZOOM]: select Zoom In [] Zoom Out [ZOOM]: select Zoom Out [] Returns the calculator to the standard viewing window Adjusts the viewing window in was:. re-centers the window at a different point in the plane;. decreases the window values b a factor of. (This makes the picture of the graph look bigger.) Does the same as Zoom In ecept that it increases the window values b a factor of. (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 displas the coordinates of the highlighted point of the graph at the bottom of the viewing window. Tables Table for equations [ nd ] [GRAPH] This displas 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 kes. Table Setup [ nd ] [WINDOW] Displas the starting value for the table as well as the increment for determining the net value. These values can be adjusted b the user. Table

4 On CD: Window Menu Overview 60 Chapter 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 onl 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 eamples eplore 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 onl 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 displas all of the important features of the given equation such as peaks valles and intercepts. Finding good viewing windows is not an eact science and there are man different windows that show a complete graph for a given equation. Eample (Graphing Equations) a) Graph the equation 0 in the standard window and in the window with -values in the interval [ 0] and -values in the interval [ 0 00]. Adjust the -scale to 0. 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 displa the graph in the new window. (What happens when the -scale is left at?) 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 the appear to be ver different. This new window does provide a complete graph because it turns On CD: Tracing the Function: The Trace Ke Section. Graphs 6 out that there are no more wiggles that occur outside the window. This knowledge comes with eperience with graphing equations and recognizing that certain families of equations have some features in common. You should alread be familiar with the characteristics of the linear famil and the quadratic famil. 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 when is and when is 6.. 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 displas the solutions at the bottom of the calculator. There are man choices for solutions. Using the arrows we find one at approimatel (.7 8.6) and another one at approimatel ( ). Remember that in most cases the calculator will give onl approimate values. Using the table feature we can displa more solutions at one time. Here we see a list of seven solutions. This also illustrates the table setup procedure. To find the -values for the given values of we can use the trace or table features. To use the trace feature enter the trace mode and tpe the value for. In this case we enter. Once we press [Enter] the machine displas the coordinates for the point with -coordinate.

5 On CD: Graphing Polnomials 6 Chapter Graphs and Calculators So = 77 when =. To find the value for when = 6. we can use the table feature. (If ou tr to use the trace method ou will get an error unless ou adjust the window.) Enter the table setup and set the value for TblStart = 6 and Tbl= 0.. This produces the following table. From this we can see that when = 6. is approimatel 66. Note that the table can displa onl 6 characters in the column. Eample (Complete Graphs) Find the complete graph of the equation 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 tring to find a complete graph. Notice that we seem to be missing a piece of the graph when is between 0 and. 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 On CD: Zoom Menu Overview Section. Graphs 6 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. 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 to see the rest of the graph. To do this trace the graph and watch the values for. We then reach the smallest value for when is about.. 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 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 likel found a complete graph. However the top portion of the graph here is indistinguishable from the -ais since there is a big difference between the Ymin and Yma values and the Yma is close to the -ais. Sometimes a better representation of a complete graph can be made b a hand-drawn sketch than from a calculator picture. The sketch ma not be drawn to scale but it might provide a clearer displa of the graphs important features.

6 6 Chapter Graphs and Calculators Eample shows that finding a good window is an art form more than it is a science. For each problem that ou encounter ou will need to first have an idea of the general shape of the graph. This will come with practice and eperience with algebra. Everone should know that quadratic equations all have a similar tpe of graph. The same is true for linear equations. More complicated polnomial graphs have a general shape which is determined b the degree (or highest power) of the polnomial. Rational eponential 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 approimate solutions to equations. It provides the answers as decimals and has a limited accurac for the numbers it can displa. Second: You should alwas tr to find a complete graph of the equation unless the contet of the problem suggests otherwise. We will use calculators to help us solve man problems and we need to know when it is appropriate to focus on a particular portion of the graph rather than the entire graph. Eercises. For problems 6 find the distance between the two points and the midpoint of the line segment joining them.. ( 6) and (6 ). ( 8 ) and (8 ). (7 8) and (7 ). ( 9) and ( 9). ( ) and (7 9) 6. ( ) and ( ) 7. Two points 0 and 0 8. Two points 0 and on the -ais. 0 on the -ais. For problems 9 solve. 9. Find the perimeter of the triangle with vertices at the points ( ) ( ) and ( ). 0. Find the point that is one quarter the distance from point (9 ) to ( ). Section. Graphs 6. A foot b 6 foot map is divided into squares that are -inch b -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 (9 ) and town B is located at point ( 60) what are the coordinates of the town that is half-wa between the two towns? b. If town C is located at the point (7 9) find the distance in miles from town C to A. c. Which of the towns A or B is closer to C? For problems draw the graph of the equation b hand (making a table of at least 6 entries) and then graph the equation in the standard viewing window..... For problems 6 7 determine if the given ordered-pairs are points on the graph of the given equation. Eplain our reasoning a. (0 7) c. ( 08) b. ( 69) d. ( 6 886) a. 6 c. 0 b. (0 6) d. (7.8.) For problems 8 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 or. 8. Find when = = and = Find when = 7 =. and = Find when = and =. For problems graph the equations in the given windows. Which window best displas a complete graph of the equation? Justif our answer.. Equation: 8 7 a. Window: Standard Window b. Window: (Note: is read as is in.) c. Window:

7 66 Chapter Graphs and Calculators. Equation: 00 a. Window: Standard Window b. Window:.. c. Window:... Equation: (.) (.) a. Window: Standard Window b. Window: c. Window: For problems 6 find a viewing window that produces a complete graph of the equation. There are man correct answers. Remember that the graph needs to displa all of the intercepts (both and ) as well as all of the peak and valles. (Hint: You ma want to use the table feature to help ou.) A cardiac test measures the concentration of a de seconds after a known amount is injected into a vein near the heart. In a normal heart Below is a graph in the standard viewing window. After adjusting the viewing window which are possible graphs of the same equation? Give our reasoning as to wh it is or isn t the same graph. a. b. Section. Graphs 67 c. d. e. f. For problems 8 graph the two equations and state whether the following equations have the same graph. 8. and and. (Note: The means absolute value function and is found on the calculator under the [Math] Num menu as abs.) log 0 and log log 0. and.. and For problem 9 if possible determine whether the two equations are equal. In order for the equations to be equal each -value must produce the same - value for both equations. Justif our answers.. ( ) 6 and Given two points and draw an associated right triangle and use the Pthagorean Theorem to prove the Distance Formula: d.

8 68 Chapter Graphs and Calculators. Solving Equations Graphicall Part : The Zero Method When we solve an equation involving an unknown quantit we are tring to find all of the values for that make the statement true. Take for instance the equation 0. This equation states that a certain number when changed b multipling b and then subtracting b ends up at the point 0. To determine all the values with that propert we can use algebra: add to both sides to get = ; divide both sides b to find. Using geometr on the number line that solution can be stated as follows: Suppose a certain number is moved times as far from the origin and then shifted left units and ends up at 0. Then the original number was. The geometric description in English seems more difficult and complicated than taking the algebraic steps. The equation = 0 can also be understood using -dimensional pictures. In the previous section we introduced the etra variable and graphed the equation =. To solve the equation = 0 we can think of this as asking which points on that graph have = 0. Since the set of points where = 0 is the -ais the solutions to the equation are eactl the -values of the points on the graph that lie on the -ais. These are called the -intercepts of the graph of the equation. This idea motivates the graphical method introduced in this section. The -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 that makes = 0. For eample the roots of the graph in Figure are = 6 = and =. 6 Figure Section. The Zero Method 69 To find the zeros algebraicall we set = 0 and solve for. To do this graphicall we use an accurate graph of the equation and find where it crosses the -ais. Eample (Finding Roots) Find the roots of 0 algebraicall. Since we are asked to find the roots algebraicall set 0 equal to This is just solving a quadratic equation. 0 ( )( 7 ) We factor the right-side. 0 or 0 7 One of the factors must equal zero. = or = 7 Solve for. Here the roots occur when = and = 7. Unfortunatel algebra techniques are not alwas 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 approimate values for the roots. This can be done b 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 approimate the solution.) Remember since we are using the calculator it is likel that there will be some loss of accurac when finding a root. Eample (Finding Roots) Find the roots of the given quantities graphicall. I. To do this problem we enter the polnomial equation into the calculator and then press the [GRAPH] ke to get a picture of the graph in the standard viewing window. Note: This is a complete graph and it has two roots.

9 On CD: Finding the Zeros (Roots) of a Polnomial 70 Chapter Graphs and Calculators Finding the roots: To find the roots we use the built-in root finder. On the TI-8 we must enter the CALC menu which is accessed b pressing [nd] and then [Trace]. In the CALC menu the TI-8 uses the word zero instead of root so we select option. With that selection the calculator asks which root we would like to find. It does this b asking three questions: the first two establish an interval and the third approimates the location. Answers to these questions can be entered b using the arrow kes or b tping numerical values before pressing [ENTER]. In this eample 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 that the calculator must check when finding the answer. Using the arrow kes we told the calculator that the root is in the interval [ ]. This seems correct in this case since the values for change from negative to positive. The Guess? tells the calculator where to start when computing the root. On CD: Scientific Notation and Your Calculatror s Use of E Section. The Zero Method 7 After entering the information above the calculator displas the root or zero value = Using the same process we can find that the other root occurs when = Since the equation in this eample was a quadratic we could have used algebraic methods finding the eact solutions to 0. In the net eample the algebra option is much harder but the graphical method provides fairl accurate solutions. II. 6 First enter the equation 6 into the calculator and find a suitable viewing window that will displa the complete graph. The standard window displas -intercepts. This graph continues ver steepl to the left and right (beond the viewing window). This is seen b using the trace feature and checking the -values. Therefore it seems likel that this is a complete graph and that the graph does not have an more -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 =.06 = and =.097. Note that for the last root the calculator found the value for when 0 which is a value ver close to zero. It is not eactl zero since calculators usuall make small rounding errors.

10 On CD: Finding Solutions of Equations 7 Chapter Graphs and Calculators Calculators seldom give trul eact answers but the are accurate enough for man purposes. The root or zero method to solve equations begins in the same wa we would use when algebraicall solving a quadratic equation: we must make one side of the equation 0. In the net eamples 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. Eample (Solving Equations) Solve the equation. Round our answer to two decimal places. I. 0 We are asked to find the values for that make the epression equal to 0. (Note that there is no need to use algebra first in this situation since the epression is alread equal to zero.) This is the same as finding the roots of the equation. To do this we enter the equation into the calculator graph and approimate the roots b using the root finder. Calculator work provides roots near to the values when = and = Rounding to two decimal places we have the solutions = 0.9 and = 0.9. (Solve the equation algebraicall to see that the eact answers are and answers?). Do these agree with the calculator Section. The Zero Method 7 II. 7 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. 0 7 We subtract 7 and 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 7 crosses the -ais. Enter the equation into the calculator find a complete graph and use the zero feature to find the -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? Eperimentation with the trace feature or table feature suggest that the graph in both directions does not bend back to towards the -ais. However a true skeptic would not be convinced b this evidence!) This tells us that the root occurs near to =.066. After rounding the solution to 7 7 is approimatel =.07. If we were to check this solution b plugging it into the original equation we would find that the right-side and left-side are not eactl equal. This is due to the rounding of the answer. In this case the equation had onl one solution.

11 7 Chapter Graphs and Calculators III. 7 We rewrite this equation using algebra. 7 0 Subtract from both sides. This problem is now reduced to analzing the graph of the equation 7. 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 = After rounding the approimate solution to the original equation is =.6. Remarks:. Does the graph actuall stop there in the third Quadrant? Where is the endpoint? Hint: When is Can ou find the eact solution algebraicall and check that it matches the estimated solution =.6? Eercises. For problems find the zeros if the eist of the equation both graphicall and algebraicall.... Section. The Zero Method When using the calculator to find the zeros Joe Student entered the following information. For problems and eplain what Joe did wrong and wh it did not work.. Joe wanted to find the zeros of so he entered the following equation in his calculator and found the answers given below.

12 76 Chapter Graphs and Calculators. To find the zeros of 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 tring 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 tring to find this root he entered the following information into the calculator and got an error message: Section. The Zero Method 77 For problems 7 determine graphicall b using the zero method the number of solutions to the equation. You do not need to find the roots For problems 8 use the zero finder to find the approimate solution(s) to the equation. (Round our solutions to decimal place accurac.)

13 78 Chapter Graphs and Calculators. Solving Equations Graphicall Part : The Intersection Method In this section we discuss another graphical method to solve equations. This is the intersection method used for a sstem of two equations. Before we discuss this graphical approach we review the algebraic methods used to solve a sstem of equations. Two common algebraic methods to solve a sstem of equations are elimination and substitution. You ma be familiar with using these methods for linear equations. 7 0 Linear Sstem Non-linear Sstem Figure A non-linear sstem is a sstem in which at least one of the equations is not a linear equation. (See figure.) We want to be able to move beond solving linear sstems to those that are non-linear. Since the substitution method is more useful than the elimination method for solving non-linear sstems 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 epression into the other equation. This produces an equation involving onl one unknown. Eample (Substitution) Solve the following sstems of equations using the substitution method. I. Linear Sstem First solve one of the equations for one of the unknowns. The first equation seems simplest to handle. Add and subtract from both sides. Net substitute that formula for in the other equation. ( ) Substitute into the second equation. Distribute the. Section. The Intersection Method 79 7 Combine like terms. 7 7 Add to both sides. = Divide b 7. We have found the value for. Remember that there were two unknowns. We have found the -value but still need to compute. To do this substitute the value that we found in back into one of the original equations. () Substitute = into the first equation. Subtract from both sides. Multipl both sides b. This gives us the solution ( ) = ( ). Remember to check that the ordered-pair is a solution to both equations. II. Non-linear Sstem 7 Even though this is a non-linear sstem 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 alread solved for we will just plug this value for into the other equation. 7 Substitute into the second equation. 7 Rearrange the terms. 0 Subtract 7 from both sides. ( )( ) 0 Factor. 0 or 0 One of the factors must equal zero. = or = Solve for. For each of these possible values for we need to find the corresponding values for. To do this substitute each value of back into one of the original equations and solve for. = : Substitute into the first equation. = : ( ) Substitute into the first equation. This gives us two ordered-pair solutions ( ) and ( ). When solving a sstem 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 the intersect. The graph intersection feature in the calculator provides us with a graphical approach to solving sstems of equations. Here are some eamples using this graphical approach.

14 On CD: Finding Solutions of Equations On CD: Finding the Intersection of Two Polnomials 80 Chapter Graphs and Calculators Eample (Graphical Approach) Solve the following sstems of equations using graphs. I. First input the two equations into the calculator. This requires that we use algebra to solve both of the equations for and. We know that there is onl one point of intersection since this is a linear sstem (two intersecting lines). We then need to find a suitable viewing window that displas the point of intersection. The picture above displas 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 sstem of equations we use the intersection finder of the calculator to locate the point of intersection. Enter the calc menu and select option. Then select the two graphs and offer a guess as to where the two equations intersect. Section. The Intersection Method 8 After completing this process the calculator returns an answer of ( 6 ). This sstem can also be solved algebraicall. II. (Previousl done as Eample II) 7 Solve both equations for and enter the formulas in the calculator. The second equation becomes 7. Since this is a non-linear sstem there ma 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 onl one point of intersection at a time.

15 8 Chapter Graphs and Calculators This gives us that the two solutions to the sstem of equations are ( ) and ( ). These answers match the solutions found algebraicall in Eample 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 b rewriting that equation as a sstem of two equations. The following eamples demonstrate this technique. Eample (Intersection Method) Solve the following equations using the intersection method. I. To solve this single equation we make it into a sstem b introducing a new variable and setting each side of the equation equal to : and. The -coordinates of the ordered-pair solutions to this sstem correspond to solutions of the original equation. To solve a sstem on the calculator we graph the two equations and find the intersections. We enter these equations into the calculator as Y X X and Y X. Looking at the standard window we see three points of intersection. Eperience with such graphs tells us that there are no more intersection points. It is not necessar to adjust the viewing window but in order to get a more usable picture we will adjust the -ais values down b units. Section. The Intersection Method 8 Unlike solving a sstem of equations we need to give onl the values for as the solutions instead of an ordered-pair. Because of this we find that the solutions are approimatel =.7 = 0.7 and =. Note that we could have solved this problem using the root method b rewriting the original equation as 6 0. (This equation can also be solved algebraicall. Can ou find the eact solutions using algebra?) II. t t 7 9 Like problem I to solve this we introduce another variable and solve the sstem. However when entering the values into the calculator we will let t be represented b. Also note that this problem is not eas to solve using the algebra tools that we have available. (Can ou find the eact 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 -ais values up b units.

16 8 Chapter Graphs and Calculators Remember that we let represent t and we are tring to find the values of t that satisf t t 7 9. These values are approimatel t = t =.66 and t =.66. We do not need to give an ordered-pair solution since our original problem was a single equation. III. 7 6 This problem will also involve setting up and solving a sstem of equations. This problem has no simple solution b 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 approimatel = and =.099. Since we discussed algebraic methods and graphical methods to solve a sstem of equations now is a good time to mention that it is possible to use a miture of algebra and graphing to solve a sstem of equations. Eample uses an algebraic/graphing hbrid approach to solve the problem originall done in eample II and re-done in eample II. Eample (Hbrid Method) Solve the following sstem of equations. 7 Section. The Intersection Method 8 To use the root method first we must use substitution and algebra to rewrite the equation 7 Substitute into the second equation. 7 Rearrange the terms. 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 onl the -coordinates of our ordered-pair solution so we need to substitute each value into one of the original equations to find. = : ( ) = : () This gives us the ordered-pairs of ( ) and ( ). Note that the hbrid approach involves using less algebra than the algebraic approach (Eample II) and doesn t require us to adjust the viewing window to find the second point of intersection (Eample II). As we have just seen in the first three sections of the book graphing calculators allow us to solve more tpes of equations then we could do just b using algebra. However to quote a recent movie with great power comes great responsibilit. Now that ou can use the calculator as a tool to solve equations ou need to be aware that it is not alwas the best option. The calculator sometimes causes ou to spend more time than necessar on a problem or it might not give reliable answers. This is where ou will need to eercise responsibilit. Here are

17 86 Chapter Graphs and Calculators some eamples showing that caution is sometimes necessar when working with a calculator. Eample (Potential Pitfalls) I. Solve using the intersection method. As we will see this innocent looking non-linear sstem is particularl 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 clearl show what is happening for the positive values of. 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.) Section. The Intersection Method 87 This is not a misprint. When ou tr to find the second intersection point which occurs when = the calculator returns to ou the point (0 0) as the result. Wh is this? Well at this point the two curves meet but do not cross. The are tangent to each other at this intersection point. The point ( ) is a solution to the sstem of equations but the calculator cannot find it. This is one of those times when ou need to choose our method of solving the problem wisel. This problem is not too difficult to solve using algebra (or even the hbrid method). II. Solve the given problem using the root method. 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 tr to use the root finder to determine the zeros. When tring 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 -values change sign. This doesn t happen with this graph since the values for are alwas nonnegative (greater than or equal to zero). There is a wa to solve this problem using the root method though. We recall the following fact: a 0 onl when a = 0. So we solve 0 b graphing and finding the roots.

18 88 Chapter Graphs and Calculators This shows that the two solutions to the original equation are approimatel =.7096 and =.77. III. Solve the following equation. 0 To do this problem we need to enter the equation into the calculator and view the graph. We find the two visible roots b using the zero or root finder. This gives us two solutions at approimatel =.98 and Section. The Intersection Method 89 = This graph looks ver similar to a graph of a quadratic equation ecept for the fact that it appears to stop. Does this graph ehibit the same behavior as the graph from Eample III in section.? In this case the graph that we see doesn t tell the whole stor. The algebraic formula helps eplain the difficult. From we see that a value of is defined whenever 0 that is whenever. We get an idea of the behavior b looking at the table starting at =. and going b steps of 0.0. The sign change of -values between =. and =. indicates that there is a zero in that interval. The [CALC] feature of the calculator computes that zero more accuratel. (Can ou find a window making the calculator displa this graph more accuratel?) This eample shows that calculator graphs can sometimes be misleading and incomplete. Eercises. For problems 6 solve the sstems of equations algebraicall b substitution and also b graphing. Check our answers to be sure that the are the same for both methods. For each problem decide which method provides a more efficient method of solution. Eplain the reasons for our choices.

19 90 Chapter Graphs and Calculators.. 6 r. r s s For problems 9 0 find approimate solutions to the equations using the intersection method. (Round our answers to decimal places.) For problems 8 solve. You ma use an method that works. You ma want to think about the problem before ou choose our strateg. When appropriate round our answers to decimal places Section. The Intersection Method For problems 9 and 0 find an eact (No decimals allowed!) solution for the equation in the given interval. You can use a calculator to find an approimate solution. This decimal value should then be converted to the eact value e.g Applications. According to data from the National Center for Educational Statistics and the College Board the average cost of tuition and fees (in thousands of dollars) at public four-ear institutions in ear is approimated b the equation where = 0 corresponds to 990. If this model continues to be accurate in what ear will tuition and fees reach $000?. According to data from the US Department of Health and Human Services the cumulative number of AIDS cases (in thousands) diagnosed in the United States during is approimated b where = 0 corresponds to 980. In what ear did the cumulative number of cases reach 0000?

20 9 Chapter Graphs and Calculators Chapter Review Midpoint and Distance For problems find the distance between the two points and find the midpoint of the line segment joining them.. (9 7) and (7 9). (..) and (.7 79). (7 0) and ( 0 0). (0.7) and (0 9) For problems & 6 find the midpoint of the two -intercepts of the graph of the equation Complete Graphs For problems 7 & 8 draw a complete graph of the equation. Find and label all of the zeros Sstems of Equations For problems 9 & 0 solve the sstem of equations For problems solve if possible The profit earned in a month b the law firm Dewe Cheatum and Howe is modeled b the equation where is the number of cases handled. Knowing that the law firm can handle at most 0 cases a month how man cases do the need to handle to earn $6. a month? A. Answers to Odd-Numbered Eercises Chapter Section. page 6 Possible graphs: b c d. d =.6 *Other interpretations possible based on good reasoning. 9. Yes. d = 6 7. No. d = d 0. Possibl True. Can t prove b looking at the graph. Section. page a. (.) b..06 miles. No zeros c. B. = 0 or. 7. = or. 9. = or.. = or. Joe forgot parentheses in 7. a. Yes the numerator. b. No c. Yes. d. No = 7 = = = 8 No value of eists. b. = 0 or appro..07. b Section. page 89. points: Must show the following 6. (-.8 0) (.08 0) (0 0) (8..7). r 7. Not possible graphs: a e f s

21 . 7. ( 0) ( 0) 9. =.0.. =.9 or.9. =. or.8 7. = 0 or Chapter Review page ( 0) & ( 0): (. 0) 7. Roots at and 6 9. (. 9.07). Answers to Odd-Numbered Eercises A cases