Q (x 1, y 1 ) m = y 1 y 0
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1 . 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 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 ) and Q (, ) 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 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 ( ) P Q
3 . Linear Functions P 5. m, which is undefined 0 Q P 6. m 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.., we found the slope to be. We can 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 Linear and Quadratic Functions Using more formal notation, given points ( 0, 0 ) and (, ), we use the Greek letter delta to write 0 and 0. In most scientific circles, the smbol means change in. Hence, we ma write m, 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... At 6 AM, it is F; at 0 AM, it is 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 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. Mathematics is often used to describe, or model, real world phenomena. 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. We will discuss this more thoroughl in Section.5. 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 5 We have m 0 0 m ( 0 ) 0 0 m ( 0 ) m ( 0 ) + 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 m ( 0 ) + 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 ( ). 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 m ( 0 ) + 0 ( ( )) 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) is the equation m + b. Note that if we have slope m 0, we get the equation 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?) and 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 6 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, we get f() 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 (with slope m 0) and a -intercept of (0, b). Now think back to Section.7., specificall Definition.8 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() Solution.. f(). f(). 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()) (, ). f() f()
7 . Linear Functions 7. At first glance, the function f() does not fit the form in Definition. but after some rearranging we get f() +. We identif m 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, we get 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, we graph the line +. 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() 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, (, ), and 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 (wh?) and f() + + ( + ) ( + ) The following eample uses linear functions to model some basic economic relationships.
8 8 Linear and Quadratic Functions Eample..5. The cost, C, in dollars, to produce 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() to get C(0) 80(0) 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 set the cost, C(), equal to 5000, and solve for C() 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.5. In this contet, represents the number of PortaBos produced. It makes no sense to produce a negative quantit of game sstems. 5. To find C(0), we replace ever occurrence of with 0 in the formula for C() to get C(0) 80(0) This means it costs $50 to produce 0 PortaBos. The $50 is often called the fied or start-up cost of this venture. (What might contribute to this cost?) 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 9 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. In this case, C() is the cost in dollars, while measures the number of PortaBos so m C $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 of each sstem, p, in dollars. 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.5 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 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, we have m p We now have determined p().5 + b. To determine b, we can use our given data again. Using p(0) 0, we substitute 0 into p().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. Incidentall, this equation is sometimes called the price-demand 6 equation for this venture. 6 Or simpl the demand equation
10 0 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. Another constraint, then, is that the price, p() 0. Solving results in.5 50 or 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) 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, Solving, we get.5 00 or 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: f f(b) f(a) 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, or in other words, its rate of change over an interval is the same as its average rate of change.
11 . Linear Functions f() (b, f(b)) (a, f(a)) 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 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 rates of change, and this is one of the central themes of Calculus. 7 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. The revenue of 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. 7 Here we go again...
12 Linear and Quadratic Functions. Since R p, we substitute p() from Eample..6 to get R() ( ) Using Definition., we get the average rate of change is R R(50) R(0) 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 R R(00) R(50) In other words, for each additional PortaBo sold, the revenue increases b $5. Note 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 R R(50) R(00) 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, first introduced in Section.5. If we wish to compute the average rate of change of a function f over the interval [, + h], then we would have f f( + h) f() f( + h) f() ( + h) h As we have indicated, the rate of change of a function (average or otherwise) is of great importance in Calculus. 8 8 So, we are not torturing ou with these for nothing.
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