Unit 7 Graphs and Graphing Utilities - Classwork

A. Graphing Equations Unit 7 Graphs and Graphing Utilities - Classwork Our first job is being able to graph equations. We have done some graphing in trigonometry but now need a general method of seeing a complete graph of any equation. To enter equations, you go to your y = menu and enter the equation. Be aware of keys and menus on your TI-84 that allow math operations: ^ = eponentiation - squaring - square root - reciprocal Math - cube Math 4 - cube root Math Num - absolute value When we view a graph, the first thing we do is to graph it in the calculator s standard window. The standard window on the calculator is 0 0 and 0 y 0. To get that window, you press Zoom 6 : ZStandard. That sets the value of min = -0, ma = 0, ymin = -0, yma = 0. For instance, to graph the equation: y = 0 7 in the standard window: You have the ability of changing the viewing window by going to the Window menu, adjusting the values of min, ma, ymin, and yma. For instance, to view this graph in the window 0 and y, adjust the window and then press Graph. Xscl and Yscl refer to the tick marks on the aes. In this case, both are. We don t usually take the time to adjust them. We now define a complete graph: A complete graph shows all the behaviors of the graph. Behaviors include all turns of the graph high and low points as well as intercepts, where the graph crosses the - and y- aes. Finding a complete graph is sometimes a matter of trial and error. There is no one window that gives a complete graph. For instance, in the graph above, it is clear that the graph goes lower so we have to adjust ymin. It is also clear that the graph goes further to the left so we have to adjust XMIN. So here is a complete graph and the window that created it. Again, let s be clear: ) There is no one best complete window and ) it is a trial and error process: Here is another possible complete window graph. The better complete windows use up most of the viewing screen. The solution above is better than the one to the right. 7. Graphing and Graphing Utilities - - www.mastermathmentor.com - Illegal to post on Internet

Finally, note that the graphs above are fatter and thinner. When we adjust the window, we change the size of the graph. The standard window [-0, 0], [-0, 0] has a basic problem because the screen is wider than the height giving distortion. The way of fiing that problem is going to the Zoom 6 and then immediately to the Zoom 5: Square view. This is the graph and the window. Note that the ratio of the screen is roughly to. If we really wanted a complete curve drawn correctly to scale, we would have to change both the min, ma and ymin, yma so that they are in the same proportions (doubling them, tripling them, etc.) Normally, we don t do these steps unless we want true accuracy. Mostly we are just interested in the shape of the graph and where it crosses the -ais.. Graph the equations given in class according to the prescribed windows. a. Equation: y = Standard [, ] [ 4,0] Complete Graph [ 0,0] [ 0,0] b. Equation: y = 7 Standard [, ] [ 9,] Complete Graph [,] [ 5,] c. Equation: y = + 5 Standard [ 0,] [ 0,6] Complete Graph [,] [,6 ] 7. Graphing and Graphing Utilities - - www.mastermathmentor.com - Illegal to post on Internet

d. Equation: y = ( 4) Standard [ 0,] [ 0,0] Complete Graph [,7 ] [ 0,5] e. Equation: y = 4 + 9 Standard [, ] [ 0,] Complete Graph [ 6,] [ 90,40] f. Equation: y = Standard [, ] [ 0,] Complete Graph [, ] [,9 ] g. Equation: y = cos 5 Standard [ 0,π ] [ 7, ] Complete Graph [ π,π] [ 7, ] 7. Graphing and Graphing Utilities - - www.mastermathmentor.com - Illegal to post on Internet

h. Equation: y = + Standard [ 0,4 ] [ 0,0] Complete Graph [ 4,6] [ 5,] i. Equation: y = ± 5 Window to fill the screen completely Window to make it to scale Standard [ 5,5] [ 5,5] [ 7.5,7.5] [ 5,5] B. Checking symmetry for a graph Symmetry means mirror image. There are three types of symmetry: -ais symmetry, y-ais symmetry, and origin symmetry. A curve that is symmetric to the -ais is the same on either side of the -ais. Anything in quadrant is reflected into quadrant IV and vice versa. Anything in quadrant II is reflected to quadrant III and vice versa. A curve that is symmetric to the y-ais is the same on either side of the y-ais. Anything in quadrant is reflected into quadrant II and vice versa. Anything in quadrant III is reflected to quadrant IV and vice versa. A curve that is symmetric to the origin is the same on either side of the origin. Anything in quadrant is reflected into quadrant III and vice versa. Anything in quadrant II is reflected to quadrant IV and vice versa. -ais symmetry y-ais symmetry origin symmetry 7. Graphing and Graphing Utilities - 4 - www.mastermathmentor.com - Illegal to post on Internet

Symmetry check for -ais symmetry: replacing every occurrence of y with y yields an equivalent equation. Symmetry check for y-ais symmetry: replacing every occurrence of with - yields an equivalent equation. Symmetry check for origin symmetry: replacing every occurrence of with AND y with -y yields an equivalent equation. If an equation is symmetric both to the -ais and y-ais, it is automatically symmetric to the origin. But if an equation is symmetric to the origin, it is not necessarily symmetric to either the or y ais.. Check the symmetry of the following and verify using the calculator. a. y = 5 b. y = 4 c. + y = 0 d. 4 y = 0 y ais symmetry No symmetry Origin symmetry ais symmetry e. y = f. y = 5 g. y = 5 h. y = + No symmetry y ais symmetry No symmetry No symmetry C. Lines in the Plane Students in Precalculus have seen lines in algebra and algebra so we will just review the material. Important formulas: The slope m of the non-vertical line passing through the points ( y ) and ( y ) is m = y y = Rise Run Slope intercept form of a line: y = m + b where m is the slope and b is the y-intercept Point slope form of a line: y y = m( ) where m is the slope and ( y ) is a point on the line Horizontal lines are in the form: y = a. Vertical lines are in the form = a. The general form of a line is A + By + C = 0 where A, B, and C are integers. Intercept form: If the line passes through ( a,0) and ( 0,b), the equation is: a + y b = Two lines are parallel if and only if their slopes are equal: m = m. Two lines are perpendicular if and only if their slopes are negative reciprocals of each other: m = m 7. Graphing and Graphing Utilities - 5 - www.mastermathmentor.com - Illegal to post on Internet

. Find the slope of the line passing through the pair of points: a. ( 4,) and (, 7) b. ( 0, 7) and (, 7) c. ( 4, ) and ( 4, ) d. ( ) and ( 4, ) m = 4 m = 0 No slope m = 40. Find the slope and y-intercept of the given line. a. y = 0 b. 5 + y = 0 c. y = 0 y = y = 5 + 4 y 6 = 0 m =,b = m = 5,b = m = 4,b =. Find the equation of the line with the given slope passing through the given point. Verify on the calculator. a. m =, ( 4, ) b. m = 5, (,6) c. m =, 4, d. m =, 4, y += ( 4) y 6 = 5( + y + ) = ( 4 ) y = + 4 y = 9 y = 5 4 y = 5 4. Find the equation of the line passing through the given points. Verify on the calculator a. (,4), ( 4, ) b. (,), (,9 ) c. (,4), 7, y 4 = 5( ) y = 5 +9 y = ( + ) y = + 7 5. Find the equation in general form given the intercepts: a. -intercept 4,0 y 4 = 7 8 ( ) y = 7 8 + 5 8 ( ) d., 4 y = + 6, ( 4 ), (, 5 6 ) y + 4 = 9 4 y = 9 4 + 84 ( ) and y-intercept ( 0, ) b. -intercept (,0 ) and y-intercept ( 0, 4 ) 4 + y = y 4 = 0 + y = 9 8y + 6 = 0 4 6. Find the equation of the line through the given point a) parallel to the given line and b) perpendicular to the given line. Verify by calculator. a. Point : (,), line : y = b. Point : (,5), line : 4y = c. Point :, 4 Parallel: y = ( ) y = Parallel: y 5 = ( + ) y = + 6 Parallel: ( ), line : = y y + 4 = y = Perpendicular: y = ( ) y = + Perpendicular: y 5 = ( + ) y = + Perpendicular: y + 4 = y = + 4 7. Graphing and Graphing Utilities - 6 - www.mastermathmentor.com - Illegal to post on Internet

D. Functions Functions are the mainstay of all mathematics and it is imperative you understand its definition. relation a set of ordered pairs, y domain allowable values of in the relation range allow values of y in the relation function A relation in which for every in the domain, there is one and only one y in the range (no repeating value) vertical line test If any vertical line intersects the relation in more than one point, the relation is not a function. function notation y = f ( ) even function when f ( ) = f ( ) (symmetric to y-ais) odd function when f ( ) ( ) = f ( ) (symmetric to origin) mapping a pictorial way of illustrating an equation E. a b c 4 a b c 4 function! not a function Which of the following does not represent a function? d a. w y z b. w y z c. w y z d. w y z Which of the following does not represent a function? b { } { locker } b. { telephone number } { students in a school} { } { license plate number } d. { U.S citizen who works } { social security # } a. students in a school c. cars in a parking lot ( ) =, and we want to find the value of y when =, we simply ask for Function notation: If y = f f ( ). It asks for the value of the function f when =. 7. Graphing and Graphing Utilities - 7 - www.mastermathmentor.com - Illegal to post on Internet

Using the calculator to find values of functions at specific values of a) Graphing a curve: e: f ( ) = 4 (use Y = ) b) find f (.5) - press nd CALC a) b) b) alternate method of finding f.5 ( ).- press VARS c) using a table to find f(many values) - press nd TBLSET, then nd TABLE ( ) = 4 + 5, find a) f ( 4) b) f ( 5) c) f ( a) d) f ( + ). If f. g 5 50 4a 8a + 5 + ( ) =, find + ( ) b) g ( π) c) g ( ) d) g ( ) a) g 7 E. Domain and Range π π π + + + 5 domain range allowable values of in the function. Assume the domain of a function is all real numbers with the following eceptions: - any value which creates a zero in the denominator - any value which creates a square root (or even root of a negative number. Any graphs with holes, vertical asymptotes, or gaps restrict the domain. allowable values of y in the function. Assume the range of a function is all real numbers with the following eceptions: We usually are more interested in the domain than the range. - squares always create positive numbers. - square (or even) roots are always positive. - absolute values always create positive numbers 7. Graphing and Graphing Utilities - 8 - www.mastermathmentor.com - Illegal to post on Internet

When analyzing domain and range, we can eamine the functions from a graphical point of view or an algebraic point of view.. Domain (, ) Range(, ). Domain (, ) Range(,). Domain (, ) Range(,) 4. Domain ( 5, ) Range(, )) 5. Domain (, ) Range( 0, ) 6. Domain (, ) Range(,0) We can also analyze domain from an algebraic point of view. The important thing to remember is that we assume the domain is all real numbers unless there is something to restrict it. Those restrictions usually fall into the categories of fractions with a variable in the denominator or the even roots of epressions. Find the domain of the following: 7. y = 4 + 8. f ( ) = 5 9. f ( ) = 5 (. ) ±5 7. Graphing and Graphing Utilities - 9 - www.mastermathmentor.com - Illegal to post on Internet

0. g ( ) = 5 0. h( ) = + 5. f ( ) = + 5 (. ) 5, 0. f ( ) = 0 4. f ( ) = + + 5. f ( ) = + 7 49 (. ), 7, 7 F. Piecewise Functions and Restricting the Domain Before we eamine how to restrict a domain on the calculator, we have to understand the concept of Boolean operators. First, on your calculator, find the TEST feature. To get to this screen, press, nd MATH. You should get the first screen. This screen opens the door to restricting domain. CLEAR and type 5 > 4. To do this, type 5 nd MATH : > 4 ENTER. Your screen should look like the screen on the right. The way the calculator handles this statement is: true statements are valued as while false statements are valued as 0. Since 5 > 4, you end up with. Now try <. nd MATH 5 ENTER. Since < is a false statement, the calculator gives you a zero. What should this give you? 0-8 = /.5. Try it. Okay, now we are ready for a restricted domain function. Let s eamine y =, > 0. To put this into the calculator, go to your Y = screen and X/(X nd MATH :> 0). It looks like this: Go to TBLSET and set your table like this: Now go to TABLE. Let s interpret this. The calculator starts with = - so it evaluates Y=(-)/(->0). Since ->0 is false and therefore equal to zero, you are evaluating the statement -6/0 which is an error. Note that you will continue to get errors until =. When =, you are evaluating Y=()/(>0). Since >0 is a true statement and therefore equal to one, you are finding Y=/ =. Notice for being positive numbers, you are always dividing by. Now graph the function using ZOOM 6. Values that are errors are not graphed. So you end up with y =, > 0. +, So, now we are ready to try a piecewise function. Let s try the first problem: y =, > We need to put the function in as two separate functions, Y and Y. Put in the function like this and graph: 7. Graphing and Graphing Utilities - 0 - www.mastermathmentor.com - Illegal to post on Internet

Note that you needed to enclose your epressions in parentheses. Note that this curve is not continuous because there is a clear break at the value of =. Graph these:, <. y = 4,. y =, 0, > 0 ( ) Range (-,4 ] Domain (, ) Range (, ] Domain, Continuous YES NO Continuous YES NO + 7, < To put in a function like this: y =, < 5, Y are straightforward., you need three functions. Y and To put in Y with its compound statement, you may not do this: While this does not generate an error, it is not what you want. Do this: You can find the operator and in the nd MATH LOGIC menu. This is the graph of that piece. So the entire function looks like this and the graph to the right. The problem of continuity needs to be done using calculus limit techniques. Do not trust your eyes using the calculator. When graphs are nearly vertical, the calculator has trouble graphing it. ZOOMING in may or may not confirm the continuity of this function at =. 7. Graphing and Graphing Utilities - - www.mastermathmentor.com - Illegal to post on Internet

Finally, when TRACING, realize that there are three separate functions here and if you are tracing at an -value in which the curve is not defined, no y-value will show. Try these., 0. y = 0, < 0 +, < 4. f ( ) = 5, 5 5, 5 < 5 + 6, < 5 ( ) Range (-, ) Domain (, ) Range (-.5] Domain, Continuous YES NO Continuous YES NO 5. In problem # 4 above, find the following: a) f 7 ( ) b. f ( 5) c. f ( 5) d. f ( 8) 0 0 6. A new 0 week diet claims that it will take weight off a person according to the following formula. In the first weeks, it will take off 7 pounds. In the net 4 weeks, it will take off.5 pounds a week. In the net 4 weeks, it will take off a pound a week. Complete the Chart. Week 4 6 8 0 Pounds off.5 7 0 5 7 Write a piecewise function that describes the amount of weight taken off as a function of weeks and graph it over the length of the diet..5, y = 7 +.5( ), 6s +( 6), 6 Domain [ 0,0] Range [ 0,7] Continuous How long would it take to lose 4 pounds? 7 weeks 7. Graphing and Graphing Utilities - - www.mastermathmentor.com - Illegal to post on Internet

7. An injury lawyer charges according to the following: His partner charges this way: First hour or part cost $50 The net 9 hours cost $50/hour After that, the charge is $5 an hour $500 for the first 0 hours $00 an hour for anything over that Complete the Chart Hours 5 0 8 Lawyer 50 50 650,400,650,400 Lawyer 500 500 500 500 900,00 Write a piecewise function that describes both lawyers and graph them over 5 hours. Lawyer 50, y = 50 +50( ), < 0,400 +5( 0), 0 Lawyer 500, 0 y = 500 + 00( 0), >0 Both domains are (-, ), both ranges are ( 0, ) When is lawyer cheaper to hire? 4 hours and under or hours and above When is lawyer cheaper to hire? between 4 and hours G. Absolute Value Curves You have all learned the meaning of the absolute value. The absolute value of an epression makes the epression positive it strips of the negative sign. However, the true definition of an absolute value is merely a piecewise function: =, 0, < 0 or epression = epression,epression 0 epression,epression < 0 Absolute value curves of linear functions form the distinctive V-shaped curve. 7. Graphing and Graphing Utilities - - www.mastermathmentor.com - Illegal to post on Internet

So, when we graph y =, we are merely graphing, 0, < 0 If we graph y = we are looking at y =, 0 or y =, ( ), < 0, < If we graph y = +, we are looking at +, + 0 +, y = or y = ( +), +< 0, < For the following epressions, write the absolute value epressions as piecewise epressions, graph and verify on the calculator.. y = +. y = 0. y = 5 + 4. f ( ) = +, y =, < H. Translations of Graphs Given y = f ( ),a > 0 f f f a f + a a f f 0, 5 y = 0, < 5 5 +, 5 y = 5, < 5, y =, < ( ) + a vertical shift up - translates the graph a units upwards ( ) a vertical shift down - translates the graph a units downwards ( ) horizontal shift right - translates the graph a units to the right ( ) horizontal shift left - translates the graph a units to the left ( ) Vertical stretch - if a >, the graph is narrower, if a <, it is wider ( ) Reflection - flips the graph across the -ais ( ) Anything below the -ais is reflected across the -ais. f y = y = + y = y = ( ) y = ( + ) y = y =.5 y = 7. Graphing and Graphing Utilities - 4 - www.mastermathmentor.com - Illegal to post on Internet

For the following problems, draw a transformation of the given graph.. f ( ) f ( ) + f ( ) f ( ) f ( +) f ( ) f ( ) f ( ) f ( ) + f + ( ) f ( ) f ( ) f ( ) f ( ) + f ( + ). f ( ) f ( ) f ( ) +.5 f ( 0.5) f ( +.5) f ( ) f ( ) f ( ) f + ( ) f + ( ) + f ( ) f ( ) f ( ) f ( ) + f ( + ) 7. Graphing and Graphing Utilities - 5 - www.mastermathmentor.com - Illegal to post on Internet

. f ( ) f ( ) + 4 f ( ) f ( 4) f ( + ) f ( ) f ( ) f ( ) f + ( ) f + ( ) + f ( ) f ( ) + 4 f ( ) f ( +) f ( + ) + I. Composition of Functions The concept of composition of functions is to start with a number or variable, apply function one to that variable, and take the answer and apply that answer to function two. It can be shown pictorially. Composition of functions f o g = f [ g( ) ] f o g g ( ) f [ g( ) ] g f A simple eample of composition of functions is when you go to the cafeteria and select a meal: Function g is ordered pairs in the form of (student, meal). Function f is in the form of (meal, cost). So f g [ ( )] will be in the form of (student, cost). Every student who purchases a meal has a cost associated with that meal. The cashier tells the student the cost the meal itself defined the cost, but in reality is no longer important. That is what you are doing when you find a function composition.. f ( ) = +, g ( ) =, find the following: [ ( )] b) g f ( ) a) f g ( ) = 8 ( ) = 79 g f 8 [ ] c) f [ f ( ) ] d) g [ g ( ) ] ( ) =4 ( ) = 4 f g 4 ( ) = 4 ( ) = f f 4 ( ) = 8 ( ) = g g 8 7. Graphing and Graphing Utilities - 6 - www.mastermathmentor.com - Illegal to post on Internet

finding composite functions: f ( ) = + and g( ) =, find f ( g( ) ) and g( f ( ) enter graphs into Y and Y VARS Y - VARS : Function : Y ( VARS Y - VARS : Function : Y (. f ( ) = + 4, g ( ) = +, find the following: ( )( ) b) ( g o f )( ) c) f g ( ) a) f o g ( ) = ( ) =0 g f. f ( ) = 5 +, g ( ) = + a) g f ( ) f ( ) =9 g( 4) = 0, find the following: + [ ] d) g [ f ( ) ] f ( +) + 4 + g( + 4 ) + 4 [ ] b) f [ g ( ) ] c) f [ g ( ) ] d) g [ f ( ) ] f ( ) = 8 g( 8) = 7 ( ) = ( ) = 7 g f + f + 5 + + = + 9 + + g( 5 + ) 5 + + ( 5 + ) + = 5 + 0 + 7 4. If f ( ) = and g( ) =, which of the following is largest at =? a) f g b) g f c) f o g d) g o f J. Inverse Functions 8 4 4 The concept of an inverse is probably the most misunderstood concept in mathematics. Try this. To your knowledge, write the inverse of the following epressions: 5 Boy Dog Hot dog... Write what you believe to be the definition of an inverse: 7. Graphing and Graphing Utilities - 7 - www.mastermathmentor.com - Illegal to post on Internet

[ ( )] = the inverse function undoes function f it is the mirror image of f ( ) across the line y =. to find f, switch and y and solve for y do not confuse inverse with reciprocal. Nouns ( epressions) have reciprocals while verbs (functions) have inverses. inverse the function f such that f f the reciprocal to 5 is 5 while the inverse to y = 5 is y = 5. determining if the inverse is a function use vertical line test on the inverse or (easier), use the horizontal line test on the function itself. You are given f ( ) as a set of ordered pairs. Find f ( ) and determine if f ( ) is a function: a) f [( ), (,6), (, ), ( 0,4) ] b) f ( ) =, [( ), ( 4,), ( 5,4), ( 0,0) ] c) f ( ) =,4 [( ), (,5), (,6) ] ( ) =,4 YES NO YES. Find which of the following mappings is a function and whose inverse is also a function. a. y z b. y z c. y z d. y z Not a function Both Function, Inverse not Not a function. For the following functions f ( ), find the inverse f ( ) and determine if f ( ) is also a function. a. y = b. y = 6 c. y = + Inv : = y Inv : = y Inv : = 6y y + y = + inverse a function y = 6 inverse a function y = inverse a function d. y = e. y = + f. y = Inv : = y y = ± + inverse not a function Inv : = y + y = + inverse a function Inv: = y y y inverse not a function as the function fails vert. line test 7. Graphing and Graphing Utilities - 8 - www.mastermathmentor.com - Illegal to post on Internet

Unit 7 Graphs and Graphing Utilities - Homework A. Graph the equations given in class according to the prescribed windows. a. Equation: y = 7 Standard [, ] [ 0,0] Complete Graph [ 0,0] [ 0,0] b. Equation: y = 4 + 5 Standard [ 0,] [ 0,7] Complete Graph [, ] [ 0,7] c. Equation: y =5 5 Standard [ 5,5] [ 0,40] Complete Graph [ 0,6] [ 0,40] d. Equation: y = 5 + 4 + Standard [ 0,0] [ 0,0] Complete Graph [ 0,0] [ 5,0] e. Equation: y = 4 + 6 7. Graphing and Graphing Utilities - 9 - www.mastermathmentor.com - Illegal to post on Internet

Standard [ 0,5] [ 0,40] Complete Graph [ 5,5] [ 0,0] f. y = 6sin cos Standard [ 0,π ] [ 0,0] Complete Graph [ π,π] [ 6,6] ( ) ( 5 + ) g. Equation: y = 0 Standard [ 0, ] [.,.5] Complete Graph [., ] [.,.5] i. Equation: y = ± 7 Window to make it fill screen Window to make it to scale Standard [ 4,] [, ] [ 5,4 ] [, ] 7. Graphing and Graphing Utilities - 0 - www.mastermathmentor.com - Illegal to post on Internet

B. Check the symmetry of the following and verify using the calculator. a. y = + 7 b. + y = 6 c. y = d. 4 y = 0 No symmetry y ais symmetry origin symmetry ais symmetry e. 4y = f. y = 5 g. = y h. y = 4 6 No symmetry No symmetry ais symmetry y ais symmetry C. Find the slope of the line passing through the pair of points: a. ( 4, ) and ( 6,7) b. (, ) and (, ) c. ( 5, ) and 5, ( ) d. (, 5 ) and (, ) m = 4 m = 0 No slope m = 40 C. Find the slope and y-intercept of the given line. a. + y 4 = 0 b. 7 y + 5 = 0 c. 4 y + = 0 y = = 4 m =,b = 4 y = 7 + 5 m = 7,b = 5 y = 6 + 8 m = 6,b = 8 C. Find the equation of the line with the given slope passing through the given point. Verify on the calculator. a. m =, (, ) b. m = 0, (, 5 ) c. m =, ( 6, ) d. m = 4,, 8 y + = ( y + = ) 6 y + ( ) 8 = 4 + y = 5 y = + y = 8y + = 6 + + y = 4 C4. Find the equation of the line passing through the given points. Verify on the calculator a. ( 5,), (, 4 ) b. (, ), (,7 ) c. (, ),, ( ) d. ( ), 6,, ( ) y =( 5) y += ( ) y = ( 8 + ) y + = 5 y = y = + 5 y = 8 + 8 y = 5 7. Graphing and Graphing Utilities - - www.mastermathmentor.com - Illegal to post on Internet

C5. Find the equation in general form given the intercepts: a. -intercept ( 8,0) and y-intercept ( 0, 6) b. -intercept (,0 ) and y-intercept ( 0, ) 8 + y 6 = 4y 4 = 0 + y = 4 + y = 0 C6. Find the equation of the line through the given point a) parallel to the given line and b) perpendicular to the given line. Verify by calculator. a. Point : (, ), line : y = + b. Point : ( 4,), line : + y = c. Point : (, ), line : = y + Parallel: y + = ( ) y = 5 Parallel: y = + 4 ( ) y = 4 Parallel: y + = ( ) y = y + = ( ) y = ( + 4 ) y + = ( ) Perpendicular: y = 5 Perpendicular: y = + 4 Perpendicular: y = + 7 D. If f ( ) = +, find the following. Verify using the calculator when possible. a) f ( ) b) f ( ) f ( ) c) f ( ) d) f ( ) 5 0 8 4 + 4 + 6 4 D. If g ( ) = +0 ( 0), find the following. Verify using the calculator when possible. a) g ( 6) b) g ( 9) c) g ( 5 ) d) g ( ) 8 0 5 + 5 + 9 + E. For the following problems, find the domain and range by inspection: a. Domain (, ) Range(, ) 7. Graphing and Graphing Utilities - - www.mastermathmentor.com - Illegal to post on Internet

b. Domain (, ) Range(, ) c. Domain (, ) Range(, ) d. Domain (, ) Range(, ) e. Domain (, ) Range( 60, ) f. Domain ( 6,6) Range( 6,6) g. Domain : Range : y = h. Domain (, ) Range(,) i. Domain (, ) Range : all integers 7. Graphing and Graphing Utilities - - www.mastermathmentor.com - Illegal to post on Internet

j. Domain (, ) Range (, ) E. Find the domain of the following: c. f ( ) = + 4 00 (, ) 4 ±0 a. y = + 9 b. f ( ) = + f. g ( ) = 0 4 g. h( ) = 4 + 50 h. f ( ) = +6 5 (, ) 8, 5 i. f ( ) = j. f ( ) = 8 4 k. f ( ) = 6 (, ) 0,± 6, 6 F. Graph the following: a. f ( ) =, 0, < 0 b. f ( ) =, 5, < ( ) Range [, ) Domain (, ) Range [, ) Domain Domain, Continuous: YES NO Continuous YES NO c. g ( ) = 4, 5 +, < d. ( ) = h sin, 0 4 4 cos, > 0 ( ) Range (,4 ] Domain (, ) Range [,4] Domain, Continuous: YES NO Continuous YES NO 7. Graphing and Graphing Utilities - 4 - www.mastermathmentor.com - Illegal to post on Internet

4, e. f ( ) =, < f., < ( ) = h 4 +, 4 5,0 < 4 + 5, < 0 Domain (, ) Range (, ], Domain, Continuous: YES NO Continuous YES NO ( ) Range (, ) g. In problem f above, find the following: ) h( 5) = 6 ) h( ) =. h( ) = 4. h ( ) = 46 5. [ h( ) ] = 4 F. You have a choice of phone plans. Plan : Plan : Pay $5 a month Pay 0 cents a minute for up to 0 minutes Pay 0 cents a minute up to 0 minutes Pay 5 cents a minute for all minutes over 0 Pay 8 cents a minute for all minutes over 0 Write piecewise functions for plan f ( ) = f Complete the chart: 5 +., 0 8 +.08( 0), > 0 ( ) and plan g ( ) ( ) = g., 0 4 +.5( 0), >0 Minutes 0 0 0 0 60 00 0 80 00 Plan 5 6 7 8 0.40.60 5.0 40 49.60 Plan 0 4 6 0 4 5 What is your recommendation for the best (cheapest) plan and why? Plan is better for minutes less than 80, and then plan is better. G. Write the following absolute value epressions as piecewise epressions and graph to verify.. y =.5. y = + 9. y = 6 4 4. f ( ) = 8 5.5,.5 y =.5, <.5 + 9, 4.5 y = 9, < 4.5 6 4, y = 4 6, < 8 + 5, 5 8 y = 8 5, < 5 8 7. Graphing and Graphing Utilities - 5 - www.mastermathmentor.com - Illegal to post on Internet

H. For the following problems, draw the transformations on the given graphs.. f ( ) f ( ) f ( ) +.5 f ( 0.5) f ( +.5) f ( ) f ( ) f ( ) f + ( ) f + ( ) + f ( ) f ( ) f ( ) f ( ) + f ( + ). f ( ) f ( ) + f ( ) f ( 4) f ( + ) f ( ) f ( ) f ( ) f + ( ) f + ( ) + f ( ) f ( ) + f ( ) f ( 4) f ( + ) + 7. Graphing and Graphing Utilities - 6 - www.mastermathmentor.com - Illegal to post on Internet

I. Given f a. f g ( ) =, g ( ) =, find the following: ( ) ( ) [ ] b. g [ f ] c. f [ f ( ) ] d. g [ g ( ) ] ( ) = f ( ) = f ( ) = g( ) =4 f ( ) = f ( ) = 7 g( ) = 5 g( 5) = 44 g I. Given f a. f g g ( ) = cos, g ( ) =, find the following: ( ) ( ) [ ] b. g [ f π ] c. f [ g ( ) ] d. g [ f ( ) ] ( ) = 0 ( ) = f 0 ( ) = ( ) = f π g ( ) ( ) f cos ( ) g cos cos I. Given f ( ) = sin, g( ) =, find the following: a. f [ g( 4π )] b. g f ( ) = π g 4π ( ) = 0 f π π 6 π f = 6 4 g = 4 [ ] d. g [ f ( ) ] c. f g ( ) f ( ) sin g( sin ) sin I4. If f ( ) is shown by the graph to the right. with each unit equal to, which is the smallest? a. f f 0 ( ( )) b. f ( f ( ) ) c. f ( f ( ) ) d. f ( f ( 4 )) 4 0 4 4 I5. For which pairs of functions does f g ( ( ))? a. f ( ) =, g( ) = b. f ( ) =, g( ) =, 0 c. f ( ) = 4 +, g ( ) = d. none of them 4 J. For the following functions, find the inverse and solve for y if possible. Determine if the inverse is also a function a. y = 6 b. y = 8 + 4 c. y = 5 Inv : = 6y Inv : = 8y + 4 Inv : = y 5 y = y = 4 y = ± + 5 6 8 Inverse not a function Inverse a function Inverse a function d. y = + e. y = + 4 + 4 f. y =, > 0 Inv : = y + Inv : = y + 4 y + 4 Inv : = y y = Inverse a function = ( y + ) y = ± Inverse not a function y = Inverse a function 7. Graphing and Graphing Utilities - 7 - www.mastermathmentor.com - Illegal to post on Internet

g. y = Inv : = + y y + + y = or Inverse a function h. y = 4 Inv : = y 4 y = + 4 Inverse a function i. y = Inv : = y y y = ± Inverse not a function J. Which of the following functions has its inverse also being a function? a. b. c. d. J. Which of the following graphs is the inverse to the graph at the right? a. b. c. d. J4. If f ( ) = + and g( ) =, which is the smallest value at =? [ ( )] b. g f ( ) a. f g [ ] c. f o g ( ) ( ) d. g o f ( ) = 0 ( ) = g f 0 ( ) = 7 ( ) = 4 f g 7 f [ g( ) ] = 5 inv : = y 5 y = + 5 y ( ) = 4 g[ f ( ) ] = inv : = y y = + y( ) =.5 7. Graphing and Graphing Utilities - 8 - www.mastermathmentor.com - Illegal to post on Internet