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1105G7sf11 - MACllOS Names K t 3 Group work 7 November 8...

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Unformatted text preview: MACllOS . Names: K t: 3 Group work 7 November 8, 2011 Exponential Functions in Terms of Percent Change In linear formulas, the amount of change stays constant, while in exponential formulas, the factor of change stays constant. The constant amount of change in linear formulas is called the slope, with a positive slope indicating increase and a negative slope indicating decrease. The constant factor of change in exponential formulas is called the growth/decay factor. A factor greater than i represents growth, and a factor between 0 and 1 represents decay. ' Another common way to describe exponential formulas is that the percent change stays constant. The difference between the growth/decay factor and the number 1 represents the rate at which the quantity will grow/decay. For example, a growth factor of a = 1.25 represents a 25% increase in whatever unit of time is relevant, and a decay factor of a = 0.75 represents a 25% decrease. This constant percent is called the growth/decay rate. If the exponential pattern represents growth we will describe the rate as positive, and if the pattern is decay we will describe the rate as negative. So for 25 0/0 growth we use R a 25% , and for 25% decay we use R fl —25% . This is the same as slope, where a positive slope represents linear increase and a negative slope represents linear decrease. In this ciass we will use R to represent the growth/decay rate in percent form, and r to represent the rate as a decimai. R R 2 35% <—) r = 0.35 _ 100 R = 100r r = 9.086 <—> n = 8.6% —) Ex: ,. The relevant formulas for moving back and forth between growth factors and growth rates are: 1' = a -— 1 and a = 1 + 1‘ Examples: Factors —) Rates (Use 1' = a—l) 2! =1.35 —> r = it Hi: 0.35 --) R a 35% (growth) a = 0.854 —) r = a —1 = “0.146 —) R = ~14.6% (decay) Examples: Rates —> Factors (Use a =1+r) R = 29.6% (growth) —> r = 0.296 —> a = 1 + r = 1.296 R = —14% (decay) —) r = —0.14 —) a = 1+ r a 0.86 Given that Q has an initial value of 100 itnd increases/decreases according to the following patterns. Derive appropriate linear or exponential formulas for eac . Let trepresent the input, and Q0 represent the initial value. If linear, the form will be Q = Q0 + mt . if exponential, the form will be Q m Q0(a)t . As t increases by l, the following happens to Q: l)Qincreasesby6. 4; linear, up”, -_-. Q, 1)Q= \00 Mai 2)Qdecreasesby6. a linear , Slope: ~69 2)Q= \OO " 6+ 3)Qincreasesbyafactorofd ..-, ex Pmmii 01L ) 0.: Q, 3)Q= moi-9‘: t 4)Qdecreasesbyafactoroflf6.-§ e95 Pmen'l'ioti ’ 4, .: 4; 4)Q= \OO(_é){—‘ "30%)” 5)Qincreasesby6%. -) Exponential, 0L:- |+ .05 in“, 5)Q= lOO (L06)? 6)Qdecreasesby6%. -7 e1: PfinmtimQ.’0~'=-l-.o 6 : 0.qq6)Q= lofljfll‘m: B'h'HO rW——“*-~—l Suppose a population has been growing exponentially since 1985. In 1990 the population was 20, 000 people, and m 2000 the population was 30, 000 people. a) What is the annual growth factor? (4 decimal place accuracy) _ a) I. I I L“ H vr lass mac aooo ,OF—w + o 5 1.5 0“" 30000 1 0L; “l P ‘Po 90000 “50000 b) What is the annual growth rate? (2 decimal place accuracy) b) Ll . I Ll 74 r=a~1 . r: Ifloulu -1=o cum 2000*“ e H.012, c) What was the population in 1985? (nearest whole number) 0) l E , 3 Q 9 P: 'P (a)4L aoooo: Po Cl mobs ' Macaw)" tw i233"; " ”:33? d) Construct an exponential formula that models this population. Your answer will be of the form P = P0 (a)I , where t represents the number of years that have elapsed since £985. 'l: Pa: “0338 (DP: $1.le LOW") 0c '- l- OH 1H 6) Use your formula to predict the population in 2015. (nearest whole number) I e) l 3 4 Has: {=0 30 - 3015: {=- 30 P: [63380,0'401) 3- SS) Bq f) Draw a graph that represents the population of this city from years 1985-2015. Clearly label the points (with coordinates) that correspond to years 1985, 1990, 2000, and 2015. 10 15 ‘ 20 25 less Mo 3000 77 _ 30:5 g) Change your formula so that the variable t measures years since 2000. J: l’o= 30000 ' . g) l’= 300000.0m) e. Lol’lll‘i ...
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