Linear and Quadratic Functions

Transcription

1 Chapter Linear and Quadratic Functions. Linear Functions We now begin the stud of families of functions. Our first famil, linear functions, are old friends as we shall soon see. Recall from Geometr that two distinct points in the plane determine a unique line containing those points, as indicated below. P ( 0, 0 ) Q (, ) To give a sense of the steepness of the line, we recall that we can compute the slope of the line using the formula below. Equation.. The slope m of the line containing the points P ( 0, 0 )andq (, )is: provided 0. m = 0 0, A couple of notes about Equation. are in order. First, don t ask wh we use the letter m to represent slope. There are man eplanations out there, but apparentl no one reall knows for sure. Secondl, the stipulation 0 ensures that we aren t tring to divide b zero. The reader is invited to pause to think about what is happening geometricall; the anious reader can skip along to the net eample. Eample... Find the slope of the line containing the following pairs of points, if it eists. Plot each pair of points and the line containing them. See or for discussions on this topic.

2 5 Linear and Quadratic Functions. P (0, 0), Q(, ). P (, ), Q(, ). P (, ), Q(, ). P (, ), Q(, ) 5. P (, ), Q(, ) 6. P (, ), Q(., ) Solution. In each of these eamples, we appl the slope formula, Equation... m = 0 0 = = Q P. m = ( ) = = P Q P. m = ( ) = 6 = Q. m = ( ) = 0 7 =0 P Q

3 . Linear Functions 5 P 5. m = =, which is undefined 0 Q P 6. m =. = 0. = 0 Q A few comments about Eample.. are in order. First, for reasons which will be made clear soon, if the slope is positive then the resulting line is said to be increasing. If it is negative, we sa the line is decreasing. A slope of 0 results in a horizontal line which we sa is constant, and an undefined slope results in a vertical line. Second, the larger the slope is in absolute value, the steeper the line. You ma recall from Intermediate Algebra that slope can be described as the ratio rise run. For eample, in the second part of Eample.., wefoundtheslopetobe.wecan interpret this as a rise of unit upward for ever units to the right we travel along the line, as shown below. over up Some authors use the unfortunate moniker no slope when a slope is undefined. It s eas to confuse the notions of no slope with slope of 0. For this reason, we will describe slopes of vertical lines as undefined.

4 5 Linear and Quadratic Functions Using more formal notation, given points ( 0, 0 )and(, ), we use the Greek letter delta Δ to write Δ = 0 and Δ = 0. In most scientific circles, the smbol Δ means change in. Hence, we ma write m = Δ Δ, which describes the slope as the rate of change of with respect to. Rates of change abound in the real world, as the net eample illustrates. Eample... Suppose that two separate temperature readings were taken at the ranger station on the top of Mt. Sasquatch: at 6 AM the temperature was F and at 0 AM it was F.. Find the slope of the line containing the points (6, ) and (0, ).. Interpret our answer to the first part in terms of temperature and time.. Predict the temperature at noon. Solution.. For the slope, we have m = 0 6 = 8 =.. Since the values in the numerator correspond to the temperatures in F, and the values in the denominator correspond to time in hours, we can interpret the slope as = = F hour, or F per hour. Since the slope is positive, we know this corresponds to an increasing line. Hence, the temperature is increasing at a rate of F per hour.. Noon is two hours after 0 AM. Assuming a temperature increase of F per hour, in two hours the temperature should rise F. Since the temperature at 0 AM is F, we would epect the temperature at noon to be + = 6 F. Now it ma well happen that in the previous scenario, at noon the temperature is onl F. This doesn t mean our calculations are incorrect, rather, it means that the temperature change throughout the da isn t a constant F per hour. As discussed in Section.., mathematical models are just that: models. The predictions we get out of the models ma be mathematicall accurate, but ma not resemble what happens in the real world. In Section., we discussed the equations of vertical and horizontal lines. Using the concept of slope, we can develop equations for the other varieties of lines. Suppose a line has a slope of m and contains the point ( 0, 0 ). Suppose (, ) is another point on the line, as indicated below. (, ) ( 0, 0 )

5 . Linear Functions 55 Equation. ields m = 0 0 m ( 0 ) = 0 0 = m ( 0 ) We have just derived the point-slope form of a line. Equation.. The point-slope form of the line with slope m containing the point ( 0, 0 )is the equation 0 = m ( 0 ). Eample... Write the equation of the line containing the points (, ) and (, ). Solution. In order to use Equation. we need to find the slope of the line in question so we use Equation. to get m = Δ Δ = ( ) =. We are spoiled for choice for a point ( 0, 0 ). We ll use (, ) and leave it to the reader to check that using (, ) results in the same equation. Substituting into the point-slope form of the line, we get 0 = m ( 0 ) = ( ( )) = ( +) = = + 7. We can check our answer b showing that both (, ) and (, ) are on the graph of = + 7 algebraicall, as we did in Section... In simplifing the equation of the line in the previous eample, we produced another form of a line, the slope-intercept form. This is the familiar = m + b form ou have probabl seen in Intermediate Algebra. The intercept in slope-intercept comes from the fact that if we set =0, we get = b. In other words, the -intercept of the line = m + b is (0,b). Equation.. The slope-intercept form of the line with slope m and -intercept (0,b)isthe equation = m + b. Notethatifwehaveslopem =0,wegettheequation = b which matches our formula for a horizontal line given in Section.. The formula given in Equation. can be used to describe all lines ecept vertical lines. All lines ecept vertical lines are functions (Wh is this?) so we have finall reached a good point to introduce linear functions. We can also understand this equation in terms of appling transformations to the function I() =. See the Eercises.

6 56 Linear and Quadratic Functions Definition.. A linear function is a function of the form f() =m + b, where m and b are real numbers with m 0. The domain of a linear function is (, ). For the case m =0,wegetf() =b. These are given their own classification. Definition.. A constant function is a function of the form f() =b, where b is real number. The domain of a constant function is (, ). Recall that to graph a function, f, we graph the equation = f(). Hence, the graph of a linear function is a line with slope m and -intercept (0,b); the graph of a constant function is a horizontal line (a line with slope m =0)anda-intercept of (0,b). Now think back to Section.6., specificall Definition.0 concerning increasing, decreasing and constant functions. A line with positive slope was called an increasing line because a linear function with m>0 is an increasing function. Similarl, a line with a negative slope was called a decreasing line because a linear function with m<0 is a decreasing function. And horizontal lines were called constant because, well, we hope ou ve alread made the connection. Eample... Graph the following functions. Identif the slope and -intercept.. f() =. f() =. f() =. f() = Solution.. To graph f() =, we graph =. This is a horizontal line (m = 0) through (0, ).. The graph of f() = is the graph of the line =. Comparison of this equation with Equation. ields m = and b =. Hence, our slope is and our -intercept is (0, ). To get another point on the line, we can plot (,f()) = (, ).

7 . Linear Functions 57 f() = f() =. At first glance, the function f() = does not fit the form in Definition. but after some rearranging we get f() = = = +.Weidentifm = and b =. Hence, our graph is a line with a slope of and a -intercept of ( 0, ). Plotting an additional point, we can choose (,f()) to get (, ).. If we simplif the epression for f, weget f() = = ( )( +) ( ) = +. If we were to state f() = +, we would be committing a sin of omission. Remember, to find the domain of a function, we do so before we simplif! In this case, f has big problems when =, and as such, the domain of f is (, ) (, ). To indicate this, we write f() = +,. So, ecept at =,wegraphtheline = +. The slope m = and the -intercept is (0, ). A second point on the graph is (,f()) = (, ). Since our function f is not defined at =, we put an open circle at the point that would be on the line = +when =,namel(, ). f() = f() =

8 58 Linear and Quadratic Functions The last two functions in the previous eample showcase some of the difficult in defining a linear function using the phrase of the form as in Definition., since some algebraic manipulations ma be needed to rewrite a given function to match the form. Keep in mind that the domains of linear and constant functions are all real numbers (, ), so while f() = simplified to a formula f() = +, f is not considered a linear function since its domain ecludes =. However, we would consider f() = + + to be a constant function since its domain is all real numbers (Can ou tell us wh?) and f() = + + = ( + ) ( + ) = The following eample uses linear functions to model some basic economic relationships. Eample..5. The cost C, indollars,toproduce PortaBo game sstems for a local retailer is given b C() = for 0.. Find and interpret C(0).. How man PortaBos can be produced for $5,000?. Eplain the significance of the restriction on the domain, 0.. Find and interpret C(0). 5. Find and interpret the slope of the graph of = C(). Solution.. To find C(0), we replace ever occurrence of with 0 in the formula for C() toget C(0) = 80(0) + 50 = 950. Since represents the number of PortaBos produced, and C() represents the cost, in dollars, C(0) = 950 means it costs $950 to produce 0 PortaBos for the local retailer.. To find how man PortaBos can be produced for $5,000, we solve C() = 5000, or = Solving, we get = = Since we can onl produce a whole number amount of PortaBos, we can produce 85 PortaBos for $5,000.. The restriction 0 is the applied domain, as discussed in Section... In this contet, represents the number of PortaBos produced. It makes no sense to produce a negative quantit of game sstems. 5 The similarit of this name to PortaJohn is deliberate. 5 Actuall, it makes no sense to produce a fractional part of a game sstem, either, as we saw in the previous part of this eample. This absurdit, however, seems quite forgivable in some tetbooks but not to us.

9 . Linear Functions 59. We find C(0) = 80(0) + 50 = 50. This means it costs $50 to produce 0 PortaBos. As mentioned on page 8, this is the fied, or start-up cost of this venture. 5. If we were to graph = C(), we would be graphing the portion of the line = for 0. We recognize the slope, m = 80. Like an slope, we can interpret this as a rate of change. Here, C() is the cost in dollars, while measures the number of PortaBos so m = Δ Δ = ΔC Δ =80=80 = $80 PortaBo. In other words, the cost is increasing at a rate of $80 per PortaBo produced. This is often called the variable cost for this venture. The net eample asks us to find a linear function to model a related economic problem. Eample..6. The local retailer in Eample..5 has determined that the number of PortaBo game sstems sold in a week is related to the price p in dollars of each sstem. When the price was $0, 0 game sstems were sold in a week. When the sstems went on sale the following week, 0 sstems were sold at $90 a piece.. Find a linear function which fits this data. Use the weekl sales as the independent variable and the price p as the dependent variable.. Find a suitable applied domain.. Interpret the slope.. If the retailer wants to sell 50 PortaBos net week, what should the price be? 5. What would the weekl sales be if the price were set at $50 per sstem? Solution.. We recall from Section. the meaning of independent and dependent variable. Since is to be the independent variable, and p the dependent variable, we treat as the input variable and p as the output variable. Hence, we are looking for a function of the form p() =m + b. To determine m and b, we use the fact that 0 PortaBos were sold during the week when the price was 0 dollars and 0 units were sold when the price was 90 dollars. Using function notation, these two facts can be translated as p(0) = 0 and p(0) = 90. Since m represents the rate of change of p with respect to, wehave m = Δp 90 0 = Δ 0 0 = 0 0 =.5. We now have determined p() =.5 + b. To determine b, we can use our given data again. Using p(0) = 0, we substitute =0intop() =.5 + b and set the result equal to 0:.5(0) + b = 0. Solving, we get b = 50. Hence, we get p() = We can check our formula b computing p(0) and p(0) to see if we get 0 and 90, respectivel. You ma recall from page 8 that the function p() is called the price-demand (or simpl demand) function for this venture.

10 60 Linear and Quadratic Functions. To determine the applied domain, we look at the phsical constraints of the problem. Certainl, we can t sell a negative number of PortaBos, so 0. However, we also note that the slope of this linear function is negative, and as such, the price is decreasing as more units are sold. Thus another constraint on the price is p() 0. Solving results in.5 50 or 500 = Since represents the number of PortaBos sold in a week, we round down to 66. As a result, a reasonable applied domain for p is [0, 66].. The slope m =.5, once again, represents the rate of change of the price of a sstem with respect to weekl sales of PortaBos. Since the slope is negative, we have that the price is decreasing at a rate of $.50 per PortaBo sold. (Said differentl, ou can sell one more PortaBo for ever $.50 drop in price.). To determine the price which will move 50 PortaBos, we find p(50) =.5(50)+50 = 5. That is, the price would have to be $5. 5. If the price of a PortaBo were set at $50, we have p() = 50, or,.5+50 = 50. Solving, we get.5 = 00 or =66.6. This means ou would be able to sell 66 PortaBos a week if the price were $50 per sstem. Not all real-world phenomena can be modeled using linear functions. Nevertheless, it is possible to use the concept of slope to help analze non-linear functions using the following. Definition.. Let f be a function defined on the interval [a, b]. The average rate of change of f over [a, b] is defined as: Δf f(b) f(a) = Δ b a Geometricall, if we have the graph of = f(), the average rate of change over [a, b] is the slope of the line which connects (a, f(a)) and (b, f(b)). This is called the secant line through these points. For that reason, some tetbooks use the notation m sec for the average rate of change of a function. Note that for a linear function m = m sec,orinotherwords,itsrateofchangeoveranintervalis the same as its average rate of change. = f() (a, f(a)) (b, f(b)) The graph of = f() and its secant line through (a, f(a)) and (b, f(b)) The interested reader ma question the adjective average in the phrase average rate of change. In the figure above, we can see that the function changes wildl on [a, b], et the slope of the secant line onl captures a snapshot of the action at a and b. This situation is entirel analogous to the

11 . Linear Functions 6 average speed on a trip. Suppose it takes ou hours to travel 00 miles. Your average speed is 00 miles hours = 50 miles per hour. However, it is entirel possible that at the start of our journe, ou traveled 5 miles per hour, then sped up to 65 miles per hour, and so forth. The average rate of change is akin to our average speed on the trip. Your speedometer measures our speed at an one instant along the trip, our instantaneous rate of change, and this is one of the central themes of Calculus. 6 When interpreting rates of change, we interpret them the same wa we did slopes. In the contet of functions, it ma be helpful to think of the average rate of change as: change in outputs change in inputs Eample..7. Recall from page 8, the revenue from selling units at a price p per unit is given b the formula R = p. Suppose we are in the scenario of Eamples..5 and..6.. Find and simplif an epression for the weekl revenue R() as a function of weekl sales.. Find and interpret the average rate of change of R() over the interval [0, 50].. Find and interpret the average rate of change of R() as changes from 50 to 00 and compare that to our result in part.. Find and interpret the average rate of change of weekl revenue as weekl sales increase from 00 PortaBos to 50 PortaBos. Solution.. Since R = p, we substitute p() = from Eample..6 to get R() = ( ) = Since we determined the price-demand function p() is restricted to 0 66, R() is restricted to these values of as well.. Using Definition., we get that the average rate of change is ΔR Δ = R(50) R(0) 50 0 = = 75. Interpreting this slope as we have in similar situations, we conclude that for ever additional PortaBo sold during a given week, the weekl revenue increases $75.. The wording of this part is slightl different than that in Definition., but its meaning is to find the average rate of change of R over the interval [50, 00]. To find this rate of change, we compute 6 Here we go again... ΔR Δ = R(00) R(50) = =5.

12 6 Linear and Quadratic Functions In other words, for each additional PortaBo sold, the revenue increases b $5. Note that while the revenue is still increasing b selling more game sstems, we aren t getting as much of an increase as we did in part of this eample. (Can ou think of wh this would happen?). Translating the English to the mathematics, we are being asked to find the average rate of change of R over the interval [00, 50]. We find ΔR Δ = R(50) R(00) = = 5. This means that we are losing $5 dollars of weekl revenue for each additional PortaBo sold. (Can ou think wh this is possible?) We close this section with a new look at difference quotients which were first introduced in Section.. If we wish to compute the average rate of change of a function f over the interval [, + h], then we would have Δf f( + h) f() f( + h) f() = = Δ ( + h) h As we have indicated, the rate of change of a function (average or otherwise) is of great importance in Calculus. 7 Also, we have the geometric interpretation of difference quotients which was promised to ou back on page 8 a difference quotient ields the slope of a secant line. 7 So we are not torturing ou with these for nothing.

13 . Linear Functions 6.. Eercises In Eercises - 0, find both the point-slope form and the slope-intercept form of the line with the given slope which passes through the given point.. m =, P(, ). m =, P( 5, 8). m =, P( 7, ). m =, P(, ) 5. m = 5, P(0, ) 6. m = 7, P(, ) 7. m =0, P(, 7) 8. m =, P(0, ) 9. m = 5, P(, ) 0. m = 678, P(, ) In Eercises - 0, find the slope-intercept form of the line which passes through the given points.. P (0, 0), Q(, 5). P (, ), Q(, ). P (5, 0), Q(0, 8). P (, 5), Q(7, ) 5. P (, 5), Q(7, 5) 6. P (, 8), Q(5, 8) 7. P (, ( ),Q 5, 7 ) 8. P (, 7 ) (,Q, ) 9. P (, ),Q (, ) 0. P (, ),Q (, ) In Eercises - 6, graph the function. Find the slope, -intercept and -intercept, if an eist.. f() =. f() =. f() =. f() =0 5. f() = + 6. f() = 7. Find all of the points on the line = + which are units from the point (, ). 8. Jeff can walk comfortabl at miles per hour. Find a linear function d that represents the total distance Jeff can walk in t hours, assuming he doesn t take an breaks. 9. Carl can stuff 6 envelopes per minute. Find a linear function E that represents the total number of envelopes Carl can stuff after t hours, assuming he doesn t take an breaks. 0. A landscaping compan charges $5 per cubic ard of mulch plus a deliver charge of $0. Find a linear function which computes the total cost C (in dollars) to deliver cubic ards of mulch.

14 6 Linear and Quadratic Functions. A plumber charges $50 for a service call plus $80 per hour. If she spends no longer than 8 hours a da at an one site, find a linear function that represents her total dail charges C (in dollars) as a function of time t (in hours) spent at an one given location.. A salesperson is paid $00 per week plus 5% commission on her weekl sales of dollars. Find a linear function that represents her total weekl pa, W (indollars)intermsof. What must her weekl sales be in order for her to earn $75.00 for the week?. An on-demand publisher charges $.50 to print a 600 page book and $5.50 to print a 00 page book. Find a linear function which models the cost of a book C as a function of the number of pages p. Interpret the slope of the linear function and find and interpret C(0).. The Topolog Tai Compan charges $.50 for the first fifth of a mile and $0.5 for each additional fifth of a mile. Find a linear function which models the tai fare F as a function ofthenumberofmilesdriven,m. Interpret the slope of the linear function and find and interpret F (0). 5. Water freezes at 0 Celsius and Fahrenheit and it boils at 00 C and F. (a) Find a linear function F that epresses temperature in the Fahrenheit scale in terms of degrees Celsius. Use this function to convert 0 C into Fahrenheit. (b) Find a linear function C that epresses temperature in the Celsius scale in terms of degrees Fahrenheit. Use this function to convert 0 F into Celsius. (c) Is there a temperature n such that F (n) =C(n)? 6. Legend has it that a bull Sasquatch in rut will howl approimatel 9 times per hour when it is 0 F outside and onl 5 times per hour if it s 70 F. Assuming that the number of howls per hour, N, can be represented b a linear function of temperature Fahrenheit, find the number of howls per hour he ll make when it s onl 0 F outside. What is the applied domain of this function? Wh? 7. Economic forces beond anone s control have changed the cost function for PortaBos to C() = Rework Eample..5 with this new cost function. 8. In response to the economic forces in Eercise 7 above, the local retailer sets the selling price of a PortaBo at $50. Remarkabl, 0 units were sold each week. When the sstems went on sale for $0, 0 units per week were sold. Rework Eamples..6 and..7 with this new data. What difficulties do ou encounter? 9. A local pizza store offers medium two-topping pizzas delivered for $6.00 per pizza plus a $.50 deliver charge per order. On weekends, the store runs a game da special: if si or more medium two-topping pizzas are ordered, the are $5.50 each with no deliver charge. Write a piecewise-defined linear function which calculates the cost C (in dollars) of p medium two-topping pizzas delivered during a weekend.

15 . Linear Functions A restaurant offers a buffet which costs $5 per person. For parties of 0 or more people, a group discount applies, and the cost is $.50 per person. Write a piecewise-defined linear function which calculates the total bill T of a part of n people who all choose the buffet.. A mobile plan charges a base monthl rate of $0 for the first 500 minutes of air time plus achargeof5 for each additional minute. Write a piecewise-defined linear function which calculates the monthl cost C (in dollars) for using m minutes of air time. HINT: You ma want to revisit Eercise 7 in Section.. The local pet shop charges per cricket up to 00 crickets, and 0 per cricket thereafter. Write a piecewise-defined linear function which calculates the price P, in dollars, of purchasing c crickets.. The cross-section of a swimming pool is below. Write a piecewise-defined linear function which describes the depth of the pool, D (in feet) as a function of: (a) the distance (in feet) from the edge of the shallow end of the pool, d. (b) the distance (in feet) from the edge of the deep end of the pool, s. (c) Graph each of the functions in (a) and (b). Discuss with our classmates how to transform one into the other and how the relate to the diagram of the pool. d ft. 7 ft. s ft. ft. 8ft. 0 ft. 5 ft. In Eercises - 9, compute the average rate of change of the function over the specified interval.. f() =, [, ] 5. f() =, [, 5] 6. f() =, [0, 6] 7. f() =, [, ] 8. f() = +, [5, 7] 9. f() = + 7, [, ]

16 66 Linear and Quadratic Functions In Eercises 50-5, compute the average rate of change of the given function over the interval [, + h]. Here we assume [, + h] is in the domain of the function. 50. f() = 5. f() = 5. f() = + 5. f() = The height of an object dropped from the roof of an eight stor building is modeled b: h(t) = 6t + 6, 0 t. Here, h is the height of the object off the ground in feet, t seconds after the object is dropped. Find and interpret the average rate of change of h over the interval [0, ]. 55. Using data from Bureau of Transportation Statistics, the average fuel econom F in miles per gallon for passenger cars in the US can be modeled b F (t) = t +0.5t + 6, 0 t 8, where t is the number of ears since 980. Find and interpret the average rate of change of F over the interval [0, 8]. 56. The temperature T in degrees Fahrenheit t hours after 6 AM is given b: T (t) = t +8t +, 0 t (a) Find and interpret T (), T (8) and T (). (b) Find and interpret the average rate of change of T over the interval [, 8]. (c) Find and interpret the average rate of change of T from t =8tot =. (d) Find and interpret the average rate of temperature change between 0 AM and 6 PM. 57. Suppose C() = represents the costs, in hundreds, to produce thousand pens. Find and interpret the average rate of change as production is increased from making 000 to 5000 pens. 58. With the help of our classmates find several other real-world eamples of rates of change that are used to describe non-linear phenomena. (Parallel Lines) Recall from Intermediate Algebra that parallel lines have the same slope. (Please note that two vertical lines are also parallel to one another even though the have an undefined slope.) In Eercises 59-6, ou are given a line and a point which is not on that line. Find the line parallel to the given line which passes through the given point. 59. = +,P(0, 0) 60. = 6 +5,P(, )

17 . Linear Functions = 7, P(6, 0) 6. =,P(, ) 6. =6,P(, ) 6. =,P( 5, 0) (Perpendicular Lines) Recall from Intermediate Algebra that two non-vertical lines are perpendicular if and onl if the have negative reciprocal slopes. That is to sa, if one line has slope m and the other has slope m then m m =. (You will be guided through a proof of this result in Eercise 7.) Please note that a horizontal line is perpendicular to a vertical line and vice versa, so we assume m 0 and m 0. In Eercises 65-70, ou are given a line and a point which is not on that line. Find the line perpendicular to the given line which passes through the given point. 65. = +,P(0, 0) 66. = 6 +5,P(, ) 67. = 7, P(6, 0) 68. =,P(, ) 69. =6,P(, ) 70. =,P( 5, 0) 7. We shall now prove that = m + b is perpendicular to = m + b if and onl if m m =. To make our lives easier we shall assume that m > 0andm < 0. We can also move the lines so that their point of intersection is the origin without messing things up, so we ll assume b = b =0. (Take a moment with our classmates to discuss wh this is oka.) Graphing the lines and plotting the points O(0, 0), P (,m ) and Q(,m )givesus the following set up. P O Q The line = m will be perpendicular to the line = m if and onl if OPQ is a right triangle. Let d be the distance from O to P,letd be the distance from O to Q and let d be the distance from P to Q. Use the Pthagorean Theorem to show that OPQ is a right triangle if and onl if m m = b showing d + d = d if and onl if m m =.

18 68 Linear and Quadratic Functions 7. Show that if a b, the line containing the points (a, b) and(b, a) is perpendicular to the line =. (Coupled with the result from Eample..7 on page, wehavenowshownthatthe line = is a perpendicular bisector of the line segment connecting (a, b) and(b, a). This means the points (a, b) and(b, a) are smmetric about the line =. We will revisit this smmetr in section 5..) 7. The function defined b I() = is called the Identit Function. (a) Discuss with our classmates wh this name makes sense. (b) Show that the point-slope form of a line (Equation.) can be obtained from I using a sequence of the transformations defined in Section.7.

19 . Linear Functions 69.. Answers. +=( ) = 0. += ( +7) = 8. 8= ( +5) =. = ( +) = = 5 ( 0) 6. = = ( +) = = 0 = 7 9. = 5( ) = = ( 0) = 0. + = 678( +) = = 5. =. = = = 5 6. = 8 7. = = = 0. =. f() = slope: m = -intercept: (0, ) -intercept: (, 0). f() = slope: m = -intercept: (0, ) -intercept: (, 0)

20 70 Linear and Quadratic Functions. f() = slope: m =0 -intercept: (0, ) -intercept: none. f() =0 slope: m =0 -intercept: (0, 0) -intercept: {(, 0) is a real number} 5. f() = + slope: m = -intercept: ( 0, ) -intercept: (, 0) 6. f() = slope: m = -intercept: ( 0, ) -intercept: (, 0) 7. (, ) and ( 5, 7 ) 5 8. d(t) =t, t E(t) = 60t, t C() =5 + 0, 0.. C(t) =80t + 50, 0 t 8.. W () = , 0 She must make $5500 in weekl sales.. C(p) =0.05p +.5 The slope 0.05 means it costs.5 per page. C(0) =.5 means there is a fied, or start-up, cost of $.50 to make each book.. F (m) =.5m +.05 The slope.5 means it costs an additional $.5 for each mile beond the first 0. miles. F (0) =.05, so according to the model, it would cost $.05 for a trip of 0 miles. Would this ever reall happen? Depends on the driver and the passenger, we suppose.

21 . Linear Functions 7 5. (a) F (C) = 9 5 C + (b) C(F )= 5 9 (F ) = 5 9 F 60 9 (c) F ( 0) = 0 = C( 0). 6. N(T )= 5 T + and N(0) = 5 howls per hour. Having a negative number of howls makes no sense and since N(07.5) = 0 we can put an upper bound of 07.5 F on the domain. The lower bound is trickier because there s nothing other than common sense to go on. As it gets colder, he howls more often. At some point it will either be so cold that he freezes to death or he s howling non-stop. So we re going to sa that he can withstand temperatures no lower than 60 F so that the applied domain is [ 60, 07.5]. { 9. C(p) = 6p +.5 if p 5 5.5p if p 6 { 5n if n 9 0. T (n) =.5n if n 0 { 0 if 0 m 500. C(m) = (m 500) if m>500 { 0.c if c 00. P (c) = + 0.(c 00) if c>00. (a) (b) (c) D(d) = D(s) = 8 if 0 d 5 d + if 5 d 7 if 7 d 7 if 0 s 0 s if 0 s 8 if s = D(d) 0 7 = D(s)

22 7 Linear and Quadratic Functions ( ) = ( ) = = h + h ( )( + h ) 5 5 = 5 ( ) ( ) =0 (() +() 7) (( ) +( ) 7) ( ) ( + h) h + 5. The average rate of change is h() h(0) 0 =. During the first two seconds after it is dropped, the object has fallen at an average rate of feet per second. (This is called the average velocit of the object.) F (8) F (0) 55. The average rate of change is 8 0 =0.7. During the ears from 980 to 008, the average fuel econom of passenger cars in the US increased, on average, at a rate of 0.7 miles per gallon per ear. 56. (a) T () = 56, so at 0 AM ( hours after 6 AM), it is 56 F. T (8) = 6, so at PM (8 hours after 6 AM), it is 6 F. T () = 56, so at 6 PM ( hours after 6 AM), it is 56 F. T (8) T () (b) The average rate of change is 8 =. Between 0 AM and PM, the temperature increases, on average, at a rate of F per hour. T () T (8) (c) The average rate of change is 8 =. Between PM and 6 PM, the temperature decreases, on average, at a rate of F per hour. T () T () (d) The average rate of change is = 0. Between 0 AM and 6 PM, the temperature, on average, remains constant. 57. The average rate of change is C(5) C() 5 =. As production is increased from 000 to 5000 pens, the cost decreases at an average rate of $00 per 000 pens produced (0 per pen.) 59. = 60. = = 6. = 6. = 6. = = 66. = = = 69. = 70. =0 =

23 . Absolute Value Functions 7. Absolute Value Functions There are a few was to describe what is meant b the absolute value of a real number. You ma have been taught that is the distance from the real number to 0 on the number line. So, for eample, 5 = 5 and 5 = 5, since each is 5 units from 0 on the number line. distance is 5 units distance is 5 units Another wa to define absolute value is b the equation =. Using this definition, we have 5 = (5) = 5 = 5 and 5 = ( 5) = 5 = 5. The long and short of both of these procedures is that takes negative real numbers and assigns them to their positive counterparts while it leaves positive numbers alone. This last description is the one we shall adopt, and is summarized in the following definition. Definition.. The absolute value of a real number, denoted, isgivenb {, if <0 =, if 0 In Definition., we define using a piecewise-defined function. (See page 6 in Section..) To check that this definition agrees with what we previousl understood as absolute value, note that since 5 0, to find 5 we use the rule =, so 5 = 5. Similarl, since 5 < 0, we use the rule =, sothat 5 = ( 5) = 5. This is one of the times when it s best to interpret the epression as theoppositeof as opposed to negative. Before we begin studing absolute value functions, we remind ourselves of the properties of absolute value. Theorem.. Properties of Absolute Value: Let a, b and be real numbers and let n be an integer. a Then ˆ Product Rule: ab = a b ˆ Power Rule: a n = a n whenever a n is defined ˆ Quotient Rule: a = a, provided b 0 b b Equalit Properties: ˆ = 0 if and onl if =0. ˆ For c>0, = c if and onl if = c or = c. ˆ For c<0, = c has no solution. a See page if ou don t remember what an integer is.

24 7 Linear and Quadratic Functions The proofs of the Product and Quotient Rules in Theorem. boil down to checking four cases: when both a and b are positive; when the are both negative; when one is positive and the other is negative; and when one or both are zero. For eample, suppose we wish to show that ab = a b. We need to show that this equation is true for all real numbers a and b. Ifa and b are both positive, then so is ab. Hence, a = a, b = b and ab = ab. Hence, the equation ab = a b is the same as ab = ab which is true. If both a and b are negative, then ab is positive. Hence, a = a, b = b and ab = ab. The equation ab = a b becomes ab =( a)( b), which is true. Suppose a is positive and b is negative. Then ab is negative, and we have ab = ab, a = a and b = b. The equation ab = a b reduces to ab = a( b) which is true. A smmetric argument shows the equation ab = a b holds when a is negative and b is positive. Finall, if either a or b (or both) are zero, then both sides of ab = a b are zero, so the equation holds in this case, too. All of this rhetoric has shown that the equation ab = a b holds true in all cases. The proof of the Quotient Rule is ver similar, with the eception that b 0. The Power Rule can be shown b repeated application of the Product Rule. The Equalit Properties can be proved using Definition. and b looking at the cases when 0, in which case =, orwhen<0, in which case =. For eample, if c>0, and = c, thenif 0, we have = = c. If, on the other hand, <0, then = = c, so = c. The remaining properties are proved similarl and are left for the Eercises. Our first eample reviews how to solve basic equations involving absolute value using the properties listed in Theorem.. Eample... Solve each of the following equations.. = =. + 5= =5 5. = = Solution.. The equation = 6 is of the form = c for c>0, so b the Equalit Properties, = 6 is equivalent to =6or = 6. Solving the former, we arrive at = 7, and solving the latter, we get = 5. We ma check both of these solutions b substituting them into the original equation and showing that the arithmetic works out.. To use the Equalit Properties to solve +5 =, we first isolate the absolute value. +5 = +5 = subtract +5 = divide b From the Equalit Properties, we have +5=or +5=, and get our solutions to be = or = 7. We leave it to the reader to check both answers in the original equation.

25 . Absolute Value Functions 75. As in the previous eample, we first isolate the absolute value in the equation + 5=0 and get + = 5. Using the Equalit Properties, we have += 5 or += 5. Solving the former gives = and solving the latter gives =. As usual, we ma substitute both answers in the original equation to check.. Upon isolating the absolute value in the equation 5 + =5,weget 5 + =. At this point, we know there cannot be an real solution, since, b definition, the absolute value of anthing is never negative. We are done. 5. The equation = 6 presents us with some difficult, since appears both inside and outside of the absolute value. Moreover, there are values of for which 6 is positive, negative and zero, so we cannot use the Equalit Properties without the risk of introducing etraneous solutions, or worse, losing solutions. For this reason, we break equations like this into cases b rewriting the term in absolute values,, using Definition.. For <0, =, sofor<0, the equation = 6 is equivalent to = 6. Rearranging this gives us + 6 = 0, or (+)( ) = 0. We get = or =. Since onl = satisfies <0, this is the answer we keep. For 0, =, so the equation = 6 becomes = 6. From this, we get 6=0or( )( + ) = 0. Our solutions are =or =, and since onl = satisfies 0, this is the one we keep. Hence, our two solutions to = 6are = and =. 6. To solve + =, wefirstisolatetheabsolutevalueandget =. Since we see both inside and outside of the absolute value, we break the equation into cases. The term with absolute values here is, so we replace with the quantit ( ) in Definition. to get { ( ), if ( ) < 0 = ( ), if ( ) 0 Simplifing ields { = +, if <, if So, for <, = + and our equation = becomes + =, which gives =. Since this solution satisfies <, we keep it. Net, for, =, so the equation = becomes =. Here, the equation reduces to =, which signifies we have no solutions here. Hence, our onl solution is =. Net, we turn our attention to graphing absolute value functions. Our strateg in the net eample is to make liberal use of Definition. along with what we know about graphing linear functions (from Section.) and piecewise-defined functions (from Section.). Eample... Graph each of the following functions.. f() =. g() =. h() =. i() = +

26 76 Linear and Quadratic Functions Find the zeros of each function and the - and-intercepts of each graph, if an eist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing or constant, and find the relative and absolute etrema, if the eist. Solution.. To find the zeros of f, we set f() = 0. We get =0,which,bTheorem. gives us =0. Since the zeros of f are the -coordinates of the -intercepts of the graph of = f(), we get (0, 0) as our onl -intercept. To find the -intercept, we set = 0, and find = f(0) = 0, so that (0, 0) is our -intercept as well. Using Definition., weget f() = = {, if <0, if 0 Hence, for <0, we are graphing the line = ; for 0, we have the line =. Proceeding as we did in Section.6, weget f() =, <0 f() =, 0 Noticethatwehavean opencircle at(0, 0) in the graph when <0. As we have seen before, this is due to the fact that the points on = approach (0, 0) as the -values approach 0. Since is required to be strictl less than zero on this stretch, the open circle is drawn at the origin. However, notice that when 0, we get to fill in the point at (0, 0), which effectivel plugs the hole indicated b the open circle. Thus we get, f() = Actuall, since functions can have at most one -intercept (Do ou know wh?), as soon as we found (0, 0) as the -intercept, we knew this was also the -intercept.

27 . Absolute Value Functions 77 B projecting the graph to the -ais, we see that the domain is (, ). Projecting to the -aisgivesustherange[0, ). The function is increasing on [0, ) and decreasing on (, 0]. The relative minimum value of f is the same as the absolute minimum, namel 0 which occurs at (0, 0). There is no relative maimum value of f. There is also no absolute maimum value of f, sincethe values on the graph etend infinitel upwards.. To find the zeros of g, we set g() = =0. BTheorem., weget =0so that =. Hence, the -intercept is (, 0). To find our -intercept, we set =0sothat = g(0) = 0 =, which ields (0, ) as our -intercept. To graph g() =, weuse Definition. to rewrite g as g() = = { ( ), if ( ) < 0 ( ), if ( ) 0 Simplifing, we get g() = { +, if <, if As before, the open circle we introduce at (, 0) from the graph of = + is filled b the point (, 0) from the line =. We determine the domain as (, ) and the range as [0, ). The function g is increasing on [, ) and decreasing on (, ]. The relative and absolute minimum value of g is 0 which occurs at (, 0). As before, there is no relative or absolute maimum value of g.. Setting h() = 0 to look for zeros gives = 0. As in Eample.., we isolate the absolute value to get =sothat =or =. As a result, we have a pair of - intercepts: (, 0) and (, 0). Setting = 0 gives = h(0) = 0 =, so our -intercept is (0, ). As before, we rewrite the absolute value in h to get h() = {, if <0, if 0 Once again, the open circle at (0, ) from one piece of the graph of h is filled b the point (0, ) from the other piece of h. From the graph, we determine the domain of h is (, ) and the range is [, ). On [0, ), h is increasing; on (, 0] it is decreasing. The relative minimum occurs at the point (0, ) on the graph, and we see is both the relative and absolute minimum value of h. Also,h has no relative or absolute maimum value.

28 78 Linear and Quadratic Functions 5 g() = h() =. As before, we set i() = 0 to find the zeros of i and get + = 0. Isolating the absolute value term gives + =,soeither+ = or + =. We get = or =, so our -intercepts are (, 0) and (, 0). Substituting = 0 gives = i(0) = (0)+ =, for a -intercept of (0, ). Rewriting the formula for i() without absolute values gives { { ( ( +)), if ( +)< , if < i() = ( +), if ( +) 0 = 6 +, if The usual analsis near the trouble spot = gives the corner of this graph is (, ), and we get the distinctive shape: 5 i() = + The ( domain ] of i is (, ) [ while the range is (, ]. The function i is increasing on, and decreasing on, ). The relative maimum occurs at the point (, ) and the relative and absolute maimum value of i is. Since the graph of i etends downwards forever more, there is no absolute minimum value. As we can see from the graph, there is no relative minimum, either. Note that all of the functions in the previous eample bear the characteristic shapeofthegraph of =. We could have graphed the functions g, h and i in Eample.. starting with the graph of f() = and appling transformations as in Section.7 as our net eample illustrates.

29 . Absolute Value Functions 79 Eample... Graph the following functions starting with the graph of f() = and using transformations.. g() =. h() =. i() = + Solution. We begin b graphing f() = and labeling three points, (, ), (0, 0) and (, ). (, ) (, ) (0, 0) f() =. Since g() = = f( ), Theorem.7 tells us to add toeachofthe-values of the points on the graph of = f() to obtain the graph of = g(). This shifts the graph of = f() totheright units and moves the point (, ) to (, ), (0, 0) to (, 0) and (, ) to (, ). Connecting these points in the classic fashion produces the graph of = g(). (, ) (, ) (, ) (, ) (0, 0) f() = shift right units addtoeach-coordinate (, 0) 5 6 g() =f( ) =. For h() = =f(), Theorem.7 tells us to subtract fromeachofthe-values of the points on the graph of = f() to obtain the graph of = h(). This shifts the graph of = f() down units and moves (, ) to (, ), (0, 0) to (0, ) and (, ) to (, ). Connecting these points with the shape produces our graph of = h(). (, ) (, ) (, ) (, ) (0, ) (0, 0) f() = shift down units subtract from each -coordinate h() =f() =

30 80 Linear and Quadratic Functions. We re-write i() = + = f( +)= f( + ) + and appl Theorem.7. First, we take care of the changes on the inside of the absolute value. Instead of, we have +, so, in accordance with Theorem.7, wefirstsubtract fromeachofthe -values of points on the graph of = f(), then divide each of those new values b. This effects a horizontal shift left unit followed b a horizontal shrink b a factor of. These transformations move (, ) to (, ),(0, 0) to (, 0) and (, ) to (0, ). Net, we take care of what s happening outside of the absolute value. Theorem.7 instructs us to first multipl each -value of these new points b thenadd. Geometricall, this corresponds to a vertical stretch b a factor of, a reflection across the -ais and finall, a vertical shift up units. These transformations move (, ) to (, ), (, 0) to (, ),and(0, ) to (0, ). Connecting these points with the usual shape produces our graph of = i(). (, ) (0, 0) (, ) (, ) (, ) (0, ) f() = i() = f( +)+ = + + While the methods in Section.7 can be used to graph an entire famil of absolute value functions, not all functions involving absolute values posses the characteristic shape. As the net eample illustrates, often there is no substitute for appealing directl to the definition. Eample... Graph each of the following functions. Find the zeros of each function and the - and-intercepts of each graph, if an eist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing or constant, and find the relative and absolute etrema, if the eist.. f() =. g() = + + Solution.. We first note that, due to the fraction in the formula of f(), 0. Thus the domain is (, 0) (0, ). To find the zeros of f, we set f() = = 0. This last equation implies =0,which,fromTheorem., implies = 0. However, = 0 is not in the domain of f,

31 . Absolute Value Functions 8 which means we have, in fact, no -intercepts. We have no -intercepts either, since f(0) is undefined. Re-writing the absolute value in the function gives {, if <0 f() =, if <0 =, if >0, if >0 To graph this function, we graph two horizontal lines: = for<0and =for>0. We have open circles at (0, ) and (0, ) (Can ou eplain wh?) so we get f() = As we found earlier, the domain is (, 0) (0, ). The range consists of just two -values: {, }. The function f is constant on (, 0) and (0, ). The local minimum value of f is the absolute minimum value of f, namel ; the local maimum and absolute maimum values for f also coincide the both are. Ever point on the graph of f is simultaneousl a relative maimum and a relative minimum. (Can ou remember wh in light of Definition.? This was eplored in the Eercises in Section.6..). To find the zeros of g, we set g() = 0. The result is + + = 0. Attempting to isolate the absolute value term is complicated b the fact that there are two terms with absolute values. In this case, it easier to proceed using cases b re-writing the function g with two separate applications of Definition. to remove each instance of the absolute values, one at a time. In the first round we get g() = { { ( +) +, if ( +)< 0 ( +) +, if ( +) 0 =, if < +, if Given that = { { ( ), if ( ) < 0, if ( ) 0 = +, if <, if, we need to break up the domain again at =. Note that if <, then <, so we replace with + for that part of the domain, too. Our completed revision of the form of g ields

32 8 Linear and Quadratic Functions g() = ( +), if < + ( +), if and< + ( ), if =, if <, if < 6, if To solve g() = 0, we see that the onl piece which contains a variable is g() = for <. Solving = 0 gives =0. Since = 0 is in the interval [, ), we keep this solution and have (0, 0) as our onl -intercept. Accordingl, the -intercept is also (0, 0). To graph g, we start with < and graph the horizontal line = withanopencircleat(, ). For <, we graph the line = and the point (, ) patches the hole left b the previous piece. An open circle at (, 6) completes the graph of this part. Finall, we graph the horizontal line =6 for, and the point (, 6) fills in the open circle left b the previous part of the graph. The finished graph is 6 5 g() = + + The domain of g is all real numbers, (, ), and the range of g is all real numbers between and 6 inclusive, [, 6]. The function is increasing on [, ] and constant on (, ] and [, ). The relative minimum value of f is which matches the absolute minimum. The relative and absolute maimum values also coincide at 6. Ever point on the graph of = g() for< and > ields both a relative minimum and relative maimum. The point (, ), however, gives onl a relative minimum and the point (, 6) ields onl a relative maimum. (Recall the Eercises in Section.6. which dealt with constant functions.) Man of the applications that the authors are aware of involving absolute values also involve absolute value inequalities. For that reason, we save our discussion of applications for Section..

33 . Absolute Value Functions 8.. Eercises In Eercises - 5, solve the equation.. =6. =0. =7. = = = = 8. 5 =5 9. = + 0. = +. = +. = 5. =. = 5. = Prove that if f() = g() then either f() =g() orf() = g(). Use that result to solve the equations in Eercises = = 8. = = =5 +. = + In Eercises -, graph the function. Find the zeros of each function and the - and-intercepts of each graph, if an eist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing or constant, and find the relative and absolute etrema, if the eist.. f() = +. f() = +. f() = 5. f() = 6. f() = + 7. f() = 8. f() = f() = 0. f() = +. f() = +. f() = +. f() = + +. With the help of our classmates, find an absolute value function whose graph is given below With help from our classmates, prove the second, third and fifth parts of Theorem.. 6. Prove The Triangle Inequalit: For all real numbers a and b, a + b a + b.

34 8 Linear and Quadratic Functions.. Answers. = 6 or =6. = or =. = or =. = or = 5. = or = 0 6. no solution 7. = or = 8. = 8 or = = 0. =0or =. =. no solution. =, =0or =. = or = 5. = or = 6. = or = 9 7. = 7 or = 8. =0or = 9. = 0. = 0. = 5 or =5. f() = + f( ) = 0 -intercept (, 0) -intercept (0, ) Domain (, ) Range [0, ) Decreasing on (, ] Increasing on [, ) Relative and absolute min. at (, 0) No relative or absolute maimum f() = + No zeros No -intercepts -intercept (0, ) Domain (, ) Range [, ) Decreasing on (, 0] Increasing on [0, ) Relative and absolute minimum at (0, ) No relative or absolute maimum

35 . Absolute Value Functions 85. f() = f(0) = 0 -intercept (0, 0) -intercept (0, 0) Domain (, ) Range [0, ) Decreasing on (, 0] Increasing on [0, ) Relative and absolute minimum at (0, 0) No relative or absolute maimum f() = f(0) = 0 -intercept (0, 0) -intercept (0, 0) Domain (, ) Range (, 0] Increasing on (, 0] Decreasing on [0, ) Relative and absolute maimum at (0, 0) No relative or absolute minimum f() = + f ( 6 ) ( =0,f 8 ) =0 -intercepts ( 6, 0), ( 8, 0) -intercept (0, 8) Domain (, ) Range [, ) Decreasing on (, ] Increasing on [, ) Relative and absolute min. at (, ) No relative or absolute maimum f() = f ( ) =0 -intercepts (, 0) -intercept ( 0, ) Domain (, ) Range [0, ) Decreasing on (, ] Increasing on [, ) Relative and absolute min. at (, 0) No relative or absolute maimum

36 86 Linear and Quadratic Functions + 8. f() = + No zeros No -intercept -intercept (0, ) Domain (, ) (, ) Range {, } Constant on (, ) Constant on (, ) Absolute minimum at ever point (, ) 9. f() = No zeros No -intercept -intercept (0, ) Domain (, ) (, ) Range {, } Constant on (, ) Constant on (, ) Absolute minimum at ever point (, ) 0. Re-write { f() = + as if <0 f() = if 0 f ( ) =0 -intercept (, 0) -intercept (0, ) Domain (, ) Range [, ) Increasing on [0, ) Constant on (, 0] Absolute minimum at ever point (, ) where 0 No absolute maimum where < Absolute maimum at ever point (, ) where > Relative maimum AND minimum at ever point on the graph where > Absolute maimum at ever point (, ) where < Relative maimum AND minimum at ever point on the graph 5 Relative minimum at ever point (, ) where 0 Relative maimum at ever point (, ) where <0

37 . Absolute Value Functions 87. Re-write { f() = + as if < f() = if No zeros No -intercepts -intercept (0, ) Domain (, ) Range [, ) Decreasing on (, ] Constant on [, ) Absolute minimum at ever point (, ) where. Re-write f() = + as if < f() = + if <0 if 0 f ( ) = 0 -intercept (, 0) -intercept (0, ) Domain (, ) Range [, ] Increasing on [, 0] Constant on (, ] Constant on [0, ) Absolute minimum at ever point (, ) where. Re-write f() = + + as if < f() = 6 if < + if No zeros No -intercept -intercept (0, 6) Domain (, ) Range [6, ) Decreasing on (, ] Constant on [, ] Increasing on [, ) Absolute minimum at ever point (, 6) where No absolute maimum Relative minimum at ever point (, 6) where No absolute maimum Relative minimum at ever point (, ) where Relative maimum at ever point (, ) where > Absolute maimum at ever point (, ) where 0 Relative minimum at ever point (, ) where and at ever point (, ) where >0 Relative maimum at ever point (, ) where < and at ever point (, ) where 0 Relative maimum at ever point (, 6) where << f() =

38 88 Linear and Quadratic Functions. Quadratic Functions You ma recall studing quadratic equations in Intermediate Algebra. In this section, we review those equations in the contet of our net famil of functions: the quadratic functions. Definition.5. A quadratic function is a function of the form f() =a + b + c, where a, b and c are real numbers with a 0. The domain of a quadratic function is (, ). The most basic quadratic function is f() =, whose graph appears below. Its shape should look familiar from Intermediate Algebra it is called a parabola. Thepoint(0, 0) is called the verte of the parabola. In this case, the verte is a relative minimum and is also the where the absolute minimum value of f can be found. (, ) (, ) (, ) (, ) (0, 0) f() = Much like man of the absolute value functions in Section., knowing the graph of f() = enables us to graph an entire famil of quadratic functions using transformations. Eample... Graph the following functions starting with the graph of f() = and using transformations. Find the verte, state the range and find the - and-intercepts, if an eist.. g() =( +). h() = ( ) + Solution.. Since g() =( +) =f( +), Theorem.7 instructs us to first subtract from each of the -values of the points on = f(). This shifts the graph of = f() totheleft units and moves (, ) to (, ), (, ) to (, ), (0, 0) to (, 0), (, ) to (, ) and (, ) to (0, ). Net, we subtract fromeachofthe-values of these new points. This moves the graph down units and moves (, ) to (, ), (, ) to (, ), (, 0) to (, ), (, ) to (, ) and (0, ) to (0, ). We connect the dots in parabolic fashion to get

39 . Quadratic Functions 89 (, ) (, ) (, ) (0, ) (, ) (, ) (, ) (, ) (0, 0) f() = (, ) g() =f( +) =( +) From the graph, we see that the verte has moved from (0, 0) on the graph of = f() to (, ) on the graph of = g(). This sets [, ) as the range of g. We see that the graph of = g() crosses the -ais twice, so we epect two -intercepts. To find these, we set = g() = 0 and solve. Doing so ields the equation ( +) = 0, or ( +) =. Etracting square roots gives +=±, or = ±. Our -intercepts are (, 0) (.7, 0) and ( +, 0) ( 0.7, 0). The -intercept of the graph, (0, ) was one of the points we originall plotted, so we are done.. Following Theorem.7 once more, to graph h() = ( ) += f( ) +, we first start b adding toeachofthe-values of the points on the graph of = f(). This effects a horizontal shift right units and moves (, ) to (, ), (, ) to (, ), (0, 0) to (, 0), (, ) to (, ) and (, ) to (5, ). Net, we multipl each of our -values first b andthen add to that result. Geometricall, this is a vertical stretch bafactorof,followedba reflection about the -ais, followed b a vertical shift up unit. This moves (, ) to (, 7), (, ) to (, ), (, 0) to (, ), (, ) to (, ) and (5, ) to (5, 7). (, ) 5 (, ) (, ) (, ) (, ) 5 6 (, ) (, ) (, 7) (5, 7) (0, 0) f() = h() = f( ) + = ( ) + Theverteis(, ) which makes the range of h (, ]. From our graph, we know that there are two -intercepts, so we set = h() = 0 and solve. We get ( ) +=0

40 90 Linear and Quadratic Functions which gives ( ) =. Etracting square roots gives =± (, so that when we add to each side, we get = 6± ). Hence, our -intercepts are 6, 0 (.9, 0) and ( ) 6+, 0 (.7, 0). Although our graph doesn t show it, there is a -intercept which can be found b setting = 0. With h(0) = (0 ) += 7, we have that our -intercept is (0, 7). A few remarks about Eample.. are in order. First note that neither the formula given for g() nor the one given for h() match the form given in Definition.5. We could, of course, convert both g() andh() into that form b epanding and collecting like terms. Doing so, we find g() =( +) = + + and h() = ( ) + = + 7. While these simplified formulas for g() andh() satisf Definition.5, the do not lend themselves to graphing easil. For that reason, the form of g and h presented in Eample.. is given a special name, which we list below, along with the form presented in Definition.5. Definition.6. Standard and General Form of Quadratic Functions: Suppose f is a quadratic function. ˆ The general form of the quadratic function f is f() =a + b + c, wherea, b and c are real numbers with a 0. ˆ The standard form of the quadratic function f is f() =a( h) + k, wherea, h and k are real numbers with a 0. It is important to note at this stage that we have no guarantees that ever quadratic function can be written in standard form. This is actuall true, and we prove this later in the eposition, but for now we celebrate the advantages of the standard form, starting with the following theorem. Theorem.. Verte Formula for Quadratics in Standard Form: For the quadratic function f() =a( h) + k, wherea, h and k are real numbers with a 0, the verte of the graph of = f() is(h, k). We can readil verif the formula given Theorem. with the two functions given in Eample... After a (slight) rewrite, g() =( +) =( ( )) +( ), and we identif h = and k =. Sure enough, we found the verte of the graph of = g() tobe(, ). For h() = ( ) +, no rewrite is needed. We can directl identif h = and k = and, sure enough, we found the verte of the graph of = h() tobe(, ). To see wh the formula in Theorem. produces the verte, consider the graph of the equation = a( h) +k. When we substitute = h, weget = k, so(h, k) is on the graph. If h, then h 0so( h) is a positive number. If a>0, then a( h) is positive, thus = a( h) +k is alwas a number larger than k. This means that when a>0, (h, k) is the lowest point on the graph and thus the parabola must open upwards, making (h, k) the verte. A similar argument and rationalizing denominators! and get common denominators!

41 . Quadratic Functions 9 shows that if a<0, (h, k) is the highest point on the graph, so the parabola opens downwards, and (h, k) is also the verte in this case. Alternativel, we can appl the machiner in Section.7. Since the verte of = is (0, 0), we can determine the verte of = a( h) +k b determining the final destination of (0, 0) as it is moved through each transformation. To obtain the formula f() =a( h) + k, we start with g() = and first define g () =ag() =a. This is results in a vertical scaling and/or reflection. Since we multipl the output b a, we multipl the -coordinates on the graph of g b a, sothepoint (0, 0) remains (0, 0) and remains the verte. Net, we define g () =g ( h) =a( h). This induces a horizontal shift right or left h units moves the verte, in either case, to (h, 0). Finall, f() =g ()+k = a( h) + k which effects a vertical shift up or down k units 5 resulting in the verte moving from (h, 0) to (h, k). In addition to verifing Theorem., the arguments in the two preceding paragraphs have also shown us the role of the number a in the graphs of quadratic functions. The graph of = a( h) +k is a parabola opening upwards if a>0, and opening downwards if a<0. Moreover, the smmetr enjoed b the graph of = about the -ais is translated to a smmetr about the vertical line = h which is the vertical line through the verte. 6 This line is called the ais of smmetr of the parabola and is dashed in the figures below. verte verte a>0 a<0 Graphs of = a( h) + k. Without a doubt, the standard form of a quadratic function, coupled with the machiner in Section.7, allows us to list the attributes of the graphs of such functions quickl and elegantl. What remains to be shown, however, is the fact that ever quadratic function can be written in standard form. To convert a quadratic function given in general form into standard form, we emplo the ancient rite of Completing the Square. We remind the reader how this is done in our net eample. Eample... Convert the functions below from general form to standard form. Find the verte, ais of smmetr and an - or-intercepts. Graph each function and determine its range.. f() = +.. g() =6 Just a scaling if a>0. If a<0, there is a reflection involved. Right if h>0, left if h<0. 5 Up if k>0, down if k<0 6 You should use transformations to verif this!

42 9 Linear and Quadratic Functions Solution.. To convert from general form to standard form, we complete the square. 7 First, we verif that the coefficient of is. Net, we find the coefficient of, in this case, and take half of it to get ( ) =. This tells us that our target perfect square quantit is ( ).To get an epression equivalent to ( ), we need to add ( ) = to the to create a perfect square trinomial, but to keep the balance, we must also subtract it. We collect the terms which create the perfect square and gather the remaining constant terms. Putting it all together, we get f() = + (Compute ( ) =.) = ( + ) + (Add and subtract ( ) =to( +).) = ( + ) + (Group the perfect square trinomial.) = ( ) (Factor the perfect square trinomial.) Of course, we can alwas check our answer b multipling out f() =( ) to see that it simplifies to f() =. In the form f() =( ), we readil find the verte to be (, ) which makes the ais of smmetr =. To find the -intercepts, we set = f() = 0. We are spoiled for choice, since we have two formulas for f(). Since we recognize f() = + to be easil factorable, 8 we proceed to solve +=0. Factoring gives ( )( ) = 0 so that =or =. The-intercepts are then (, 0) and (, 0). To find the -intercept, we set = 0. Once again, the general form f() = + is easiest to work with here, and we find = f(0) =. Hence, the -intercept is (0, ). With the verte, ais of smmetr and the intercepts, we get a prett good graph without the need to plot additional points. We see that the range of f is [, ) and we are done.. To get started, we rewrite g() =6 = + 6 and note that the coefficient of is, not. This means our first step is to factor out the ( ) from both the and terms. We then follow the completing the square recipe as above. g() = +6 = ( ) ( + ) + 6 ( ) (Factor the coefficient of from and.) = ( ) = ( ) ( + + ( ) +( ) ) + 6 = ( + ) + 5 (Group the perfect square trinomial.) 7 If ou forget wh we do what we do to complete the square, start with a( h) + k, multipl it out, step b step, and then reverse the process. 8 Eperience pas off, here!

43 . Quadratic Functions 9 From g() = ( + ) + 5,wegetthevertetobe(, 5 ) and the ais of smmetr to be =. To get the -intercepts, we opt to set the given formula g() =6 =0. Solving, we get = and =,sothe-intercepts are (, 0) and (, 0). Setting =0, we find g(0) = 6, so the -intercept is (0, 6). Plotting these points gives us the graph below. We see that the range of g is (, 5 ]. 8 (, 5 ) 6 (0, 6) = 5 (0, ) (, 0) (, 0) 5 (, ) f() = + (, 0) = (, 0) g() =6 With Eample.. fresh in our minds, we are now in a position to show that ever quadratic function can be written in standard form. We begin with f() =a + b + c, assume a 0,and complete the square in complete generalit. f() = a + b + c = a ( + ba ) + c (Factor out coefficient of from and.) = a ( ba b + + = a ( ba b + + a ( = a + b ) + a a b a ) a ac b a ) + c ( b a ) + c (Group the perfect square trinomial.) (Factor and get a common denominator.) Comparing this last epression with the standard form, we identif ( h) with ( + a) b so that h = b ac b a. Instead of memorizing the value k = a, we see that f ( b ) a = ac b a. As such, we have derived a verte formula for the general form. We summarize both verte formulas in the bo at the top of the net page.

44 9 Linear and Quadratic Functions Equation.. Verte Formulas for Quadratic Functions: Suppose a, b, c, h and k are real numbers with a 0. ˆ If f() =a( h) + k, the verte of the graph of = f() isthepoint(h, k). ( ˆ If f() =a + b + c, the verte of the graph of = f() isthepoint b ( a,f b )). a There are two more results which can be gleaned from the completed-square form of the general form of a quadratic function, ( f() =a + b + c = a + b ) ac b + a a We have seen that the number a in the standard form of a quadratic function determines whether the parabola opens upwards (if a>0) or downwards (if a<0). We see here that this number a is none other than the coefficient of in the general form of the quadratic function. In other words, it is the coefficient of alone which determines this behavior a result that is generalized in Section.. The second treasure is a re-discover of the quadratic formula. Equation.5. The Quadratic Formula: If a, b and c are real numbers with a 0, then the solutions to a + b + c = 0 are = b ± b ac. a Assuming the conditions of Equation.5, the solutions to a + b + c = 0 are precisel the zeros of f() =a + b + c. Since ( f() =a + b + c = a + b ) ac b + a a the equation a + b + c = 0 is equivalent to ( a + b ) ac b + a a =0. Solving gives

45 . Quadratic Functions 95 ( a + b ) ac b + a a ( a + b a [ ( a + b a a = 0 ) ac b = a ) ] = a ( b ) ac a ( + b ) = b ac a a + b a = ± b ac a etract square roots + b a = ± = b a ± b ac a b ac = b ± b ac a In our discussions of domain, we were warned against having negative numbers underneath the square root. Given that b ac is part of the Quadratic Formula, we will need to pa special attention to the radicand b ac. It turns out that the quantit b ac plas a critical role in determining the nature of the solutions to a quadratic equation. It is given a special name. Definition.7. If a, b and c are real numbers with a 0, then the discriminant of the quadratic equation a + b + c = 0 is the quantit b ac. The discriminant discriminates between the kinds of solutions we get from a quadratic equation. These cases, and their relation to the discriminant, are summarized below. Theorem.. Discriminant Trichotom: Let a, b and c be real numbers with a 0. ˆ If b ac < 0, the equation a + b + c = 0 has no real solutions. ˆ If b ac = 0, the equation a + b + c = 0 has eactl one real solution. ˆ If b ac > 0, the equation a + b + c = 0 has eactl two real solutions. The proof of Theorem. stems from the position of the discriminant in the quadratic equation, and is left as a good mental eercise for the reader. The net eample eploits the fruits of all of our labor in this section thus far. a

46 96 Linear and Quadratic Functions Eample... Recall that the profit (defined on page 8) for a product is defined b the equation Profit = Revenue Cost, or P () =R() C(). In Eample..7 the weekl revenue, in dollars, made b selling PortaBo Game Sstems was found to be R() = with the restriction (carried over from the price-demand function) that The cost, in dollars, to produce PortaBo Game Sstems is given in Eample..5 as C() = for 0.. Determine the weekl profit function P ().. Graph = P (). Include the - and-intercepts as well as the verte and ais of smmetr.. Interpret the zeros of P.. Interpret the verte of the graph of = P (). 5. Recall that the weekl price-demand equation for PortaBos is p() = , where p() is the price per PortaBo, in dollars, and is the weekl sales. What should the price per sstem be in order to maimize profit? Solution.. To find the profit function P (), we subtract P () =R() C() = ( ) ( ) = Since the revenue function is valid when 0 66, P is also restricted to these values.. To find the -intercepts, we set P () =0andsolve = 0. The mere thought of tring to factor the left hand side of this equation could do serious pschological damage, so we resort to the quadratic formula, Equation.5. Identifing a =.5, b = 70, and c = 50, we obtain = b ± b ac a = 70 ± 70 (.5)( 50) (.5) = 70 ± 8000 = 70 ± 0 70 ( ) ( ) and To find the -intercept, we set We get two -intercepts:, 0, 0 = 0 and find = P (0) = 50 for a -intercept of (0, 50). To find the verte, we use the fact that P () = is in the general form of a quadratic function and appeal to Equation.. Substituting a =.5 andb = 70, we get = 70 (.5) = 70.

47 . Quadratic Functions 97 To find the -coordinate of the verte, we compute P ( ) 70 = 000 and find that our verte is ( 70, 000 ). The ais of smmetr is the vertical line passing through the verte so it is the line = 70. To sketch a reasonable graph, we approimate the -intercepts, (0.89, 0) and (., 0), and the verte, (56.67, ). (Note that in order to get the -intercepts and the verte to show up in the same picture, we had to scale the -ais differentl than the -ais. This results in the left-hand -intercept and the -intercept being uncomfortabl close to each other and to the origin in the picture.) The zeros of P are the solutions to P () = 0, which we have found to be approimatel 0.89 and.. As we saw in Eample.5., these are the break-even points of the profit function, where enough product is sold to recover the cost spent to make the product. More importantl, we see from the graph that as long as is between 0.89 and., the graph = P () isabovethe-ais, meaning = P () > 0 there. This means that for these values of, a profit is being made. Since represents the weekl sales of PortaBo Game Sstems, we round the zeros to positive integers and have that as long as, but no more than game sstems are sold weekl, the retailer will make a profit.. From the graph, we see that the maimum value of P occurs at the verte, which is approimatel (56.67, ). As above, represents the weekl sales of PortaBo sstems, so we can t sell game sstems. Comparing P (56) = 666 and P (57) = 666.5, we conclude that we will make a maimum profit of $ if we sell 57 game sstems. 5. In the previous part, we found that we need to sell 57 PortaBos per week to maimize profit. To find the price per PortaBo, we substitute = 57 into the price-demand function to get p(57) =.5(57) + 50 = 6.5. The price should be set at $6.50. Our net eample is another classic application of quadratic functions. Eample... Much to Donnie s surprise and delight, he inherits a large parcel of land in Ashtabula Count from one of his (e)strange(d) relatives. The time is finall right for him to pursue his dream of farming alpaca. He wishes to build a rectangular pasture, and estimates that he has enough mone for 00 linear feet of fencing material. If he makes the pasture adjacent to a stream (so no fencing is required on that side), what are the dimensions of the pasture which maimize the area? What is the maimum area? If an average alpaca needs 5 square feet of grazing area, how man alpaca can Donnie keep in his pasture?

48 98 Linear and Quadratic Functions Solution. It is alwas helpful to sketch the problem situation, so we do so below. river w pasture l w We are tasked to find the dimensions of the pasture which would give a maimum area. We let w denote the width of the pasture and we let l denote the length of the pasture. Since the units given to us in the statement of the problem are feet, we assume w and l are measured in feet. The area of the pasture, which we ll call A, is related to w and l b the equation A = wl. Sincew and l are both measured in feet, A has units of feet, or square feet. We are given the total amount of fencing available is 00 feet, which means w + l + w = 00, or, l +w = 00. We now have two equations, A = wl and l +w = 00. In order to use the tools given to us in this section to maimize A, we need to use the information given to write A as a function of just one variable, either w or l. This is where we use the equation l +w = 00. Solving for l, we find l = 00 w, and we substitute this into our equation for A. WegetA = wl = w(00 w) = 00w w.we now have A as a function of w, A(w) = 00w w = w + 00w. Before we go an further, we need to find the applied domain of A so that we know what values of w make sense in this problem situation. 9 Since w represents the width of the pasture, w>0. Likewise, l represents the length of the pasture, so l = 00 w >0. Solving this latter inequalit, we find w<00. Hence, the function we wish to maimize is A(w) = w +00w for 0 <w<00. Since A is a quadratic function (of w), we know that the graph of = A(w) is a parabola. Since the coefficient of w is, we know that this parabola opens downwards. This means that there is a maimum value to be found, and we know it occurs at the verte. Using the verte formula, we find w = 00 ( ) = 50, and A(50) = (50) + 00(50) = Since w = 50 lies in the applied domain, 0 <w<00, we have that the area of the pasture is maimized when the width is 50 feet. To find the length, we use l = 00 w and find l = 00 (50) = 00, so the length of the pasture is 00 feet. The maimum area is A(50) = 5000, or 5000 square feet. If an average alpaca requires 5 square feet of pasture, Donnie can raise = 00 average alpaca. We conclude this section with the graph of a more complicated absolute value function. Eample..5. Graph f() = 6. Solution. Using the definition of absolute value, Definition., wehave { ( f() = 6 ), if 6 < 0 6, if 6 0 The trouble is that we have et to develop an analtic techniques to solve nonlinear inequalities such as 6 < 0. You won t have to wait long; this is one of the main topics of Section.. 9 Donnie would be ver upset if, for eample, we told him the width of the pasture needs to be 50 feet.

49 . Quadratic Functions 99 Nevertheless, we can attack this problem graphicall. To that end, we graph = g() = 6 using the intercepts and the verte. To find the -intercepts, we solve 6 = 0. Factoring gives ( )( +) = 0 so = or =. Hence, (, 0) and (, 0) are -intercepts. The -intercept (0, 6) is found b setting =0. Toplottheverte,wefind = b a = () =,and = ( ( ) ) 6= 5 = 6.5. Plotting, we get the parabola seen below on the left. To obtain points on the graph of = f() = 6, we can take points on the graph of g() = 6 and appl the absolute value to each of the values on the parabola. We see from the graph of g that for or, the values on the parabola are greater than or equal to zero (since the graph is on or above the -ais), so the absolute value leaves these portions of the graph alone. For between and, however, the values on the parabola are negative. For eample, the point (0, 6) on = 6 would result in the point (0, 6 ) =(0, ( 6)) = (0, 6) on the graph of f() = 6. Proceeding in this manner for all points with -coordinates between and results in the graph seen below on the right = g() = 6 = f() = 6 If we take a step back and look at the graphs of g and f in the last eample, we notice that to obtain the graph of f from the graph of g, we reflect a portion of the graph of g about the -ais. We can see this analticall b substituting g() = 6 into the formula for f() and calling to mind Theorem. from Section.7. { g(), if g() < 0 f() = g(), if g() 0 The function f is defined so that when g() is negative (i.e., when its graph is below the -ais), the graph of f is its refection across the -ais. This is a general template to graph functions of the form f() = g(). From this perspective, the graph of f() = can be obtained b reflecting the portion of the line g() = which is below the -ais back above the -ais creating the characteristic shape.

50 00 Linear and Quadratic Functions.. Eercises In Eercises - 9, graph the quadratic function. Find the - and-intercepts of each graph, if an eist. If it is given in general form, convert it into standard form; if it is given in standard form, convert it into general form. Find the domain and range of the function and list the intervals on which the function is increasing or decreasing. Identif the verte and the ais of smmetr and determine whether the verte ields a relative and absolute maimum or minimum.. f() = +. f() = ( +). f() = 8. f() = ( +) + 5. f() = 6. f() = f() = f() = f() = 00 In Eercises 0 -, the cost and price-demand functions are given for different scenarios. For each scenario, ˆ Find the profit function P (). ˆ Find the number of items which need to be sold in order to maimize profit. ˆ Find the maimum profit. ˆ Find the price to charge per item in order to maimize profit. ˆ Find and interpret break-even points. 0. The cost, in dollars, to produce I d rather be a Sasquatch T-Shirts is C() = + 6, 0 and the price-demand function, in dollars per shirt, is p() =0, The cost, in dollars, to produce bottles of 00% All-Natural Certified Free-Trade Organic Sasquatch Tonic is C() =0 + 00, 0 and the price-demand function, in dollars per bottle, is p() =5, The cost, in cents, to produce cups of Mountain Thunder Lemonade at Junior s Lemonade Stand is C() = 8 + 0, 0 and the price-demand function, in cents per cup, is p() =90, The dail cost, in dollars, to produce Sasquatch Berr Pies is C() = + 6, 0and the price-demand function, in dollars per pie, is p() = 0.5, 0.. The monthl cost, in hundreds of dollars, to produce custom built electric scooters is C() = , 0 and the price-demand function, in hundreds of dollars per scooter, is p() = 0, We have alread seen the graph of this function. It was used as an eample in Section.6 to show how the graphing calculator can be misleading.

51 . Quadratic Functions 0 5. The International Silver Strings Submarine Band holds a bake sale each ear to fund their trip to the National Sasquatch Convention. It has been determined that the cost in dollars of baking cookies is C() = and that the demand function for their cookies is p =0.0. How man cookies should the bake in order to maimize their profit? 6. Using data from Bureau of Transportation Statistics, the average fuel econom F in miles per gallon for passenger cars in the US can be modeled b F (t) = t +0.5t + 6, 0 t 8, where t is the number of ears since 980. Find and interpret the coordinates of the verte of the graph of = F (t). 7. The temperature T, in degrees Fahrenheit, t hours after 6 AM is given b: T (t) = t +8t +, 0 t What is the warmest temperature of the da? When does this happen? 8. Suppose C() = represents the costs, in hundreds, to produce thousand pens. How man pens should be produced to minimize the cost? What is this minimum cost? 9. Skipp wishes to plant a vegetable garden along one side of his house. In his garage, he found linear feet of fencing. Since one side of the garden will border the house, Skipp doesn t need fencing along that side. What are the dimensions of the garden which will maimize the area of the garden? What is the maimum area of the garden? 0. In the situation of Eample.., Donnie has a nightmare that one of his alpaca herd fell into the river and drowned. To avoid this, he wants to move his rectangular pasture awa from the river. This means that all four sides of the pasture require fencing. If the total amount of fencing available is still 00 linear feet, what dimensions maimize the area of the pasture now? What is the maimum area? Assuming an average alpaca requires 5 square feet of pasture, how man alpaca can he raise now?. What is the largest rectangular area one can enclose with inches of string?. The height of an object dropped from the roof of an eight stor building is modeled b h(t) = 6t + 6, 0 t. Here, h is the height of the object off the ground, in feet, t seconds after the object is dropped. How long before the object hits the ground?. The height h in feet of a model rocket above the ground t seconds after lift-off is given b h(t) = 5t + 00t, for 0 t 0. When does the rocket reach its maimum height above the ground? What is its maimum height?. Carl s friend Jason participates in the Highland Games. In one event, the hammer throw, the height h in feet of the hammer above the ground t seconds after Jason lets it go is modeled b h(t) = 6t +.08t + 6. What is the hammer s maimum height? What is the hammer s total time in the air? Round our answers to two decimal places.

52 0 Linear and Quadratic Functions 5. Assuming no air resistance or forces other than the Earth s gravit, the height above the ground at time t of a falling object is given b s(t) =.9t + v 0 t + s 0 where s is in meters, t is in seconds, v 0 is the object s initial velocit in meters per second and s 0 is its initial position in meters. (a) What is the applied domain of this function? (b) Discuss with our classmates what each of v 0 > 0, v 0 = 0 and v 0 < 0 would mean. (c) Come up with a scenario in which s 0 < 0. (d) Let s sa a slingshot is used to shoot a marble straight up from the ground (s 0 =0)with an initial velocit of 5 meters per second. What is the marble s maimum height above the ground? At what time will it hit the ground? (e) Now shoot the marble from the top of a tower which is 5 meters tall. When does it hit the ground? (f) What would the height function be if instead of shooting the marble up off of the tower, ou were to shoot it straight DOWN from the top of the tower? 6. The two towers of a suspension bridge are 00 feet apart. The parabolic cable attached to the tops of the towers is 0 feet above the point on the bridge deck that is midwa between the towers. If the towers are 00 feet tall, find the height of the cable directl above a point of the bridge deck that is 50 feet to the right of the left-hand tower. 7. Graph f() = 8. Find all of the points on the line = which are units from (, ). 9. Let L be the line = +. Find a function D() which measures the distance squared from a point on L to (0, 0). Use this to find the point on L closest to (0, 0). 0. With the help of our classmates, show that if a quadratic function f() =a + b + c has two real zeros then the -coordinate of the verte is the midpoint of the zeros. In Eercises - 6, solve the quadratic equation for the indicated variable.. 0 =0for. = for. m =for. = for 5. = for 6. gt + v 0 t + s 0 =0fort (Assume g 0.) The weight of the bridge deck forces the bridge cable into a parabola and a free hanging cable such as a power line does not form a parabola. We shall see in Eercise 5 in Section 6.5 what shape a free hanging cable makes.

53 . Quadratic Functions 0.. Answers. f() = + (this is both forms!) No -intercepts -intercept (0, ) Domain: (, ) Range: [, ) Decreasing on (, 0] Increasing on [0, ) Verte (0, ) is a minimum Ais of smmetr = f() = ( +) = -intercept (, 0) -intercept (0, ) Domain: (, ) Range: (, 0] Increasing on (, ] Decreasing on [, ) Verte (, 0) is a maimum Ais of smmetr = f() = 8=( ) 9 -intercepts (, 0) and (, 0) -intercept (0, 8) Domain: (, ) Range: [ 9, ) Decreasing on (, ] Increasing on [, ) Verte (, 9) is a minimum Ais of smmetr = f() = ( +) += + -intercepts (, 0) and ( +, 0) -intercept (0, ) Domain: (, ) Range: (, ] Increasing on (, ] Decreasing on [, ) Verte (, ) is a maimum Ais of smmetr =

54 0 Linear and Quadratic Functions 5. f() = ( =( ) -intercepts ) ( 6 +, 0 and ) 6, 0 -intercept (0, ) Domain: (, ) Range: [, ) Increasing on [, ) Decreasing on (, ] Verte (, ) is a minimum Ais of smmetr = 6. f() = + 7= ( ) 7 No -intercepts -intercept (0, 7) Domain: (, ) Range: (, 7 ] Increasing on (, ] Decreasing on [, ) Verte (, 7 ) is a maimum Ais of smmetr = f() = + += ( + ) + No -intercepts -intercept (0, ) Domain: (, ) Range: [, ) Increasing on [, ) Decreasing on (, ] Verte (, ) is a minimum Ais of smmetr =

55 . Quadratic Functions f() = +5 += ( 5 6) + 7 ( 5 -intercepts ) ( , 0 and ) 7 6, 0 -intercept (0, ) Domain: (, ) Range: (, 7 ] Increasing on (, 5 ] 6 Decreasing on [ 5 6, ) Verte ( 5 6, 7 ) is a maimum Ais of smmetr = f() = 00 =( ) 000 ) -intercepts and ( intercept (0, ) Domain: (, ) Range: [ , ) Decreasing on (, 00 ] ( ) Increasing on [ 00, ) Verte ( 00, 000 ) 0000 is a minimum Ais of smmetr = ˆ P () = +8 6, for 0 5. ˆ 7 T-shirts should be made and sold to maimize profit. ˆ The maimum profit is $7. ˆ The price per T-shirt should be set at $6 to maimize profit. ˆ The break even points are = and =, so to make a profit, between and T-shirts need to be made and sold.. ˆ P () = +5 00, for 0 5 ˆ Since the verte occurs at =.5, and it is impossible to make or sell.5 bottles of tonic, maimum profit occurs when either or bottles of tonic are made and sold. ˆ The maimum profit is $56. ˆ The price per bottle can be either $ (to sell bottles) or $ (to sell bottles.) Both will result in the maimum profit. ˆ The break even points are = 5 and = 0, so to make a profit, between 5 and 0 bottles of tonic need to be made and sold. You ll need to use our calculator to zoom in far enough to see that the verte is not the -intercept.

56 06 Linear and Quadratic Functions. ˆ P () = +7 0, for 0 0 ˆ cups of lemonade need to be made and sold to maimize profit. ˆ The maimum profit is 9 or $.9. ˆ The price per cup should be set at 5 per cup to maimize profit. ˆ The break even points are = and = 0, so to make a profit, between and 0 cups of lemonade need to be made and sold.. ˆ P () = , for 0 ˆ 9 pies should be made and sold to maimize the dail profit. ˆ The maimum dail profit is $.50. ˆ The price per pie should be set at $7.50 to maimize profit. ˆ The break even points are = 6 and =, so to make a profit, between 6 and pies need to be made and sold dail.. ˆ P () = , for 0 70 ˆ 0 scooters need to be made and sold to maimize profit. ˆ The maimum monthl profit is 800 hundred dollars, or $80,000. ˆ The price per scooter should be set at 80 hundred dollars, or $8000 per scooter. ˆ The break even points are =0and = 50, so to make a profit, between 0 and 50 scooters need to be made and sold monthl cookies 6. The verte is (approimatel) (9.60,.66), which corresponds to a maimum fuel econom of.66 miles per gallon, reached sometime between 009 and 00 (9 0 ears after 980.) Unfortunatel, the model is onl valid up until 008 (8 ears after 908.) So, at this point, we are using the model to predict the maimum fuel econom at PM (8 hours after 6 AM.) pens should be produced for a cost of $ feet b 6 feet; maimum area is 8 square feet feet b 50 feet; maimum area is 500 feet; he can raise 00 average alpacas.. The largest rectangle has area.5 square inches.. seconds.. The rocket reaches its maimum height of 500 feet 0 seconds after lift-off.. The hammer reaches a maimum height of approimatel.6 feet. The hammer is in the air approimatel.6 seconds.

57 . Quadratic Functions (a) The applied domain is [0, ). (d) The height function is this case is s(t) =.9t +5t. The verte of this parabola is approimatel (.5,.8) so the maimum height reached b the marble is.8 meters. It hits the ground again when t.06 seconds. (e) The revised height function is s(t) =.9t +5t + 5 which has zeros at t.0 and t.6. We ignore the negative value and claim that the marble will hit the ground after.6 seconds. (f) Shooting down means the initial velocit is negative so the height functions becomes s(t) =.9t 5t Make the verte of the parabola (0, 0) so that the point on the top of the left-hand tower wherethecableconnectsis( 00, 00) and the point on the top of the right-hand tower is (00, 00). Then the parabola is given b p() = Standing 50 feet to the right of the left-hand tower means ou re standing at = 50 and p( 50) = So the cable is feet above the bridge deck there. ( 7. = ) ( 7 8., ,, ) D() = +(+) =5 ++, D is minimized when = 5,sothepointon =+ closest to (0, 0) is ( 5, ) 5. = ± 0. = ±( ). = m ± m +. = ± =± 6. t = v 0 ± v 0 +gs 0 g

58 08 Linear and Quadratic Functions. Inequalities with Absolute Value and Quadratic Functions In this section, not onl do we develop techniques for solving various classes of inequalities analticall, we also look at them graphicall. The first eample motivates the core ideas. Eample... Let f() = andg() =5.. Solve f() =g().. Solve f() <g().. Solve f() >g().. Graph = f() and = g() on the same set of aes and interpret our solutions to parts through above. Solution.. To solve f() =g(), we replace f() with andg() with5toget = 5. Solving for, weget =.. The inequalit f() <g() is equivalent to < 5. Solving gives <or(, ).. To find where f() >g(), we solve > 5. We get >, or (, ).. To graph = f(), we graph =, which is a line with a -intercept of (0, ) and a slope of. The graph of = g() is = 5 which is a horizontal line through (0, 5) = g() = f() To see the connection between the graph and the Algebra, we recall the Fundamental Graphing Principle for Functions in Section.6: thepoint(a, b) is on the graph of f if and onl if f(a) =b. In other words, a generic point on the graph of = f() is(, f()), and a generic

59 . Inequalities with Absolute Value and Quadratic Functions 09 point on the graph of = g() is(, g()). When we seek solutions to f() =g(), we are looking for values whose values on the graphs of f and g are the same. In part, we found = is the solution to f() =g(). Sure enough, f() = 5 and g() = 5 so that the point (, 5) is on both graphs. In other words, the graphs of f and g intersect at (, 5). In part, we set f() < g() and solved to find <. For <, the point (, f()) is below (, g()) since the values on the graph of f are less than the values on the graph of g there. Analogousl, in part, we solved f() >g() and found >. For >, note that the graph of f is above the graph of g, sincethe values on the graph of f are greater than the values on the graph of g for those values of = f() 6 5 = g() 6 5 = g() = f() f() <g() on(, ) f() >g() on(, ) The preceding eample demonstrates the following, which is a consequence of the Fundamental Graphing Principle for Functions. Graphical Interpretation of Equations and Inequalities Suppose f and g are functions. ˆ The solutions to f() =g() arethe values where the graphs of = f() and = g() intersect. ˆ The solution to f() <g() is the set of values where the graph of = f() isbelow the graph of = g(). ˆ The solution to f() >g() is the set of values where the graph of = f() above the graph of = g(). The net eample turns the tables and furnishes the graphs of two functions and asks for solutions to equations and inequalities.

60 0 Linear and Quadratic Functions Eample... The graphs of f and g are below. (The graph of = g() is bolded.) Use these graphs to answer the following questions. = g() (, ) (, ) = f(). Solve f() =g().. Solve f() <g().. Solve f() g(). Solution.. To solve f() =g(), we look for where the graphs of f and g intersect. These appear to be at the points (, ) and (, ), so our solutions to f() =g() are = and =.. To solve f() <g(), we look for where the graph of f is below the graph of g. Thisappears to happen for the values less than and greater than. Our solution is (, ) (, ).. To solve f() g(), we look for solutions to f() =g() aswellasf() >g(). We solved the former equation and found = ±. To solve f() >g(), we look for where the graph of f is above the graph of g. This appears to happen between = and =,onthe interval (, ). Hence, our solution to f() g() is[, ]. = g() = g() (, ) (, ) (, ) (, ) f() <g() = f() f() g() = f()

61 . Inequalities with Absolute Value and Quadratic Functions We now turn our attention to solving inequalities involving the absolute value. following theorem from Intermediate Algebra to help us. Theorem.. Inequalities Involving the Absolute Value: Let c be a real number. ˆ For c>0, <cis equivalent to c <<c. ˆ For c>0, c is equivalent to c c. ˆ For c 0, <chas no solution, and for c<0, c has no solution. ˆ For c 0, >cis equivalent to < c or >c. ˆ For c 0, c is equivalent to c or c. ˆ For c<0, >cand c are true for all real numbers. We have the As with Theorem. in Section., we could argue Theorem. using cases. However, in light of what we have developed in this section, we can understand these statements graphicall. For instance, if c>0, the graph of = c is a horizontal line which lies above the -ais through (0,c). To solve <c, we are looking for the values where the graph of = is below the graph of = c. We know that the graphs intersect when = c, which,fromsection., we know happens when = c or = c. Graphing, we get ( c, c) (c, c) c c We see that the graph of = is below = c for between c and c, and hence we get <c is equivalent to c <<c. The other properties in Theorem. can be shown similarl. Eample... Solve the following inequalities analticall; check our answers graphicall... + >. < Solution.. From Theorem., is equivalent to or. Solving, we get or, which, in interval notation is (, ] [, ). Graphicall, we have

62 Linear and Quadratic Functions = = 5 We see that the graph of = is above the horizontal line =for< and> hence this is where >. The two graphs intersect when = and =,sowehave graphical confirmation of our analtic solution.. To solve + > analticall, we first isolate the absolute value before appling Theorem.. Tothatend,weget + > 6 or + <. Rewriting, we now have < +< sothat <<. In interval notation, we write (, ). Graphicall we see that the graph of = + is above = for values between and. = + =. Rewriting the compound inequalit < 5as < and 5 allows us to solve each piece using Theorem.. The first inequalit, < can be re-written as > so < or >. We get < or>. Our solution to the first inequalit is then (, ) (, ). For 5, we combine results in Theorems. and. to get 5 5sothat 6, or [, 6]. Our solution to < 5is comprised of values of which satisf both parts of the inequalit, so we take the intersection of (, ) (, ) and[, 6] to get [, ) (, 6]. Graphicall, we see that the graph of = is between the horizontal lines = and =5for values between and as well as those between and 6. Including the values where = and =5 intersect, we get See Definition. in Section...

63 . Inequalities with Absolute Value and Quadratic Functions 8 = 7 6 =5 5 = We need to eercise some special caution when solving + +.AswesawinEample.. in Section., when variables are both inside and outside of the absolute value, it s usuall best to refer to the definition of absolute value, Definition., to remove the absolute values and proceed from there. To that end, we have + = ( +) if < and + = + if. We break the inequalit into cases, the first case being when <. For these values of, our inequalit becomes ( +) +. Solving, we get +, so that 6, which means. Since all of these solutions fall into the categor <, we keep them all. For the second case, we assume. Our inequalit becomes + +,whichgives + + or. Since all of these values of are greater than or equal to, we accept all of these solutions as well. Our final answer is (, ] [, ). = + = + We now turn our attention to quadratic inequalities. In the last eample of Section., weneeded to determine the solution to 6 < 0. We will now re-visit this problem using some of the techniques developed in this section not onl to reinforce our solution in Section., but to also help formulate a general analtic procedure for solving all quadratic inequalities. If we consider f() = 6andg() = 0, then solving 6 < 0 corresponds graphicall to finding

64 Linear and Quadratic Functions thevaluesof for which the graph of = f() = 6 (the parabola) is below the graph of = g() = 0 (the -ais). We ve provided the graph again for reference = 6 We can see that the graph of f does dip below the -ais between its two -intercepts. The zeros of f are = and = in this case and the divide the domain (the -ais) into three intervals: (, ), (, ) and (, ). For ever number in (, ), the graph of f is above the -ais; in other words, f() > 0 for all in (, ). Similarl, f() < 0 for all in (, ), and f() > 0 for all in (, ). We can schematicall represent this with the sign diagram below. (+) 0 ( ) 0 (+) Here, the (+) above a portion of the number line indicates f() > 0 for those values of ; the( ) indicates f() < 0 there. The numbers labeled on the number line are the zeros of f, soweplace 0 above them. We see at once that the solution to f() < 0is(, ). Our net goal is to establish a procedure b which we can generate the sign diagram without graphing the function. An important propert of quadratic functions is that if the function is positive at one point and negative at another, the function must have at least one zero in between. Graphicall, this means that a parabola can t be above the -ais at one point and below the -ais at another point without crossing the -ais. This allows us to determine the sign of all of the function values on a given interval b testing the function at just one value in the interval. This gives us the following. We will give this propert a name in Chapter and revisit this concept then.

65 . Inequalities with Absolute Value and Quadratic Functions 5 Steps for Solving a Quadratic Inequalit. Rewrite the inequalit, if necessar, as a quadratic function f() on one side of the inequalit and 0 on the other.. Find the zeros of f and place them on the number line with the number 0 above them.. Choose a real number, called a test value, ineachoftheintervalsdeterminedinstep.. Determine the sign of f() for each test value in step, and write that sign above the corresponding interval. 5. Choose the intervals which correspond to the correct sign to solve the inequalit. Eample... Solve the following inequalities analticall using sign diagrams. answer graphicall. Verif our.. >. +. Solution.. To solve, we first get 0 on one side of the inequalit which ields + 0. We find the zeros of f() = + b solving + =0for. Factoring gives ( +)( ) = 0, so = or =. We place these values on the number line with 0 above them and choose test values in the intervals (, ( ),, ) and (, ). For the interval (, ), we choose = ; for (, ),wepick =0;andfor(, ), =. Evaluating the function at the three test values gives us f( ) = > 0, so we place (+) above (, ( ) ; f(0) = < 0, so ( ) goesabovetheinterval, ) ; and, f() = 7, whichmeans(+)isplacedabove(, ). Since we are solving + 0, we look for solutions to + < 0 as well as solutions for + =0. For + < 0, we need the intervals which we have a ( ). Checking the sign diagram, we see this is (, ). We know + =0when = and =, so our final answer is [, ]. To verif our solution graphicall, we refer to the original inequalit,. We let g() = and h() =. We are looking for the values where the graph of g is below that of h (the solution to g() <h()) as well as the points of intersection (the solutions to g() =h()). The graphs of g and h are given on the right with the sign chart on the left. We have to choose something in each interval. If ou don t like our choices, please feel free to choose different numbers. You ll get the same sign chart.

66 6 Linear and Quadratic Functions (+) 0 ( ) 0 (+) = 5 =. Once again, we re-write >as > 0 and we identif f() =. When we go to find the zeros of f, we find, to our chagrin, that the quadratic doesn t factor nicel. Hence, we resort to the quadratic formula to solve = 0, and arrive at =±. As before, these zeros divide the number line into three pieces. To help us decide on test values, we approimate 0. and+.. We choose =, = 0 and = as our test values and find f( ) =, which is (+); f(0) = which is ( ); and f() = which is (+) again. Our solution to > 0iswhere we have (+), so, in interval notation (, ) ( +, ). To check the inequalit > graphicall, we set g() = and h() =. We are looking for the values where the graph of g is above the graph of h. As before we present the graphs on the right and the sign chart on the left. (+) 0 ( ) 0 (+) + 0 = =. To solve +, as before, we solve + 0. Setting f() = +=0, we find the onl one zero of f, =. This one value divides the number line into two intervals, from which we choose = 0 and = as test values. We find f(0) = > 0and f() = > 0. Since we are looking for solutions to + 0, we are looking for values where +< 0aswellaswhere + = 0. Looking at our sign diagram, there are no places where +< 0 (there are no ( )), so our solution is onl = (where +=0). We write this as {}. Graphicall, we solve + b graphing g() = + and h() =. We are looking for the values where the graph of g is below the graph of h (for + < ) and where the two graphs intersect ( + = ). Notice that the line and the parabola touch at (, ), but the parabola is alwas above the line otherwise. In this case, we sa the line = is tangent to = + at (, ). Finding tangent lines to arbitrar functions is a fundamental problem solved, in general, with Calculus.

67 . Inequalities with Absolute Value and Quadratic Functions 7 (+) 0 (+) 0 = + =. To solve our last inequalit,, we re-write the absolute value using cases. For <, = ( ) =, soweget, or 0. Finding the zeros of f() =, weget = 0 and =. However, we are onl concerned with the portion of the number line where <, so the onl zero that we concern ourselves with is = 0. This divides the interval <intotwointervals: (, 0) and (0, ). We choose = and = as our test values. We find f( ) = and f ( ) = 5. Hence, our solution to 0for<is[0, ). Net, we turn our attention to the case. Here, =, so our original inequalit becomes, or 0. Setting g() =, we find the zeros of g to be = and =. Of these, onl = lies in the region, so we ignore =. Our test intervals are now [, ) and (, ). We choose = and = as our test values and find g() = andg() =. Hence, our solution to g() = 0, in this region is [, ). (+) 0 ( ) 0 ( ) 0 (+) Combining these into one sign diagram, we have that our solution is [0, ]. Graphicall, to check, we set h() = and i() = and look for the values where the graph of h isabovethethegraphofi (the solution of h() >i()) as well as the -coordinates of the intersection points of both graphs (where h() =i()). The combined sign chart is given on the left and the graphs are on the right. = (+) 0 ( ) 0 (+) 0 0 =

68 8 Linear and Quadratic Functions One of the classic applications of inequalities is the notion of tolerances. 5 Recall that for real numbers and c, thequantit c ma be interpreted as the distance from to c. Solving inequalities of the form c d for d 0 can then be interpreted as finding all numbers which lie within d units of c. Wecanthinkofthenumberd as a tolerance and our solutions as being within an accepted tolerance of c. We use this principle in the net eample. Eample..5. The area A (in square inches) of a square piece of particle board which measures inches on each side is A() =. Suppose a manufacturer needs to produce a inch b inch square piece of particle board as part of a home office desk kit. How close does the side of the piece of particle board need to be cut to inches to guarantee that the area of the piece is within a tolerance of 0.5 square inches of the target area of 576 square inches? Solution. Mathematicall, we epress the desire for the area A() tobewithin0.5 square inches of 576 as A Since A() =, we get , which is equivalent to One wa to proceed at this point is to solve the two inequalities and individuall using sign diagrams and then taking the intersection of the solution sets. While this wa will (eventuall) lead to the correct answer, we take this opportunit to showcase the increasing propert of the square root: if 0 a b, then a b. To use this propert, we proceed as follows (add 576 across the inequalities.) (take square roots.) ( = ) B Theorem., we find the solution to to be (, ] [ , ) and the solution to to be [ 576.5, ].Tosolve , we intersect these two sets to get [ 576.5, ] [ , 576.5]. Since represents a length, we discard the negative answers and get [ , 576.5]. This means that the side of the piece of particle board must be cut between and inches, a tolerance of (approimatel) inches of the target length of inches. Our last eample in the section demonstrates how inequalities can be used to describe regions in the plane, as we saw earlier in Section.. Eample..6. Sketch the following relations.. R = {(, ) :> }. S = {(, ) : }. T = {(, ) : < } 5 The underling concept of Calculus can be phrased in terms of tolerances, so this is well worth our attention.

69 . Inequalities with Absolute Value and Quadratic Functions 9 Solution.. The relation R consists of all points (, ) whose-coordinate is greater than. Ifwegraph =, then we want all of the points in the plane above the points on the graph. Dotting the graph of = as we have done before to indicate that the points on the graph itself are not in the relation, we get the shaded region below on the left.. For a point to be in S, its-coordinate must be less than or equal to the -coordinate on the parabola =. This is the set of all points below or on the parabola =. The graph of R The graph of S. Finall, the relation T takes the points whose -coordinates satisf both the conditions given in R and those of S. Thus we shade the region between = and =, keeping those points on the parabola, but not the points on =. To get an accurate graph, we need to find where these two graphs intersect, so we set =. Proceeding as before, breaking this equation into cases, we get =,. Graphing ields The graph of T

70 0 Linear and Quadratic Functions.. Eercises In Eercises -, solve the inequalit. Write our answer using interval notation > < < > 7. < 7 8. < < < < > < < > 5. > < <. The profit, in dollars, made b selling bottles of 00% All-Natural Certified Free-Trade Organic Sasquatch Tonic is given b P () = +5 00, for 0 5. How man bottles of tonic must be sold to make at least $50 in profit?. Suppose C() = 0 + 7, 0 represents the costs, in hundreds of dollars, to produce thousand pens. Find the number of pens which can be produced for no more than $ The temperature T, in degrees Fahrenheit, t hours after 6 AM is given b T (t) = t +8t+, for 0 t. When is it warmer than Fahrenheit?

71 . Inequalities with Absolute Value and Quadratic Functions 6. The height h in feet of a model rocket above the ground t seconds after lift-off is given b h(t) = 5t + 00t, for 0 t 0. When is the rocket at least 50 feet off the ground? Round our answer to two decimal places. 7. If a slingshot is used to shoot a marble straight up into the air from meters above the ground with an initial velocit of 0 meters per second, for what values of time t will the marble be over 5 meters above the ground? (Refer to Eercise 5 in Section. for assistance if needed.) Round our answers to two decimal places. 8. What temperature values in degrees Celsius are equivalent to the temperature range 50 F to 95 F? (Refer to Eercise 5 in Section. for assistance if needed.) In Eercises 9 -, write and solve an inequalit involving absolute values for the given statement. 9. Find all real numbers so that is within units of. 0. Find all real numbers so that is within units of.. Find all real numbers so that is within unit of.. Find all real numbers so that is at least 7 units awa from.. The surface area S of a cube with edge length is given b S() =6 for >0. Suppose the cubes our compan manufactures are supposed to have a surface area of eactl square centimeters, but the machines ou own are old and cannot alwas make a cube with the precise surface area desired. Write an inequalit using absolute value that sas the surface area of a given cube is no more than square centimeters awa (high or low) from the target of square centimeters. Solve the inequalit and write our answer using interval notation.. Suppose f is a function, L is a real number and ε is a positive number. Discuss with our classmates what the inequalit f() L <εmeans algebraicall and graphicall. 6 In Eercises 5-50, sketch the graph of the relation. 5. R = {(, ) : } 6. R = { (, ) :> + } 7. R = {(, ) : < +} 8. R = { (, ) : <+ } 9. R = {(, ) : << } 50. R = { (, ) : < } 5. Prove the second, third and fourth parts of Theorem.. 6 Understanding this tpe of inequalit is reall important in Calculus.

72 Linear and Quadratic Functions.. Answers. [, ]. (, ) ( 7 8 7, ). (, ). (, ] [, ) 5. No solution 6. (, ) 7. (, ] [6, ) 8. [, ) (5, 6] 9. [ 7, 6 ] 5 0. (, ) (, ). (, ] [6, ). (, 5). No Solution.. [ 7, 5 ] 5. (, 5 ) 6. (, ) 7. (, ] [, ) 8. (, ) (, ) 9. No solution 0. (, ). {}. No solution. [, ]. (0, ) ( ) ( ) ( 5., 6 + 6, (, ] [ ) ( 7, 0 0, ] [, ) 7, ] [ ) , 8. [ 7, + 7 ] [, ] 9. (, ) 0. (, ] {0} [, ). [ 6, ] [, ) ( ). (, ), + 7. P () 50 on [0, 5]. This means anwhere between 0 and 5 bottles of tonic need to be sold to earn at least $50 in profit.. C() on [, 8]. This means anwhere between 000 and 8000 pens can be produced and the cost will not eceed $ T (t) > on (8, 8+ ) (.7,.6), which corresponds to between 7: AM (.7 hours after 6 AM) to 8:8 PM (.6 hours after 6 AM.) However, since the model is valid onl for t, 0 t, we restrict our answer and find it is warmer than Fahrenheit from 7: AM to 6 PM.

73 . Inequalities with Absolute Value and Quadratic Functions 6. h(t) 50 on [0 5, ] [.9, 7.07]. This means the rocket is at least 50 feet off the ground between.9 and 7.07 seconds after lift off. 7. s(t) =.9t +0t +. s(t) > 5 on (approimatel) (.,.68). This means between. and.68 seconds after it is launched into the air, the marble is more than 5 feet off the ground. 8. From our previous work C(F )= 5 9 (F ) so 50 F 95 becomes 0 C 5. 9., [, 6] 0. +, [, ]., [, ] [, ]. 7, (, ] [, ). Solving S(), and disregarding the negative solutions ields [.550,.79]. The edge length must be within.550 and.79 centimeters. [, ]

74 Linear and Quadratic Functions

75 .5 Regression 5.5 Regression We have seen eamples alread in the tet where linear and quadratic functions are used to model a wide variet of real world phenomena ranging from production costs to the height of a projectile above the ground. In this section, we use some basic tools from statistical analsis to quantif linear and quadratic trends that we ma see in real world data in order to generate linear and quadratic models. Our goal is to give the reader an understanding of the basic processes involved, but we are quick to refer the reader to a more advanced course for a complete eposition of this material. Suppose we collected three data points: {(, ), (, ), (, )}. B plotting these points, we can clearl see that the do not lie along the same line. If we pick an two of the points, we can find a line containing both which completel misses the third, but our aim is to find a line which is in some sense close to all the points, even though it ma go through none of them. The wa we measure closeness in this case is to find the total squared error between the data points and the line. Consider our three data points and the line = +. For each of our data points, we find the vertical distance between the point and the line. To accomplish this, we need to find a point on the line directl above or below each data point - in other words, a point on the line with the same -coordinate as our data point. For eample, to find the point on the line directl below (, ), we plug =into = + andwegetthepoint(, ). Similarl, we get (, ) to correspond to (, ) and (, 5 ) for (, ). We find the total squared error E b taking the sum of the squares of the differences of the - coordinates of each data point and its corresponding point on the line. For the data and line above E =( ) +( ) + ( 5 ) = 9. Using advanced mathematical machiner, it is possible to find the line which results in the lowest value of E. This line is called the least squares regression line, or sometimes the line of best fit. The formula for the line of best fit requires notation we won t present until Chapter 9., so we will revisit it then. The graphing calculator can come to our assistance here, since it has a built-in feature to compute the regression line. We enter the data and perform the Linear Regression feature and we get and authors with more epertise in this area, Like Calculus and Linear Algebra

76 6 Linear and Quadratic Functions The calculator tells us that the line of best fit is = a + b wheretheslopeisa 0. and the -coordinate of the -intercept is b.8. (We will stick to using three decimal places for our approimations.) Using this line, we compute the total squared error for our data to be E.786. The value r is the correlation coefficient and is a measure of how close the data is to being on the same line. The closer r is to, the better the linear fit. Since r 0.7, this tells us that the line of best fit doesn t fit all that well - in other words, our data points aren t close to being linear. The value r is called the coefficient of determination and is also a measure of the goodness of fit. Plotting the data with its regression line results in the picture below. Our first eample looks at energ consumption in the US over the past 50 ears. Year Energ Usage, in Quads Eample.5.. Using the energ consumption data given above,. Plot the data using a graphing calculator. We refer the interested reader to a course in Statistics to eplore the significance of r and r. See this Department of Energ activit 5 The unit Quad is Quadrillion = 0 5 BTUs, which is enough heat to raise Lake Erie roughl F

77 .5 Regression 7. Find the least squares regression line and comment on the goodness of fit.. Interpret the slope of the line of best fit.. Use the regression line to predict the annual US energ consumption in the ear Use the regression line to predict when the annual consumption will reach 0 Quads. Solution.. Entering the data into the calculator gives The data certainl appears to be linear in nature.. Performing a linear regression produces We can tell both from the correlation coefficient as well as the graph that the regression line is a good fit to the data.. The slope of the regression line is a.87. To interpret this, recall that the slope is the rate of change of the -coordinates with respect to the -coordinates. Since the -coordinates represent the energ usage in Quads, and the -coordinates represent ears, a slope of positive.87 indicates an increase in annual energ usage at the rate of.87 Quads per ear.. To predict the energ needs in 0, we substitute = 0 into the equation of the line of best fit to get =.87(0) The predicted annual energ usage of the US in 0 is approimatel 6.8 Quads.

78 8 Linear and Quadratic Functions 5. To predict when the annual US energ usage will reach 0 Quads, we substitute = 0 into the equation of the line of best fit to get 0 = Solving for ields Since the regression line is increasing, we interpret this result as saing the annual usage in 05 won t et be 0 Quads, but that in 06, the demand will be more than 0 Quads. Our net eample gives us an opportunit to find a nonlinear model to fit the data. According to the National Weather Service, the predicted hourl temperatures for Painesville on March, 009 were given as summarized below. Time Temperature, F 0AM 7 AM 9 PM PM PM PM PM To enter this data into the calculator, we need to adjust the values, since just entering the numbers could cause confusion. (Do ou see wh?) We have a few options available to us. Perhaps the easiest is to convert the times into the hour clock time so that PM is, PM is, etc.. If we enter these data into the graphing calculator and plot the points we get While the beginning of the data looks linear, the temperature begins to fall in the afternoon hours. This sort of behavior reminds us of parabolas, and, sure enough, it is possible to find a parabola of best fit in the same wa we found a line of best fit. The process is called quadratic regression and its goal is to minimize the least square error of the data with their corresponding points on the parabola. The calculator has a built in feature for this as well which ields

79 .5 Regression 9 The coefficient of determination R seems reasonabl close to, and the graph visuall seems to be a decent fit. We use this model in our net eample. Eample.5.. Using the quadratic model for the temperature data above, predict the warmest temperature of the da. When will this occur? Solution. The maimum temperature will occur at the verte of the parabola. Recalling the Verte Formula, Equation., = b 9.6 a ( 0.).7. This corresponds to roughl : 5 PM. To find the temperature, we substitute =.7 into = to get.899, or.899 F. The results of the last eample should remind ou that regression models are just that, models. Our predicted warmest temperature was found to be.899 F, but our data sas it will warm to F. It s all well and good to observe trends and guess at a model, but a more thorough investigation into wh certain data should be linear or quadratic in nature is usuall in order - and that, most often, is the business of scientists.