ࡱ> B`EDCq` ,vbjbjqPqP 7::(%NNN8OOPRT"2T2T2T U U U$ZhŒ;^ U U;^;^2T2T^ddd;^2T2Td;^ddRÁ2TP P-ʆUN^Džt0ՂPbP<P$ UWdY{[ U U UcX U U U;^;^;^;^d5&:&: MATH 1001 (Quantitative Skills and Reasoning) Gordon College, Barnesville, GA. Quadratic Functions and Modeling In this unit, we will study quadratic functions and the relationships for which they provide suitable models. An important application of such functions is to describe the trajectory, or path, of an object near the surface of the earth when the only force acting on the object is gravitational attraction. What happens when you toss a ball straight up into the air? What about an outfielder on a baseball team throwing a ball into the infield? If air resistance and outside forces are negligible, what is the mathematical model for the relationship between time and height of the ball? Definition: Quadratic Functions A quadratic function is one of the form  EMBED Equation.3  where a, b, and c are real numbers with a `" 0. The graph of a quadratic function is called a parabola and its shape resembles that of the graph in each of the following two examples. Example 1 Figure 1 shows the graph of the quadratic function  EMBED Equation.3 .  figure 1 Observe that there is a lowest point V(2, "3) on the graph in figure 1. The point V is called the vertex of the parabola. Example 2 Figure 2 shows the graph of the quadratic function  EMBED Equation.3 .  figure 2 Again, observe that there is a highest point V(1, 5) on the graph in figure 2. This point V is also the vertex of the parabola. By completing the square, the quadratic function (example 1)  EMBED Equation.3  can be written as  EMBED Equation.3 ; the quadratic function (example 2)  EMBED Equation.3  can be written as  EMBED Equation.3 . Note that the parabola in example 1 opens upward, with vertex V(2, "3) and a vertical axis of symmetry x = 2. The parabola in example 2 opens downward with vertex V(1, 5) and a vertical axis of symmetry x = 1. Standard Form of a Quadratic Function Whose Graph (a Parabola) has a Vertical Axis The graph of  EMBED Equation.3   EMBED Equation.3  is a parabola that has vertex V(h, k) and a vertical axis. The parabola opens upward if a > 0 or downward if a < 0. If a > 0, the vertex V(h, k) is the lowest point on the parabola, and the function f has a minimum value f (h) = k. If a < 0, the vertex V(h, k) is the highest point on the parabola, and the function f has a maximum value f (h) = k. Quadratic Models and Equations Quadratic Population Models:  EMBED Equation.3 . Here, we denote the independent variable by t (time) instead of x, and the constant c by  EMBED Equation.3  because substitution of t = 0 yields  EMBED Equation.3 . We refer to  EMBED Equation.3  as the initial population. Example 3 Suppose that the future population of Stockton City t years after January 1, 2000 is described (in thousands) by the quadratic model  EMBED Equation.3 . What is the population of Stockton City on January 1, 2007? In what month of what calendar year will the population of Stockton City reach 180 thousand? Solution We only need to substitute t = 7 in the population function P(t) and calculate  EMBED Equation.3  thousand. We need to find the value of t when P(t) = 180. That is we need to solve the equation,  EMBED Equation.3  Equation (i) is an example of a quadratic equation which we can solve in a variety of ways. First, by graphing both sides of the equation (figure 3):  EMBED Equation.3   figure 3 The line  EMBED Equation.3  and the parabola  EMBED Equation.3  are shown in the calculator window "100 d" x d" 80, "100 d" y d" 300. To solve equation (i), we find the x-coordinate of the intersection point in the first quadrant. The negative solution of the intersection point in the second quadrant would be in the past. Figure 3 indicates that we have already used the intersection-finding feature to locate the point (14.047, 180). Hence, t = 14.047 years = 14 years + (0.047 x 12) months = 14 years + 0.56 month. Thus Stockton City should reach a population of 180 thousand 14 years, 0.56 month after January 1, 2000. Hence, sometime during January 2014. Alternatively, by using the Quadratic Formula: The quadratic equation  EMBED Equation.3  has solutions  EMBED Equation.3  . To use the quadratic formula, we first write equation (i) in the form  EMBED Equation.3  Here, a = 0.07, b = 4, c = "70, giving  EMBED Equation.3  The negative solution would be in the past. So, we only accept the positive solution, t = 14.047 years = 14 years, 0.56 month. The Position Function (Model) of a Particle Moving Vertically: If an object is projected straight upward at time t = 0 from a point  EMBED Equation.3 feet above ground, with an initial velocity  EMBED Equation.3 ft/sec, then its height above ground after t seconds is given by  EMBED Equation.3 . Example 4 A projectile is fired vertically upward from a height of 600 feet above the ground, with an initial velocity of 803 ft.sec. (a) Write a quadratic model for its height h(t) in feet above the ground after t seconds. (b) During what time interval will the projectile be more than 5000 feet above the ground? (c) How long will the projectile be in flight? Solution (a) Given  EMBED Equation.3 ft and  EMBED Equation.3 ft/sec,  EMBED Equation.3  (b) We need to find the values of t for which the height y(t) > 5000 feet. That is  EMBED Equation.3  We solve the inequality (i) by graphing (figure 4) both the parabola  EMBED Equation.3  and the line  EMBED Equation.3 , in the calculator window "20 d" x d" 70 and "2000 d" y d" 12000.  figure 4 The parabola in figure 4 shows the height of the projectile at time t and the horizontal line is the graph of a height of 5000 feet. The projectile is more than 5000 feet above the ground when the graph of the parabola is above the horizontal line. Using the intersection-finding feature of the calculator, we find the approximate intersection points to be t = 6.3 sec and t = 43.9 sec. Hence, the projectile is above 5000 feet when 6.3 sec < t < 43.9 sec. (c) The projectile will be in flight until its height h(t) = 0. This corresponds to the x-intercept (not the origin) in figure 4. Using the root or zero feature of the calculator, we obtain t = 50.9 sec. EXERCISES 1. The population (in thousands) for Alpha City, t years after January 1, 2004 is modeled by the quadratic function  EMBED Equation.3  In what month of what year does Alpha Citys population reach twice its initial (1/1/2004) population? 2. The population (in thousands) for Beta City, t years after January 1, 2005 is modeled by the quadratic function  EMBED Equation.3  How long will it take Beta Citys population to reach 350 thousand? 3. The population (in thousands) for Gamma City, t years after January 1, 2002 is modeled by the quadratic function  EMBED Equation.3  How long will it take Gamma Citys population to reach 500 thousand? 4. The population (in thousands) for Delta City, t years after January 1, 2003 is modeled by the quadratic function  EMBED Equation.3  In what month of what year does Alpha Citys population reach twice its initial (1/1/2003) population? 5. The population (in thousands) for Omega City, t years after January 1, 2002 is modeled by the quadratic function  EMBED Equation.3  In what month of what year does Alpha Citys population reach 200 thousand? 6. A ball is thrown straight up, from ground zero, with an initial velocity of 48 feet per second. Find the maximum height attained by the ball and the time it takes for the ball to return to ground zero. 7. From the top of a 48 feet tall building, a ball is thrown straight up with an initial velocity of 32 feet per second. 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Iowa City, IA t (years)1970198019902000P (people)46,85050,50859,73562,220 12. Arizona t (years)20002001200220032004P (thous)5,1315,3205,4735,5815,744 13. Florida t (years)19701980199020002004P (thous)6,7899,74612,93815,98217,346 14. Georgia t (years)19701980199020002004P (thous)4,5905,4636,4788,1868,825 15. Nevada t (years)19701980199020002004P (thous)4898011,2021,9982,315 16. U.S. t (years)19701980199020002004P (millions)203227249281294 Answers to Exercises 1. February, 2013. 2. 8 years, 143 days. 3. 6 years, 183 days. 4. February, 2011. 5. May, 2014. 6. 36 feet, 3 sec. 7. 64 feet, 3 sec. 8. 5 sec. 9. 7.75 sec. 10. 4.33 sec. 11.  EMBED Equation.3  12.  EMBED Equation.3  13.  EMBED Equation.3  14.  EMBED Equation.3  15.  EMBED Equation.3  16.  EMBED Equation.3  Note: Answers to questions 11-16 are approximate values. References 1. Elementary Mathematical Modeling: Functions and Graphs. Mary Ellen Davis & C. Henry Edwards. Prentice Hall, 2001. 2. Intermediate Algebra. R. David Gustafson & Peter D. Frisk. Thomson (Brooks/Cole), 2005. 3.  HYPERLINK "http://www.fairus.org" www.fairus.org 4.  HYPERLINK "http://www.gpec.org" www.gpec.org 5.  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0 Y 2 =180 FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native s[_1177146921Tc[FʆU"ʆUOle uCompObjZ\vfObjInfo]xEquation Native y_1177148238Yr`F"ʆU"ʆUOle |2j Y 1 =110+4x+0.07x 2  FMicrosoft Equation 3.0 DS Equation Equation.39qCompObj_a}fObjInfobEquation Native _1177148171eF"ʆUlʆU2Ll ax 2 +bx+c=0,a`"0 FMicrosoft Equation 3.0 DS Equation Equation.39q20 { x=Ole CompObjdffObjInfogEquation Native "b b 2 "4ac  2a FMicrosoft Equation 3.0 DS Equation Equation.39q2|A 0.07t 2 +4t"70_1177148382jFlʆU ʆUOle CompObjikfObjInfolEquation Native _1177236386oF ʆU ʆUOle CompObjnpf=0""""(ii) FMicrosoft Equation 3.0 DS Equation Equation.39q2($f t="4 4 2 "ObjInfoqEquation Native _1177149496hwtF ʆU' ʆUOle 4(0.07)("70)  2(0.07)="4 35.6  0.14=14.046954or"71.189811 FMicrosoft Equation 3.0 DS Equation Equation.39qCompObjsufObjInfovEquation Native :_1177149560yF' ʆU' ʆU2PLg y 0 FMicrosoft Equation 3.0 DS Equation Equation.39q2Qr v 0Ole CompObjxzfObjInfo{Equation Native :_1177149695^~F' ʆUʆUOle CompObj}fObjInfo FMicrosoft Equation 3.0 DS Equation Equation.39q2 y(t)="16t 2 +v 0 t+y 0Equation Native _1177150828FʆU ʆUOle CompObjf FMicrosoft Equation 3.0 DS Equation Equation.39q27 A y 0 =600 FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfoEquation Native S_1177150827F ʆU ʆUOle CompObjfObjInfoEquation Native S_1177150862F ʆUSʆU27/ v 0 =803 FMicrosoft Equation 3.0 DS Equation Equation.39q2nA y(t)=Ole CompObjfObjInfoEquation Native "16t 2 +803t+600. FMicrosoft Equation 3.0 DS Equation Equation.39q2 @ "16t 2 +803t+_1177151101FSʆUʆUOle CompObjfObjInfoEquation Native _1177152799FʆUʆUOle CompObjf600>5000""""(i) FMicrosoft Equation 3.0 DS Equation Equation.39q2}A Y 1 ="16t 2 +ObjInfoEquation Native _1177152865FʆU5ʆUOle 803t+600 FMicrosoft Equation 3.0 DS Equation Equation.39q2CXU Y 2 =5000CompObjfObjInfoEquation Native __1177157316F5ʆUʆUOle CompObjfObjInfoEquation Native  FMicrosoft Equation 3.0 DS Equation Equation.39q2@P P(t)=0.3t 2 +6t+80. FMicrosoft Equation 3.0 DS Eq_1177157632FʆUʆUOle CompObjfObjInfouation Equation.39q2P  P(t)=0.7t 2 +12t+200. FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native _1177157857FʆUʆUOle CompObjfObjInfoEquation Native _1177158038FʆUp:ʆUOle 2 0 P(t)=1.5t 2 +21t+300. FMicrosoft Equation 3.0 DS Equation Equation.39qCompObjfObjInfoEquation Native _1177158306Fp:ʆUp:ʆU2 mA P(t)=0.5t 2 +7t+90. FMicrosoft Equation 3.0 DS Equation Equation.39q2pmA P(t)=Ole CompObjfObjInfoEquation Native 0.25t 2 +5t+100. FMicrosoft Equation 3.0 DS Equation Equation.39q2@\ P(t)=46,963+807t"_1177234560Fp!ʆUp$ʆUOle CompObjfObjInfoEquation Native /_1177234383Fp$ʆU`f&ʆUOle CompObjf    #$%&),-./12345679:;<=>?@ABCE9.5t 2 ;319people;63,000people. FMicrosoft Equation 3.0 DS Equation Equation.39q2 P(t)=ObjInfoEquation Native _1177234892F`f&ʆU`f&ʆUOle  5,139+176t"6.9t 2 ;14thous;6,030thous. FMicrosoft Equation 3.0 DS Equation Equation.39qCompObj fObjInfoEquation Native #_1177234905F`f&ʆU`(ʆU2# P(t)=6,777+298t+0.37t 2 ;44thous;18,300thous. FMicrosoft Equation 3.0 DS Equation Equation.39qOle CompObjfObjInfoEquation Native 24& P(t)=4,610+59t+1.93t 2 ;59thous;9,400thous. 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