Objectives. By the time the student is finished with this section of the workbook, he/she should be able

Transcription

1 FUNCTIONS Quadratic Functions Absolute Value Functions Translations o Functions..57 Radical Functions...61 Eponential Functions...7 Logarithmic Functions Cubic Functions Piece-Wise Functions. 98 Functions Review

2 Objectives The ollowing is a list o objectives or this section o the workbook. By the time the student is inished with this section o the workbook, he/she should be able to Determine the properties o a quadratic unction in standard orm. Find the and y intercepts o a quadratic unction. Find the range and domain o a quadratic unction. Find the verte o a quadratic unction in standard orm. Graph a quadratic unction. Determine the properties o an absolute value unction in standard orm. Find the and y intercepts o an absolute value unction. Find the range and domain o an absolute value unction. Find the verte o an absolute value unction. Graph an absolute value unction. Determine the properties o a radical unction in standard orm. Find the and y intercepts o a radical unction. Find the range and domain o a radical unction. Find the point o origin o a radical unction. Graph a radical unction. Determine the properties o an eponential unction in standard orm. Find the and y intercepts o an eponential unction. Find the range and domain o an eponential unction. Find the key point o an eponential unction. Graph an eponential unction. Determine the properties o a logarithmic unction in standard orm. Find the and y intercepts o a logarithmic unction. Find the range and domain o a logarithmic unction. Find the key point o a logarithmic unction. Graph a logarithmic unction. Determine the properties o a cubic unction in standard orm. Find the and y intercepts o a cubic unction. Find the range and domain o a cubic unction. Find the verte o a cubic unction. Graph a cubic unction. Shit the graph o a unction without actually knowing the equation, i.e. graphing ( + ). Graph piece-wise unctions. 6

3 Math Standards Addressed The ollowing state standards are addressed in this section o the workbook. Algebra II 1.0 Students solve equations and inequalities involving absolute value. 8.0 Students solve and graph quadratic equations by actoring, completing the square, or using the quadratic ormula. Students apply these techniques in solving word problems. They also solve quadratic equations in the comple number system. 9.0 Students demonstrate and eplain the eect that changing a coeicient has on the graph o quadratic unctions; that is, students can determine how the graph o a parabola changes as a, b, and c vary in the equation y = a(-b) + c Students graph quadratic unctions and determine the maima, minima, and zeros o the unction Students prove simple laws o logarithms Students understand the inverse relationship between eponents and logarithms and use this relationship to solve problems involving logarithms and eponents. 11. Students judge the validity o an argument according to whether the properties o real numbers, eponents, and logarithms have been applied correctly at each step. 1.0 Students know the laws o ractional eponents, understand eponential unctions, and use these unctions in problems involving eponential growth and decay Students determine whether a speciic algebraic statement involving rational epressions, radical epressions, or logarithmic or eponential unctions is sometimes true, always true, or never true. 7

4 Quadratic Functions The translation o a unction is simply the shiting o a unction. In this section, or the most part, we will be graphing various unctions by means o shiting the parent unction. We will go over the parent unction or a variety o algebraic unctions in this section. It is much easier to see the eects dierent constants have on a particular unction i we use the parent unction. We will begin with quadratics. Observe the ollowing regarding a quadratic unction in standard orm. y = a( h) + k y = Notice that in the equation above, the h and k values are zero, while the value o a is one. This gives you the parent unction or all quadratics. Everything else is merely a manipulation o the parent unction. y = ( ) y = ( + ) The graph o the unction shits right. The graph o the unction shits let. The number inside the parenthesis makes the graph shit to the let or right. Remember P.L.N.R., Positive Let Negative Right, tells about the horizontal shit needed to graph the unction. I the unction above is ( ), the unctions below would be ( ) and ( + ) respectively. This is important to know, because in the uture, you will be required to graph unctions based solely on the picture provided. No equation will be given. You must rely solely on your knowledge o translating graphs. 8

5 y = Once again, the parent unction is illustrated above, and translations o it below. y = + y = The graph o the unction shits up. The graph o the unction shits down. In these eamples, the k value is what is changing. The value o k dictates a vertical shit o the unction. In this case, consider the parent unction as being. Given no inormation regarding the speciic equation o the unction, the equations or these two translations o y = ( ) ( ) are ( ) +, and ( ). Now, on the let we have the opposite o the parent unction. In this particular eample, the value o a, in the standard orm is -1. A negative relects the graph o the unction about the horizontal ais. Once again, i the parent unction given is reerred to as unction is ( ). ( ), this 9

6 We have seen how to graph a unction by shiting the parent unction. You may have noticed that we graphed ( ), but not ( ). The reason we did not see ( ), is because this is the graph o an even unction. That means that i a were plugged in to the unction, it would make no dierence. The outcome would be the same. However, i we are dealing with a dierent type o unction, one that was not even, ( ) would cause the graph o the unction to relect about a vertical ais. In other words, i ( ) makes a graph lip upside down, ( ) would make the graph lip rom right to let, or let to right, whatever may be the case. Lets see how diering values o a, h and k will cause various shits o the unction. y = a ( h) + k Once again, take note o the parent unction y = y = ( + ) + y ( ) = + This graph opens down, and shits let, up. This graph opens up and shits right, and up I this graph is a translation o the unction It would by written as + +. ( ) ( ) I this graph is a translation o the unction It would by written as +. ( ) ( ) We will be graphing unctions using only ( ) in the translations o unctions section. 40

7 y = a ( h) + k y = Here we will see how the value o a or the quadratic unction in standard orm aects the graph o the unction. To illustrate this, we will look at the graph o a parabola that has its verte on the origin. y = 4 y = 1 4 This graph seems very narrow, but what is actually happening, is the value o the unction is increasing very rapidly. The y values are increasing at 4 times their normal rate. The rapid increase causes the graph to appear narrow. This graph is wider than the parent unction. In this case, the y values o the unction are increasing at ¼ their normal rate, causing a more gradual increase. As you can see, i the value o the leading coeicient is a whole number, the y values o the graph will increase rapidly causing a narrow and steeper curve. In contrast, i the leading coeicient is a raction, the y values o the unction will increase mildly, causing a more gradual curve. 41

8 Describe the movement o each o the ollowing quadratic unctions. Describe how each opens and i there is any horizontal or vertical movement. Be sure to state how many spaces it moves, or eample: This graph opens down, and shits let, up. 1 A) y = ( 4) + B) y = ( + ) 8 C) y = ( ) D) 1 y = + E) y = ( + 5) + 6 F) y ( ) = G) y = ( ) + H) y = ( + 6) + 8 I) y ( ) = 4 J) y = K) y ( ) = L) 5 y = + 8 As you describe the graphs o the quadratic unctions above, you wrote that it shits to the let or right, and up or down. What is actually shiting? 4

9 Write the equation or a quadratic unction in y = a ( h) + k orm that opens down, shits let and up 7. Write the equation or a quadratic unction in y = a ( h) + k orm that opens up, shits right 4 and down. Write the equation or a quadratic unction in y = a ( h) + k orm that opens up, and only shits down 4. Write the equation or a quadratic unction in y = a ( h) + k orm that opens down and shits to the let 8 spaces. Write the equation or a quadratic unction in y = a ( h) + k orm that opens down and shits up 7. Is a quadratic unction a one-to-one unction? Why or why not? What does this tell you about the inverse o a quadratic unction? 4

10 Match the appropriate graph with its equation below. Eplain why each o your solutions is true. A B C D E F = 1 + ) ( ) 1) ( ) ( ) = ) ( ) ( ) = ) ( ) = ( ) + 1 5) ( ) = ( + ) 6) ( ) ( ) = + 44

11 Graph each o the ollowing unctions. You may need to use an ais o symmetry to graph some o these. Label the verte, y-intercept, and all -intercepts. A) ( ) = ( ) + 1 B) ( ) ( ) = C) ( ) = ( ) D) ( ) ( ) =

12 E) ( ) = + F) ( ) ( ) = G) ( ) = ( 4) H) ( ) ( ) =

13 The quadratic unction given by the equation ( ) ( ) o. = + 6 has an ais o symmetry The quadratic unction given by the equation ( ) ( ) o. = has an ais o symmetry The quadratic unction given by the equation ( ) ( ) o. = a h + k has an ais o symmetry Considering your answers to the previous questions, we can conclude that the ais o symmetry or any quadratic unction is given by the value o the. Why does the ais o symmetry look as though we are saying equals a number ( = # )? Why is it sometimes necessary to graph a quadratic unction using the ais o symmetry? 47

14 Absolute Value Functions In order to graph an absolute value unction, you will be using many o the same methods you did or quadratics. The standard orm o an absolute value unction is nearly identical to that o a quadratic unction. ( ) = a h + k The standard graph by which we translate absolute value unctions comes rom the equation o the diagonal line y =. y = This is the graph o the unction y =. In this case, the and y values o coordinates are identical. For eample, (-,-). You can see the and y values are the same. Now, lets take a look at what happens when I want the absolute value o. y = I the graph o y = above is ( ), the unction to the let is ( ). We know that the absolute value o a number cannot be negative. I we take the absolute value o ( ), it would cause the let portion o the graph above to relect above the ais. Now as you can see, all y values o the unction are positive. This is where the graph o the absolute value o comes rom. 48

15 As we look at the ollowing absolute value unctions, you will notice how similar they are to quadratic unctions. The verte o an absolute value unction is also given by (h,k). Horizontal and vertical shits are identical, as well as the eect the value o a has on the graph. The rules or inding the range and domain o an absolute value unction are also the same as a quadratic. Sometimes, an ais o symmetry must be used to graph your unction. Intercepts are ound by substituting zero or either or y, and solving or the remaining variable. ( ) = a h + k ( ) = = ( ) = + ( ) The graph o this unction shits to the right. The graph o this unction shits to the let. Once again, notice that the value o h determines the horizontal shit o the unction. I the unction is deined as ( ), the graph on the let is ( ), while the graph on the right is ( + ). 49

16 ( ) = a h + k ( ) = = + ( ) = ( ) The graph o this unction shits up. The graph o this unction shits down. The value o k, or an absolute value unction in standard orm determines the vertical shit o the unction. As beore, i the unction is simply deined as ( ), we are looking at ( ) + and ( ) respectively. ( ) = On the let we have the opposite o the parent unction. In this eample, the value o a, in the standard orm is -1. A negative relects the graph o the unction about the horizontal ais. This is read as the opposite o the absolute value o. I the parent unction given is reerred to as ( ), this unction is ( ). 50

17 ( ) = a h + k ( ) = Here we will see how the value o a in an absolute value unction in standard orm aects the graph o the unction. To illustrate this, we will look at the ollowing graphs that have their vertices on the origin. 1 ( ) = 4 ( ) = 4 This graph seems very narrow, but what is actually happening, is the value o the unction is increasing very rapidly. The y values are increasing at 4 times their normal rate. The rapid increase causes the graph to appear narrow. This graph is wider than the parent unction. In this case, the y values o the unction are increasing at ¼ their normal rate, causing a more gradual increase. I the value o the leading coeicient is a whole number, the y values o the graph will increase rapidly causing a narrow graph and more etreme slope. I the leading coeicient is a raction, the y values o the unction will increase mildly, yielding a more gradual slope. 51

18 Describe the movement o each o the ollowing absolute value unctions. Describe how the graph o the unction opens and i there is any horizontal or vertical movement. Be sure to tell identiy how many spaces it moves, or eample: This graph opens up, and shits let 6, up. A) ( ) = B) ( ) = C) ( ) = + 4 D) ( ) = + 5 E) ( ) = F) ( ) = G) ( ) 1 = + H) ( ) = + 6 I) ( ) = 4 State the range and domain or each o the ollowing. A) ( ) = B) ( ) = C) ( ) = + 4 D) ( ) = + 5 E) ( ) = F) ( ) = G) ( ) 1 = + H) ( ) = + 6 I) ( ) = 4 5

19 Find the verte o each o the ollowing absolute value unctions. 1 A) ( ) = + B) ( ) = 4 C) ( ) = D) ( ) = + E) ( ) = F) ( ) = Solve each o the ollowing absolute value equations. This is what you will need to do to ind the intercepts o absolute value unctions. Remember, irst isolate the absolute value, then set up two separate equations to ind your solutions = Set equation equal to 6 and solve by subtracting to both sides. + = 6 = + = 6 and + = 6 = 9 Set equation equal to -6 and solve by subtracting to both sides. So the two solutions are and -9. These would be the intercepts o the graph o the unction y = + 6. Watch or abnormalities. Ii the absolute value equals a negative number, you cannot create two problems. I this happens, there will be no solutions to the problem, which in terms o the graph o the unction, tells you that there are no intercepts. A) 4 = 0 B) = 0 C) = D) = 0 E) 6 5 = 0 F) 1 + = 0 5

20 Match the appropriate graph with its equation below. Eplain why each o your solutions is true. A B C D E F 1) ( ) = 4 1 ) 1 ( ) = 1 ) ( ) = + 4 4) ( ) = + 5) ( ) = 1 + 6) ( ) = + 54

21 Graph each o the ollowing unctions. You may need to use an ais o symmetry to graph some o these. Label the verte, y-intercept, and all -intercepts. Remember, to ind the intercepts o an absolute value unction you will need to set the unction equal to zero and solve an absolute value equation. A) ( ) = + B) ( ) = C) ( ) = 5 D) ( ) = 55

22 E) ( ) = F) ( ) = G) ( ) = + H) ( ) =

23 Translations o Functions We will now look at graphing a unction without actually knowing the equation. Based on the graph o a unction, it will be possible to shit, or translate the graph in any manner indicated. For eample, i given the picture o a graph and told This is the graph o the unction ( ). Proceed to irst identiy the coordinate o any verte seen. These will serve as a guide or the graph o the unction s translation. To graph the unction o ( + 6), the unction will need to shit to the let 6 spaces. To accomplish this, subtract 6 rom all values in the original unction. The results will be the coordinates or the new graph. Likewise, to graph ( 4), this unction will need to shit to the right 4 spaces, so add 4 to all values. In order to graph to graph ( ) + 5, the unction will shit up 5 spaces, requiring that 5 be added to all y values. I asked, the will unction shit down spaces, meaning subtract rom all y values. ( ) I the number is in the parenthesis, the unction is shiting using P.L.N.R.. I the number is ater the ( ), simply shit as indicated, + says shit up, - says shit down. Any number in ront o the the unction grows. When graphing, or eample, ( ) will aect the scale o the unction. This means it will aect the rate at which ( ), change the sign o all y values on the graph o the unction. This will cause the graph o the unction to lip upside down. A number other than -1 can also be used. Lets say we need to graph ( ), this means the actual curve will increase times as ast. It will thereore, be necessary to multiply all y values by. This will result in the coordinates or the new unction. I the were grouped with the such as ( ), the horizontal change is the inverse o what it appears to be. So instead o multiplying values by, divide by. When graphing ( ), take the opposite o the values o the unction. This will cause the graph o the unction to lip along a vertical ais. Combinations o these rules will be encountered throughout your study o unctions, or eample, to shit right and up 6. Just stick with the rules and the graph will be translated to its new location. I aced with a problem such as ( ) +, ollow the order o operations. Multiply all y values by irst, then add to each. Reerring to the previous two topics, quadratic unctions and absolute value unctions, you will ind reerences to these rules and eamples throughout. 57

24 The ollowing is the graph o the unction ( ). Use this to graph each unction or letters A-D. B) + ( ) A) ( ) C) ( ) + 1 D) ( ) 58

25 The ollowing is the graph o the unction ( ). Use this to graph each unction or letters E-H. F) ( ) E) ( 1) G) ( ) 4 H) ( + ) 59

26 The ollowing is the graph o the unction ( ). Use this to graph each unction or letters I-L. I) ( ) J) ( ) 1 1 K) ( ) + 1 L) ( ) 60

27 Radical Functions For radical unctions we will use the equation ( ) = a h + k to denote the standard orm o the equation. Be aware, that the variable may have a coeicient rom time to time. Follow the standard procedure to ind the and y intercepts o any radical unction. Set the or y equal to zero, depending on which one you wish to ind, and solve or the remaining variable. Finding the domain o a radical unction is a little tricky. To ind the domain o any radical unction with an even inde, set the radicand greater that or equal to zero ( ) and solve. I the radicand is a polynomial, you will need to solve the polynomial inequality by inding critical points, and testing intervals. To ind the range o the radical unction, ind y value o the point o origin, and use the constant a to determine the range o the unction. Given the radical unction ( ) = + 4, the ollowing can be determined. First ind the domain o the unction. This will give you the value needed or the point o origin. ( ) = + 4 Finding the domain You can see the domain o the unction is[ 4, ) The -4 is the value the point o origin.. Finding the range. Since the constant a is -1, the unction will go downwards. Meaning that the range is (, ]. Finding the point o origin o a radical unction. To ind the point o origin o a radical unction use the rules discussed in previous sections. The point o origin or the parent unction y = is (0, 0). This particular graph will shit let 4 and down, so the point o origin is ( 4, ). Be careul when using these rules. Make sure to ind the domain o the unction beore you attempt to ind the point o origin. Consider a unction such as y =. Since there is a positive inside the radicand, you would normally shit to the let. However, I you were to ind the domain o this unction by setting the radicand 0, You will ind the domain is actually spaces.. This says the graph is shiting to the right Finding the -intercept. Substitute 0 or y and solve or. 0 = = Finding the y-intercept. Substitute 0 or and solve or y. ( ) y = y = 4 y = This is not possible. That means there is no intercept or this unction. y = 5 The y intercept o this unction is (0, -5). 61

28 We will now look at the parent unction, and some translations o it. = a h + k ( ) ( ) = ( ) The parent unction has the point o origin at (0, 0) = ( ) = + The graph o this unction shits right. The graph o this unction shits let. ( ) = + ( ) = Here the graph shits up. The graph o this unction shits down. 6

29 ( ) = a h + k ( ) = The parent unction has the point o origin at (0, 0) ( ) = ( ) = The graph o this unction lips upside down. ( ) = 5 The graph o this unction lips rom right to let as the aects the domain o the unction. = ( ) 1 5 Scale increased by a actor o 5. This is 1/5 the normal scale. 6

30 Find the domain o each o the ollowing radical unctions in interval notation. A) ( ) = + 4 B) ( ) = C) ( ) = D) ( ) = 4 E) ( ) = F) ( ) 1 = 6 G) ( ) = H) ( ) = + 1 I) ( ) = The range o a radical unction in ( ) = a h + k orm can be ound using the value o the a term, and the y value o the point o origin. a >, the range o the unction is [ k, ). a <, the range o the unction is (, k] I 0 I 0. Find the range or each o the ollowing. A) ( ) = + 5 B) ( ) = + C) ( ) = 4 + D) ( ) = E) ( ) = 4 F) ( ) =

31 Find the point o origin or each o the ollowing radical unctions. A) ( ) = + 4 B) ( ) = C) ( ) = 4 D) ( ) = E) ( ) = F) ( ) 1 = 6 G) ( ) = H) ( ) = + 1 I) ( ) = Why is the graph o the unction ( ) = moving towards the let rather than the right? Eplain why the graph o the unction ( ) = is identical to that o ( ) =. To ind the domain o a radical unction that has an even inde, why do you need to set the radicand 0? 65

32 We will now look at the cube root unction. = a h + k ( ) ( ) = ( ) The parent unction has the point o origin at (0, 0) = ( ) = + The graph o this unction shits right. The graph o this unction shits let. ( ) = 5 As you can see on the let, the curve is just about the same or a 5 th root, verses a cubed root. This will be the same case or any radical unction where the inde is odd. This also means that any radical unction where the inde is even will look like a normal square root unction. The curves o these unctions are a little latter than a regular square root or cubed root. Vertical translations o the unction are identical to that o a regular square root unction. As you can see, the domain and range o any radical unction with an odd inde is all real numbers. 66

33 Match the appropriate graph with its equation below. Eplain why each o your solutions is true. A B C D E F 1) ( ) = + + ) ( ) = ) ( ) = + 1 4) ( ) = + 1 5) ( ) = + 6) ( ) = 67

34 Graph each o the ollowing radical unctions. Find all required inormation. A) ( ) = + Point o Origin: Y-intercept: X-intercepts: Range: Domain: B) ( ) = + 1 Point o Origin: Y-intercept: X-intercepts: Range: Domain: 68

35 C) ( ) = + 1 Point o Origin: Y-intercept: X-intercepts: Range: Domain: D) ( ) = 4 Point o Origin: Y-intercept: X-intercepts: Range: Domain: 69

36 E) ( ) = Point o Origin: Y-intercept: X-intercepts: Range: Domain: F) ( ) = + + Point o Origin: Y-intercept: X-intercepts: Range: Domain: 70

37 G) ( ) = Point o Origin: Y-intercept: X-intercepts: Range: Domain: H) ( ) = 6 Point o Origin: Y-intercept: X-intercepts: Range: Domain: Why are the graphs o y = and y = identical? 71

38 Eponential Functions This inormation was covered in a previous section o the workbook, but it won t hurt to go over it again. Standard eponential unction h = ca + k ( ) The c term is a constant that can make to graph relect about a horizontal ais or change to scale o the graph o the unction proportionately. I c is a positive value, then you will have a standard looking growth or decay curve. I c is negative, the growth or decay curve will lip upside down. We will get into the eects dierent values o h and k have on this unction shortly. What we will concentrate on here is identiying an eponential unction as being growth or decay, and inding the range, domain and key point o the unction. Eponential Growth Eponential Decay h h ( ) = ca + k ( ) = ca + k I a>1 I 0<a<1 Obviously i a = 1, we are raising 1 to various powers, and we wind up getting a horizontal line because no matter what you do, raising one to any power still yields a result o one. Pay special attention to the eponential decay unction. The statement 0 < a < 1 is saying that the value o a is a raction whose value is between zero and one. Do not make the mistake o just looking or a raction to determine whether or not the unction is decay. Make sure the value o the raction is between zero and one. The values or variables,, h k and c act to make the graph shit let/right, up/down, change the scale or will relect the unction about a horizontal ais. 7

39 Notice the key point or each o these unctions is the point ( 0,1 ). This inormation is vital. This key point will shit depending on the values o h, k and c. To ind the value o the key point, evaluate h = 0. In other words, ind the value o that would create a problem such as to the zero power. This number is the value o the key point. To ind the y value, substitute the value back in. Reer to the ollowing eample. ( ) = + 5 to ind the key point evaluate = 0 = 0 = this is the value o the key point now substitute back into the problem or ( ) ( ) ( ) ( ) 5 = = + = 1+ 5 = 6 so the key point is (,6) h = ca + k ( ) As we work to translate these unctions, use ( 0,1 ), as the deault key point to any eponential growth or decay curve that is above the horizontal asymptote where the value o c is 1. In 0,1 to assist you in shiting the other words, i the value o c is positive one, use the point ( ) unction. I the graph o the unction is below the horizontal asymptote, and the c value is 0, 1 as the key point. I the value o c is any other number, you must -1, you will use ( ) ind the key point algebraically. Consider the eample above. ( ) = + 5 The irst thing I did was notice that the graph o this unction shits right and up 5. Pay special attention to the value o k. That tells you where your new horizontal asymptote is going to be, in this case, at y = 5. Since the value o c is positive one, begin at the key point ( 0,1 ). Since the graph will shit right and up 5, simply add to the value, and 5 to the y value o the key point. This produces a new key point o (,6 ). Observe how this inormation matches the work above. The key point in these unctions acts as the verte in a parabola. It gives you a point o reerence with which to shit the unction. Make sure the correct key point is used rom the beginning, either ( 0,1 ) or ( 0, 1) number. For eample, the unction ( ) = ( ) has a key point o ( 0, ), not ( 0,1 ).. Remember, i the c term is a number other that 1 or -1, the key point is actually multiplied by that 7

40 Here we will look at how to graph these unctions by means o translation. Eponential Growth h = ca + k ( ) ( ) = ( ) 4 = + ( ) = 5 Notice that the horizontal asymptote is at y=. Since the graph o this unction is going to be above the ais, begin with the key point (0,1). This unction shits right 4 and up. The dots have been let on the graph so it would be easier to see. Adding 4 to the value o the key point, and to the y value, the new key point is at (4,). Just remember where to begin, and do not cross the horizontal asymptote. In this unction, the value o k is -5. This tells you the new horizontal asymptote will be at y = -5. Since the value o the constant c is a positive one, begin with the key point (0,1). This unction will only shit down 5 spaces. Thereore, subtract 5 rom the y value o the key point which is 1. This results in: (1 5 = -4) thereore, the new key point is at (0,-4). Graphing eponential unctions by translation is relatively simple. The most diicult part will be inding the and y intercepts as the -intercept will involve the use o logarithms. 74

41 Eponential Growth h = ca + k ( ) ( ) = ( ) = + 4 ( ) = + 1 Since the value o c in the equation o this unction is -1, we must begin with the key point o (0,-1). This is the key point, because that value o c caused the graph to relect about the horizontal asymptote. The entire unction will shit up 4, so the new horizontal asymptote is y = 4. The curve is going to shit right and up 4. By adding to the value o the key point, and 4 to the y value, the new point can be ound at (,). Notice the graph runs right though that point. Here we have something that looks like decay. The value o a in this unction is greater that one, so it should be growth. What really happened here, is the laws o eponents went to work. Is the same ( ) thing as. The power o a power rule says this 1 can be seen as ( ). This simpliies to ( ) decay curve. OK, so we begin with a decay curve that has a key point o (0,1). Add to the value, and 1 to the y value o the key point, and the new key point is (,), with a horizontal asymptote o y = 1. 1, a 75

42 Eponential Decay h = ca + k ( ) ( ) 1 = ( ) + 1 = ( ) 4 1 = + This is an eponential decay curve. The original graph will lie above the ais. Thereore, begin with the key point (0,1). The key point will shit to the let, and down, so subtract rom the value and rom the y value o the key point, and the coordinates o the new point will be at (-,-1). The graph o this unction shits down, so the horizontal asymptote o this unction is y = -. This is an eponential decay unction that is relected and lies below the horizontal asymptote. The initial key point here is (0,-1). Since this graph will shit right 4 and up, add 4 to the value o the key point and to the y value. This yields a result o (4,). Since the entire graph shited upwards spaces, the horizontal asymptote is y =. Once again, when graphing, do not cross the horizontal asymptote. The translations o these unctions are very similar to that o other unctions we have seen. A point o reerence with which to shit is all that is needed. Most important is to make sure to always use the appropriate key point to start with. Draw the horizontal asymptote irst, that way the graph o the unction does not accidentally cross it. 76

43 h = ca + k ( ) The domain o any eponential unction is (, ). The values o c and k terms will determine the range o the unction. Since the horizontal asymptote o an eponential unction is given by y=k, the value o k will determine where the horizontal asymptote o the unction lies, whereas the value o c will determine i the unction is above or below that asymptote. Be careul not to use brackets when describing the range o an eponential unction. The horizontal asymptote must not be touched, so only parenthesis may be used to describe the range in interval notation. Find the range and domain o each o the ollowing eponential unctions. A) ( ) + 6 = 4 B) ( ) 1 1 = + + = 5 C) ( ) ( ) 1 D) ( ) 5 = E) ( ) ( ) + = 5 F) ( ) = e + G) ( ) 8 5 = + 4 H) ( ) = 7 I) ( ) = + 4 J) ( ) 5 1 = + 1 K) ( ) 7 = 6 1 L) ( ) = e + 77

44 Here is an eample o inding the and y intercept o an eponential unction. Finding the intercept. ( ) = + 4 Finding the y intercept. Begin by substituting 0 or ( ) Begin by substituting 0 or. + 0 = 4 4 = + log 4 = log ( ) + log 4 = + log log 4 = log + log log 4 log = log ( ) ( ) ( ) = ( ) 0+ 4 = 4 = 9 4 = 5 Now divide both sides by log. log 4 log log = log log log 4 log = log As you can see, this unction has an intercept o approimately (-0.74,0), and a y intercept o (0,5). Find the key point to each o the ollowing unctions. A) ( ) + 4 = B) ( ) = C) ( ) = = 5 E) ( ) = D) ( ) ( ) F) ( ) =

45 Match the appropriate graph with its equation below. Eplain why each o your solutions is true. A B C D E F 1) ( ) = + 1 ) ( ) 1 = ) ( ) = ( ) 4) ( ) = + 1 5) ( ) 5 = + 4 6) ( ) = 79

46 Graph each o the ollowing eponential unctions. Be sure to label the key point o the unction. Find the intercept (i it eists) and y intercept o each unction. A) ( ) = 1 Y-intercept: X-intercepts: Range: Domain: B) ( ) = + 4 Y-intercept: X-intercepts: Range: Domain: 80

47 C) ( ) + 1 = Y-intercept: X-intercepts: Range: Domain: D) ( ) = 4 Y-intercept: X-intercepts: Range: Domain: 81

48 E) ( ) 1 = Y-intercept: X-intercepts: Range: Domain: F) ( ) ( ) = + 1 Y-intercept: X-intercepts: Range: Domain: 8

49 Logarithmic Functions = a b + c + d = a b + c + d ( ) log ( ) ( ) ln n ( ) Logarithmic unctions will be graphed in the same manner as radical unctions. It is irst necessary to ind the domain o the logarithmic unction. The range o a logarithmic unction is all real numbers, so only the domain needs to be ound. To ind the domain o a logarithmic unction evaluate b + c > 0. Remember, this is not, because you cannot take the log o zero. Once the domain is ound, it will tell in which direction the unction is moving. This inequality will also help ind the vertical asymptote or the unction. I the inside the log does not have a negative coeicient, the curve will be on the right side o the vertical asymptote. I the coeicient in ront o is 1, begin with the key point o (1,0). From that point on, treat the unction just like an eponential unction. Adding or subtracting to either the or y values to ind the new key point making the graph shit. I the inside the log has a negative coeicient, the curve will be on the let side o the vertical asymptote. I the coeicient in ront o is -1, begin with the key point o (-1,0) and shit rom there. *Once again, just like eponential growth and decay unctions, watch the value o a, as it aects the scale o the unction. I the value o a is some number other that 1 or -1, ind the key point algebraically beore you translate the unction. As the unction shits, it will be helpul to draw a broken line or both the horizontal and vertical asymptotes. It is OK to cross the horizontal asymptote, as you will ind the key point always rests on it. The vertical asymptote, however, may never be crossed. = a b + c + d = a b + c + d ( ) log ( ) ( ) ln n ( ) Solving or b + c = 0, will yield the equation or the vertical asymptote. The equation or the horizontal asymptote is y = d. Finding the domain. ( ) ( ) = log > 0 Notice the similarity in the procedures. > 4 Finding the vertical asymptote. 4 = 0 = 4 *I the variable inside the log has a coeicient other than 1 or -1, the key point will be dierent. The key point must then be ound algebraically. To ind the value o the key point solve or b + c = 1. Substitute that solution back into the problem to ind the y value. Finding the horizontal asymptote. y = There is no real work involved with inding the horizontal asymptote. Identiy the vertical shit. This is the equation o the horizontal asymptote. 8

50 = a b + c + d = a b + c + d ( ) log ( ) ( ) ln n ( ) = log ( ) The parent unction has the key point at (1, 0) ( ) = log ( ) ( ) = log ( + ) The graph o this unction shits right. Notice the key point moved to the right places to (4,0). The graph o this unction shits to the let. The new key point is (-,0). = log + ( ) = log ( ) This unction shits up. Add to the y value o the key point, and it is now at (1,). This unction shits down. Subtracting rom the y value o the key point results in (1,-). 84

51 = a b + c + d = a b + c + d ( ) log ( ) ( ) ln n ( ) = log ( ) ( ) log ( ) The parent unction has the key point at (1, 0) = ( ) = log The graph o this unction relects about the vertical asymptote. Key point is now (-1,0). The graph o this unction is relected about the horizontal asymptote. Key point is still at (1,0). ( ) log ( ) = ( ) = log Since the coeicient o is -1, this graph will be on the let side o the vertical asymptote. Begin with the key point (-1,0), and shit right because it is positive. Add to the value o the key point. The new key point is (1,0). Notice the negative portion o the graph relected above the ais. 85

52 Match the appropriate graph with its equation below. Eplain why each o your solutions is true. A B C D E F 1) ( ) = log ( ) ) ( ) log ( 1 ) = ) ( ) = log + 1 4) ( ) = log + 5) ( ) = log 6) ( ) = log 86

53 The translation o a logarithmic unction is almost identical to that o an eponential unction. Just make sure to identiy on which side o the vertical asymptote the graph o the unction will reside. This will determine which key point to begin with. Remember to draw both asymptotes to graph the unction and watch or the value o a which will aect key point. Graph each o the ollowing logarithmic unctions by inding the asymptotes and labeling the key point. Be sure to ind the intercept and y intercept (i they eist). A) ( ) = log + Y-intercept: X-intercepts: Range: Domain: B) ( ) = log ( + ) Y-intercept: X-intercepts: Range: Domain: 87

54 C) ( ) = log + 4 Y-intercept: X-intercepts: Range: Domain: D) ( ) = ln ( ) Y-intercept: X-intercepts: Range: Domain: 88

55 E) ( ) = ln ( ) Y-intercept: X-intercepts: Range: Domain: F) ( ) ( ) = log + Y-intercept: X-intercepts: Range: Domain: 89

56 G) ( ) = ln Y-intercept: X-intercepts: Range: Domain: H) ( ) = ln Y-intercept: X-intercepts: Range: Domain: 90

57 All standard logarithmic unctions (meaning a unction without absolute value symbols), must have an intercept. All standard eponential growth and decay unctions must have a y intercept. Are these two statements true? Why or why not? In order to ind the domain o the logarithmic unction ( ) ( ) evaluate + 5 > 0. Why must we use this inequality? = log + 5, we need to 4 What is the problem with relying on a graphing calculator to graph a logarithmic unction? Cubic Functions The cubic unction is similar to the cubed root. You will notice similarities in the shape o the curve. Translations are the same as any standard unction. The range and domain o any cubic unction is all real numbers. = a h + k ( ) ( ) Let us look at this as the standard orm o a cubic unction. The center o the cubic unction h, k. To ind the and y intercepts o the unction, ollow the standard is given by ( ) procedures o substituting zero or one o the values, and solving or the remaining variable. All cubic unctions in this section will be given to you in this standard orm. In the net section o the workbook, we will address polynomial unctions that are greater than nd degree. These unctions will have no standard rom with which to work. We will be graphing them by alternative means. For now we will concentrate on the unction ( ) =, and translations o this. 91

58 = a h + k ( ) ( ) ( ) = The parent unction has the point o origin at (0, 0) ( ) = ( + ) ( ) = ( ) The graph o this unction shits let. The graph o this unction shits right. ( ) = + ( ) = Graph shits up. Graph shits down. 9

59 = a h + k ( ) ( ) ( ) = The parent unction has the point o origin at (0, 0) ( ) = ( ) ( ) = 4 The graph o this unction lips upside down. ( ) = 5 The graph o this unction shits right, down 4. verte at (,-4). ( ) = 1 5 Once again, note dierence the value o a makes in terms o the scale o the graph. 9

60 Match the appropriate graph with its equation below. Eplain why each o your solutions is true. A B C D E F 1) ( ) = ) ( ) ( ) = + 1 ) ( ) = ( + ) + 1 4) ( ) 1 = 5) ( ) = 6) ( ) ( ) = + 94

61 Graph each o the ollowing cubic unctions. Label the verte, ind all intercepts, and the range and domain o each o the ollowing. Don t worry about graphing the intercept i it is too ar o the chart. = + A) ( ) ( ) Verte: Y-intercept: X-intercepts: Range: Domain: = + 1 B) ( ) ( ) Verte: Y-intercept: X-intercepts: Range: Domain: 95

62 C) ( ) = + Verte: Y-intercept: X-intercepts: Range: Domain: = + 5 D) ( ) ( ) Verte: Y-intercept: X-intercepts: Range: Domain: 96

63 = 4 + E) ( ) ( ) Verte: Y-intercept: X-intercepts: Range: Domain: = 5 F) ( ) ( ) Verte: Y-intercept: X-intercepts: Range: Domain: 97

64 Piece-Wise Functions The best way to describe a piece-wise unction is to look at a simple eample. Consider the absolute value unction. y = This is the graph o the unction y =. In this case, the and y values o coordinates are identical. For eample, (-,-). You can see the and y values are the same. Now, lets take a look at what happens when we want the absolute value o. y = I the graph o y = above is ( ), the unction to the let is ( ). We know that the absolute value o a number cannot be negative. I we take the absolute value o ( ), it would cause the let portion o the original graph to relect above the ais. This results in all y values o the unction being positive. This is where the graph o the absolute value o comes rom. It is possible to get this same graph i the linear unction y [ 0, ), and the unction y = in the interval (,0) = is graphed in the interval. What happens here is a speciic section o two dierent graphs is drawn. I these two sections are placed together on the same plane, it would result in the graph o y ( ) =. Such an equation would appear as ollows., 0 =, < 0 98

65 , 0 ( ) =, < 0 ( ) =, < 0 ( ) =, 0 Notice the graph o the unction on the let has a hole when = 0. Since the unction does not eist when is zero, because o the domain o the unction, an open circle must be used at that point. A pronounced dot is placed on the unction to the right, showing the graph o the unction in the interval 0. This is how you plot a point when you have the greater than or equal to sign in the domain o the unction. I both o these unctions are placed on the same plane, the result would be as ollows., 0 ( ) =, < 0 Since the open circle and solid dot are right on top o each other, the dot would ill in the hole on the irst curve. This results in a continuous unction. I the hole were still there, the unction would have a removable discontinuity. I there is a complete break in the curve, rom one portion o the graph to the net, it would be a discontinuous unction. The simplest way to graph a piece-wise unction is to graph the entire unction and erase the portions that are not needed. This will be done or each part o the overall graph until a complete picture has been created. Very simply, a piece-wise unction is just as it sounds, pieces o dierent unctions put together to create one graph. Some eamples o piecewise unctions can be ound in the Translations o Functions section o this chapter. 99

66 Graph each o the ollowing piece-wise unctions. A) ( ) + 5, =, < < 1 + 1, 1 B) ( ) = ( ) ( ), <, 00

67 C) ( ) ( ) +, 4 =, 4 < <, D) ( ), < 1 = +, 1 1 ( ), > 01

68 E) ( ) = + 4, < , > 1 F) ( ) = ( + 4) +, 7 < 4, 4 < , 1 < 1 ( ), 5 0

69 G) , 7 4 ( ) ( ) =, 4 < < 0, 0 4 H) log ( ), 8 < 1 ( ) = 1, 1 1 log, 1 < 9 Which i any o the piece-wise unctions you just graphed are discontinuous? 0

70 Functions Review Graph each o the ollowing unctions. You need only label the key point or verte or each. Do not worry about anything else. These problems are meant to quiz you on your knowledge o the parent unctions and the translations thereo. 1. ( ) = +. ( ) = +. ( ) = + 4. ( ) 1 = 4 5. ( ) = log 6. ( ) =

71 = ( ) = ( ) ( ) = ( ) = 9. ( ) ( ) 11. ( ) = + 1. ( ) = ln ( ) 05

72 1. ( ) + = ( ) = ( + ) 15. ( ) = log ( ) = ( ) e + = ( ) ( ) =

73 19. ( ) + 1 = + 0. ( ) = + 4 =. ( ) ( ) 1. ( ) =. ( ) = + 4. ( ) = + 07

74 5. ( ) = e + 6. ( ) = ln( + ) + 7. ( ) = 1 8. ( ) = + 9. ( ) 1 = 0. ( ) = 08

75 Checking Progress You have now completed the Functions section o the workbook. The ollowing is a checklist so that you may check your progress. Check o each o the objectives you have accomplished. The student should be able to QUADRATIC FUNCTIONS Determine the properties o a quadratic unction in standard orm. Find the and y intercepts o a quadratic unction. Find the range and domain o a quadratic unction. Find the verte o a quadratic unction in standard orm. Graph a quadratic unction. ABSOLUTE VALUE FUNCTIONS Determine the properties o an absolute value unction in standard orm. Find the and y intercepts o an absolute value unction. Find the range and domain o an absolute value unction. Find the verte o an absolute value unction. Graph an absolute value unction. RADICAL FUNCTIONS Determine the properties o a radical unction in standard orm. Find the and y intercepts o a radical unction. Find the range and domain o a radical unction. Find the point o origin o a radical unction Graph a radical unction. EXPONENTIAL FUNCTIONS Determine the properties o an eponential unction in standard orm. Find the and y intercepts o an eponential unction. Find the range and domain o an eponential unction. Find the key point o an eponential unction. Graph an eponential unction. LOGARITHMIC FUNCTIONS Determine the properties o a logarithmic unction in standard orm. Find the and y intercepts o a logarithmic unction. Find the range and domain o a logarithmic unction. Find the key point o a logarithmic unction. Graph a logarithmic unction. 09

76 Checklist continued. The student should be able to CUBIC FUNCTIONS Determine the properties o a cubic unction in standard orm. Find the and y intercepts o a cubic unction. Find the range and domain o a cubic unction. Find the verte o a cubic unction. Graph a cubic unction. PIECEWISE FUNCTIONS Shit the graph o a unction without actually knowing the equation, i.e. graphing. + ( ) Graph piece-wise unctions. 10