Chapter Three QUADRATIC FUNCTIONS

Transcription

1 Chapter Three QUADRATIC FUNCTIONS Contents 3.1 Introduction to the Family of Quadratic Functions Finding the Zeros of a Quadratic Function Concavity and Rates of Change for Quadratic Functions Finding a Formula From the Zeros and Vertical Intercept Formulas for Quadratic Functions The Verte of a Parabola The Verte Form of a Quadratic Function Finding a Formula Given the Verte and Another Point Modeling with Quadratic Functions REVIEW PROBLEMS STRENGTHEN YOUR UNDERSTANDING 130 Skills Refresher for CHAPTER 3: QUADRATIC EQUATIONS Skills for Factoring Epanding an Epression Factoring Removing a Common Factor Grouping Terms Factoring Quadratics Perfect Squares and the Difference of Squares Solving Quadratic Equations Solving by Factoring Solving with the Quadratic Formula. 134 Completing the Square Visualizing Completing the Square Deriving the Quadratic Formula

2 116 Chapter Three QUADRATIC FUNCTIONS 3.1 INTRODUCTION TO THE FAMILY OF QUADRATIC FUNCTIONS A baseball is popped straight up by a batter. The height of the ball above the ground is given by the function y = ft) = 16t 2 +47t+3, where t is time in seconds after the ball leaves the bat and y is in feet. See Figure 3.1. Note that the path of the ball is straight up and down, although the graph of height against time is a curve. The ball goes up fast at first and then more slowly because of gravity; thus the graph of its height as a function of time is concave down. y feet) y = ft) = 16t 2 +47t+3 Initial height 3 3 t seconds) Figure 3.1: The height of a ball t seconds after being popped up. Note: This graph does not represent the ball s path.) The baseball height function is an eample of a quadratic function whose standard form is y = a 2 +b+c. The graph of a quadratic function is called a parabola. Notice that the function in Figure 3.1 is concave down and has a maimum corresponding to the time at which the ball stops rising and begins to fall back to the earth. The maimum point on the parabola is called the verte. The intersection of the parabola with the horizontal ais gives the times when the height of the ball is zero; these times are the zeros of the height function. The horizontal intercepts of a graph occur at the zeros of the function. In this section we eamine the zeros and concavity of quadratic functions, and in the net section we see how to find the verte. Finding the Zeros of a Quadratic Function A natural question to ask is when the ball hits the ground. The graph suggests that y = 0 when t is approimately 3. In symbols, the question is: For what values) of t does ft) = 0? These are the zeros of the function. It is easy to find the zeros of a quadratic function if it is epressed in factored form, q) = a r) s), where a, r,s are constants, a 0. Then r and s are zeros of the function q. The baseball function factors as y = 116t+1)t 3), so the zeros of f are t = 3 and t = 1/16. We are interested in positive values of t, so the ball hits the ground 3 seconds after it was hit. For more on factoring, see the Skills Review on page 131.) Eample 1 Find the zeros of f) = 2 6. To find the zeros, setf) = 0 and solve for by factoring: 2 6 = 0 3)+2) = 0. Thus the zeros are = 3 and = 2.

3 3.1 INTRODUCTION TO THE FAMILY OF QUADRATIC FUNCTIONS 117 We can also find the zeros of a quadratic function by using the quadratic formula. See the Skills Review on page 131 to review the quadratic formula.) Eample 2 Find the zeros of f) = 2 6 by using the quadratic formula. We solve the equation 2 6 = 0. For this equation, a = 1, b = 1, and c = 6. Thus = b± b 2 4ac 2a = 1)± 1) 2 41) 6) 21) = 1± 25 = 3 or 2. 2 The zeros are = 3 and = 2, the same as we found by factoring. Since the zeros of a function occur at the horizontal intercepts of its graph, quadratic functions without horizontal intercepts such as in the net eample) have no zeros. Eample 3 Figure 3.2 shows a graph of h) = What happens if we try to use algebra to find the zeros of h? y y intercept 0, 2) h) Figure 3.2: Zeros of h) = 2 /2) 2? To find the zeros, we solve the equation = = 2 2 = 4 = ± 4. Since 4 is not a real number, there are no real solutions, sohhas no real zeros. This corresponds to the fact that the graph of h in Figure 3.2 does not cross the-ais. Concavity and Rates of Change for Quadratic Functions A quadratic function has a graph that is either concave up or concave down. Recall that for a graph that is concave up, rates of change increase as we move right. For a graph that is concave down, rates of change decrease as we move right. Eample 4 Let f) = 2. Find the average rate of change of f over the four consecutive intervals of length 2 between = 4 and = 4. What do these rates tell you about the concavity of the graph of f?

4 118 Chapter Three QUADRATIC FUNCTIONS Between = 0 and = 2, we have Average rate of change of f Between = 2 and = 4, we have Average rate of change of f = f2) f0) 2 0 = f4) f2) 4 2 = = 2. = = 6. Similarly, between = 4 and = 2, we can show the average rate of change is 6; between = 2 and = 0, the rate of change is 2. See Figure 3.3. Since the rates of change are increasing, the graph is concave up. Slope= 6 Slope= 6 f) = 2 Slope= 2 Slope= Figure 3.3: Rate of change and concavity of f) = 2 Eample 5 A high-diver jumps off a 10-meter platform. For t in seconds after the diver leaves the platform until she hits the water, her height h in meters above the water is given by h = ft) = 4.9t 2 +8t+10. The graph of this function is shown in Figure 3.4. a) Estimate and interpret the domain and range of the function, and the intercepts of the graph. b) Identify the concavity. h meters) ft) t seconds) Figure 3.4: Height of diver above water as a function of time a) The diver enters the water when her height above the water is 0. This occurs when h = ft) = 4.9t 2 +8t+10 = 0. Using the quadratic formula to solve this equation, we find the only positive solution is t = seconds. The domain is the interval of time the diver is in the air, which is approimately 0 t To find the range of f, we look for the largest and smallest outputs for h. From the graph, the diver s maimum height appears to occur at about t = 1, so we estimate the largest output value for f to be about f1) = = 13.1 meters. Thus, the range of f is approimately 0 ft) Methods to find the eact maimum height are in Section 3.2.

5 The vertical intercept of the graph is 3.1 INTRODUCTION TO THE FAMILY OF QUADRATIC FUNCTIONS 119 f0) = = 10 meters. The diver s initial height is 10 meters the height of the diving platform). The horizontal intercept is the point estimated earlier where ft) = 0. This intercept, t = seconds, gives the time after leaving the platform that the diver enters the water. b) In Figure 3.4, we see that the graph is bending downward over its entire domain, so it is concave down. This is reflected in Table 3.1, where the rate of change, h/ t, is decreasing. Table 3.1 Slope of ft) = 4.9t 2 +8t+10 t sec) h meters) Rate of change h/ t The previous two eamples illustrate that the concavity of the graph of y = a 2 + b + c is determined by the sign of the coefficient a: Ifa > 0, the parabola is concave up. Ifa < 0, the parabola is concave down. Eample 6 Determine the concavity of the graphs of the following quadratic functions: a) y = b) y = 3 2+1) 4) a) Since the coefficient of 2 is negative, this function has a graph which is concave down. b) Epanding we have y = 3 2+1) 4) = ) = Since the coefficient of 2 is positive, the graph of this function is concave up. Finding a Formula From the Zeros and Vertical Intercept If we know the zeros and the vertical intercept of a quadratic function, we can use the factored form to find a formula for the function. Eample 7 Find the equation of the parabola in Figure 3.5 using the factored form. f) Figure 3.5: Finding a formula for a quadratic from the zeros

6 120 Chapter Three QUADRATIC FUNCTIONS Since the parabola has -intercepts at = 1 and = 3, its formula can be written as f) = a 1) 3). To find a, we substitute = 0,y = 6, giving 6 = a 3 a = 2. Thus, the formula is f) = 2 1) 3). Multiplying out gives f) = Formulas for Quadratic Functions The function f in Eample 7 can be written in at least two different ways. The standard form f) = shows that the parabola opens upward, since the coefficient of 2 is positive; the constant 6 gives the vertical intercept. The factored form f) = 2 1) 3) shows that the parabola crosses the-ais at = 1 and = 3. In general, we have the following: The graph of a quadratic function is a parabola. The standard form for a quadratic function is y = a 2 +b+c, where a, b, c are constants, a 0. The parabola is concave up opens upward) ifa > 0 or concave down opens downward) ifa < 0, and intersects they-ais at c. The factored form, when it eists, is y = a r) s), where a, r, s are constants, a 0. The parabola intersects the-ais at = r and = s. In the net section we look at another form for quadratic functions which shows how to find the verte of the graph. Eercises and Problems for Section 3.1 Skill Refresher Multiply and write the epressions in Eercises S1 S2 without parentheses. Gather like terms. S1. t 2 +1 ) 50t 25t ) 2t S2. A 2 B 2) 2 For Eercises S3 S8, factor completely if possible. S3. u 2 2u S S S6. s+2t) 2 4p 2 S S8. y 3 y 2 12y Solve the equations in Eercises S9 S10. S = 0 S10. 2w 2 +w 10 = 0

7 3.1 INTRODUCTION TO THE FAMILY OF QUADRATIC FUNCTIONS 121 Eercises Are the functions in Eercises 1 8 quadratic? If so, write the function in the formf) = a 2 +b+c. 1. gt) = 3t 2) f) = 2 3)5 ) 3. wn) = nn 3)n 2) n 2 n 8) 4. ht) = 16t 3)t+1) 5. Rq) = 1 q 2q2 +1) 2 6. K) = Tn) = 5+ 3n 4 n 4 8. rv) = v v 3 +πv Find the zeros of Qr) = 2r 2 6r 36 by factoring. 10. Find the zeros ofq) = using the quadratic formula. 11. Solve for using the quadratic formula and demonstrate your solution graphically: a) = 32 b) = In Eercises 12 19, find the zeros if any) of the function algebraically. 12. y = 2 )3 2) 13. y = y = Nt) = t 2 7t y = Qr) = 2r 2 6r y = y = 12 In Eercises 20 23, is the graph of the quadratic function concave up or concave down? 20. y = y = 5 3) y = 31 2) 4) 23. y = 25 1)4 ) Problems 24. Use the quadratic formula to find the time at which the baseball in Figure 3.1 on page 116 hits the ground. In Problems 25 26, find a formula for the quadratic function whose graph has the given properties. 25. Ay-intercept of y = 7 and the only zero at = Ay-intercept of y = 7 and -intercepts at = 1, Is there a quadratic function with zeros = 1, = 2 and = 3? 28. Graph a quadratic function which has all the following properties: concave up, y-intercept is 6, zeros at = 2 and = Can you graph a quadratic function which has all the following properties: concave down, y-intercept is 6, zeros at = 2 and = 3? 30. Determine the concavity of the graph of f) = 4 2 between = 1 and = 5 by calculating average rates of change over intervals of length The graph of a quadratic function, f), passes through the points 1,2), 3,4) and 5,2). Is the graph of f) concave up or concave down? Give a reason for your answer. 32. The quadratic function, f), has no zeros and its graph passes through the point1,1). Is the graph off) concave up or concave down? Give a reason for your answer. 33. Without a calculator, graph the following function by factoring and plotting zeros: y = 4c+ 2 +4c 2 for c > Without a calculator, graphy = by factoring and plotting zeros. In Problems 35 36, find a formula for the parabola , 0) 0, 1) y 36. 3,0) 6, 0) y 0,5) 2,0) 37. Using the factored form, find the formula for the parabola whose zeros are = 1 and = 5, and which passes through the point 2,6). 38. Find two different quadratic functions with zeros = 1, = 2.

8 122 Chapter Three QUADRATIC FUNCTIONS 39. Writey = )4 20) in the form y = k r) s) and give the values of k,r,s. 40. A ball is thrown into the air. Its height in feet)tseconds later is given byht) = 80t 16t 2. a) Evaluate and interpret h2). b) Solve the equation ht) = 80. Interpret your solutions and illustrate them on a graph of ht). 41. A snowboarder slides up from the bottom of a half-pipe and comes down again, sliding with little resistance on the snow. Her height above the top edge of the pipetseconds after starting up the side is 4.9t 2 +14t 5 meters. a) What is her height at t = 0? b) After how many seconds does she reach the top edge? Return to the edge of the pipe? c) How long is she in the air? 42. Let Vt) = t 2 4t + 4 represent the velocity after t seconds of an object in meters per second. a) What is the object s initial velocity? b) When is the object not moving? c) Identify the concavity of the velocity graph. 43. Table 3.2 gives the approimate number of cell phone subscribers, S, in millions, in the US. 1 If is in years since 2005, show that this data can be approimated by the quadratic functionp) = What does this model predict for the year 2015? How good is this model for predicting the future? Table 3.2 Year S millions) Letf) = 2 and g) = a) Graph f and g in the window 10 10, 10 y 10. How are the two graphs similar? How are they different? b) Graph f and g in the window 10 10, 10 y 100. Why do the two graphs appear more similar on this window than on the window from part a)? c) Graph f and g in the window 20 20, 10 y 400, the window 50 50, 10 y 2500, and the window , 2500 y 250,000. Describe the change in appearance of f and g on these three successive windows. 45. A relief package is dropped from a moving airplane. The package has an initial forward horizontal velocity and follows a quadratic graph path instead of dropping straight down). Figure 3.6 shows the height of the package, h, in km, as a function of the horizontal distance, d, in meters, as it drops. a) From what height was the package released? b) How far away from the spot above which it was released does the package hit the ground? c) Write a formula for hd). [Hint: The package starts falling at the highest point on the graph.] h, height km) Airplane Package dropped here Figure d, horizontal distance meters) 3.2 THE VERTEX OF A PARABOLA In Section 3.1, we looked at the eample of a baseball popped upward by a batter. The height of the ball above the ground is given by the quadratic function y = ft) = 16t 2 +47t+3, where t is time in seconds after the ball leaves the bat, and y is in feet. See Figure 3.7. The point on the graph with the largest y value appears to be approimately 1.5,37.5). See Eample 5 on page 126 for the eact values.) This means that the baseball reaches its maimum height of about 37.5 feet about 1.5 seconds after being hit. The maimum point on the parabola is called the verte. For quadratic functions the verte shows where the function reaches either its maimum value or, in the case of a concave-up parabola, its minimum value. The verte of a parabola can be determined eactly if the quadratic function is written in the 1 Last accessed January 13, 2014.

9 y feet) verte , 37.5) y = ft) = 16t 2 +47t THE VERTEX OF A PARABOLA 123 Initial height 3 3 t seconds) form Figure 3.7: Height of baseball at timet y = a h) 2 +k. This form arises when we shift the parabola y = 2 horizontally by h and vertically by k, moving the verte from 0,0) toh,k). Eample 1 a) Sketch f) = +3) 2 4, and indicate the verte. b) Estimate the coordinates of the verte from the graph. c) Eplain how the formula forf can be used to obtain the minimum of f. d) Eplain how vertical and horizontal shifts can be used to obtain the coordinates of the verte. a) Figure 3.8 shows a sketch of f); the verte gives the minimum value of the function. b) The verte of f appears to be about at the point 3, 4). c) Note that + 3) 2 is always positive or zero, so + 3) 2 takes on its smallest value when +3 = 0, that is, at = 3. At this point +3) 2 4 takes on its smallest value, f 3) = 3+3) 2 4 = 0 4 = 4. Thus, we see that the verte is eactly at the point 3, 4) and the minimum is 4. d) The functionf) = +3) 2 4 is a transformation of the functionf) = 2, with an inside change of 3 and an outside change of 4. This indicates that f) = 2 has been shifted 3 units to the left and 4 units down. Thus, the verte is 3, 4). f) Figure 3.8: A graph off) = +3) 2 4 Notice that if we select-values that are equally spaced to the left and the right of the verte, the y-values of the function are equal. For eample, f 2) = f 4) = 3. The graph is symmetric about a vertical line that passes through the verte. This line is called the ais of symmetry. The function in Eample 1 has ais of symmetry = 3. The Verte Form of a Quadratic Function Writing the function f in Eample 1 in the form f) = +3) 2 4 enabled us to find the verte of the graph and the location and value for the minimum of the function.

10 124 Chapter Three QUADRATIC FUNCTIONS In general, we have the following: The verte form of a quadratic function is y = a h) 2 +k, where a, h, k are constants, a 0. The graph of this quadratic function has verte h,k) and ais of symmetry = h. A quadratic function can always be epressed in both standard form and verte form. Eample 2 a) Convert the quadratic f) = +3) 2 4 in Eample 1 to standard form. b) Eplain how the vertical intercept can be found from the standard form. c) Find the zeros of f. d) Eplain how the ais of symmetry of the parabola is related to the zeros. a) Epanding +3) 2 and gathering like terms converts f to standard form: f) = = Since the coefficient of 2 isa = 1, the parabola opens upward. b) The vertical intercept isf0) = 5. In the standard formy = a 2 +b+c, they-intercept is the constant c. c) To find the zeros, we factor f and write it as f) = +1)+5). This form shows that the zeros are = 1 and = 5. d) The ais of symmetry is = 3, which is the vertical line through the midpoint of the zeros: 1 5)/2 = 3. As we see in Eample 2, if a quadratic can be written in factored form, its ais of symmetry is at the midpoint of the zeros. To convert from verte to standard form, we epand the squared term and gather like terms. In the net eample, we convert from from standard form to verte form by completing the square. 2 Eample 3 Put each quadratic function into verte form by completing the square and then graph it. a) s) = b) t) = a) To complete the square, find the square of half of the coefficient of the -term, 6/2) 2 = 9. Add and subtract this number after the -term: so s) = , }{{} Perfect square s) = 3) 2 1. The verte of s is 3, 1) and the ais of symmetry is the vertical line = 3. The parabola opens upward. See Figure A more detailed eplanation of this method is in the Skills Review on page 136.

11 3.2 THE VERTEX OF A PARABOLA 125 = 3/ = 3 3 Figure 3.9: s) = Figure 3.10: t) = b) To complete the square, first factor out 4, the coefficient of 2, giving t) = ). Now add and subtract the square of half the coefficient of the -term, 3/2) 2 = 9/4, inside the parentheses. This gives t) = ) }{{ 4 } 4 +2 Perfect square t) = ) ) t) = ) The verte of t is 3/2,1), the ais of symmetry is = 3/2, and the parabola opens downward. See Figure Finding a Formula Given the Verte and Another Point If we know the verte of a quadratic function and one other point, we can use the verte form to find its formula. Eample 4 Find the formula for the quadratic function graphed in Figure y m) 0,5) 3, 2) = 3 Figure 3.11: Finding a formula for a quadratic from the verte Since the verte is given, we use the formm) = a h) 2 +k to finda, h, andk. The verte is 3,2), soh = 3 and k = 2. Thus, so m) = a 3)) 2 +2, m) = a+3) 2 +2.

12 126 Chapter Three QUADRATIC FUNCTIONS To find a, use the y-intercept 0,5). Substitute = 0 and y = m0) = 5 into the formula for m) and solve for a: Thus, the formula is 5 = a0+3) = 9a a = 1 3. m) = )2 +2. If we want the formula in standard form, we multiply out: m) = Modeling with Quadratic Functions Eample 5 In applications, it is often useful to find the maimum or minimum value of a quadratic function. The net eample returns to the baseball that started this chapter. Problem 25 on page 128 asks you to derive the verte form of the baseball height function; here we see what this form can tell us. For t in seconds, the height of a baseball in feet is given both in standard and in verte form by y = ft) = 16t 2 +47t+3 = 16 t 47 ) Find the maimum height reached by the baseball and the time at which that height is reached. From the verte form, we see the verte is at the point 47 32, ) = 1.469,37.516). This means that the ball reaches it maimum height of feet at t = seconds. Note that these values are close to the graphical estimates 1.5,37.5) made at the beginning of this section. Eample 6 A city decides to make a park by fencing off a section of riverfront property. Funds are allotted for 80 meters of fence. The area enclosed will be a rectangle, but only three sides will be enclosed by a fence the other side will be bounded by the river. What is the maimum area that can be enclosed? Two sides are perpendicular to the bank of the river and have equal length, which we call h. The other side is parallel to the bank of the river; its length isb. See Figure The area,a, of the park is the product of the lengths of adjacent sides, soa = bh. Since the fence is 80 meters long, we have 2h+b = 80 b = 80 2h. River h h b Figure 3.12: A park net to a river

13 Thus, A = bh = 80 2h)h A = 2h 2 +80h. 3.2 THE VERTEX OF A PARABOLA 127 The function A = 2h h is quadratic. Since the coefficient of h 2 is negative, the parabola opens downward and we have a maimum at the verte. The factored form of the quadratic is A = 2hh 40), so the zeros areh = 0 andh = 40. The verte of the parabola occurs on its ais of symmetry, midway between the zeros, at h = 20. Substituting h = 20 gives the maimum area: Eercises and Problems for Section 3.2 Skill Refresher A = ) = 40 20) = 800 meters 2. For Eercises S1 S4, complete the square for each epression. S1. y 2 12y S2. s 2 +6s 8 S3. c 2 +3c 7 S4. 4s 2 +s+2 In Eercises S5 S7, solve by completing the square. In Eercises S8 S10, solve using factoring, completing the square, or the quadratic formula. S8. 3t 2 +4t+9 = 0 S9. n 2 +4n 3 = 2 S10. 2q 2 +4q 5 = 8 S5. r 2 6r +8 = 0 S6. n 2 = 3n+18 S7. 5q 2 8 = 2q Eercises For the quadratic functions in Eercises 1 2, state the coordinates of the verte, the ais of symmetry, and whether the parabola opens upward or downward. 1. f) = 3 1) g) = +3) Find the verte and ais of symmetry of the graph of vt) = t 2 +11t Find the verte and ais of symmetry for the parabola whose equation is y = Sketch the quadratic functions given in standard form. Identify the values of the parameters a, b, and c. Label the zeros, ais of symmetry, verte, andy-intercept. a) g) = 2 +3 b) f) = In Eercises 6 9, find a formula for the parabola. 8. 0, 4) y 2,0) 9. 10, 8) 6,5) For Eercises 10 13, convert the quadratic functions to verte form by completing the square. Identify the verte and the ais of symmetry. 10. f) = g) = pt) = 2t t wz) = 3z 2 +9z 2 y 6. y 7. 6, 9) 9 y In Eercises 14 16, write the quadratic function in its standard, verte, and factored forms. 1, 0) 1 3, 0) 0, 3) y = ft) = 5t gs) = s 5)2s+3)

14 128 Chapter Three QUADRATIC FUNCTIONS In Eercises 17 24, write a formula and graph the transformation of mn) = 1 2 n y = mn) y = mn+1) 19. y = mn) y = mn 3.7) 21. y = mn) y = mn+2 2) 23. y = mn+3) y = mn 17) 159 Problems 25. Find the verte form of ft) = 16t 2 +47t Show that y = has no real zeros. y 4 f) = 2 4 y g 27. Find the value ofk so that the graph ofy = 3) 2 +k passes through the point 6,13). 28. The parabolay = a 2 +k has verte0, 2) and passes through the point 3,4). Find its equation Figure Figure Using the verte form, find a formula for the parabola with verte 2,5) that passes through the point 1,2). In Problems 30 34, find a formula for the quadratic function whose graph has the given properties. 30. Verte at 4,2) and y-intercept of y = Verte at 4,2) and y-intercept of y = Verte at 4,2) and zeros at = 3, Verte at 7,3) and passing through the point 3,7). 34. A single-intercept at = 1/2 and ay-intercept at Let f be a quadratic function whose graph is concave up with a verte at 1, 1), and a zero at the origin. a) Graph y = f). b) Determine a formula for f). c) Determine the range of f. d) Find any other zeros. 36. Find the verte ofy = eactly. Graph the function, labeling all intercepts. 37. Find the verte ofy = eactly. Graph the function, labeling all intercepts. 38. If we know a quadratic function f has a zero at = 1 and verte at 1,4), do we have enough information to find a formula for this function? If your answer is yes, find it; if not, give your reasons. 39. Figure 3.13 shows f) = 2. Define g by shifting the graph of f to the right 2 units and down 1 unit; see Figure Find a formula for g in terms of f. Find a formula for g in terms of. 40. The function g) is obtained by shifting the graph of y = 2. If g3) = 16, give a possible formula for g when a) g is the result of applying only a horizontal shift to y = 2. b) g is the result of applying only a vertical shift to y = 2. c) g is the result of applying a horizontal shift right 2 units and an appropriate vertical shift of y = Graph y = and y = 2. Use a shift transformation to eplain the relationship between the two graphs. 42. Letf) = 2 and let g) = 3) a) Give the formula for g in terms of f, and describe the relationship between f andg in words. b) Isg a quadratic function? If so, find its standard form and the parameters a, b, and c. c) Graphg, labeling all important features. 43. If you have a string of length 50 cm, what are the dimensions of the rectangle of maimum area that you can enclose with your string? Eplain your reasoning. What about a string of length k cm? 44. A ballet dancer jumps in the air. The height,ht), in feet, of the dancer at time t, in seconds since the start of the jump, is given by 3 ht) = 16t 2 +16Tt, wheret is the total time in seconds that the ballet dancer is in the air. a) Why does this model apply only for 0 t T? b) When, in terms of T, does the maimum height of the jump occur? 3 K. Laws, The Physics of Dance Schirmer, 1984).

15 REVIEW EXERCISES AND PROBLEMS FOR CHAPTER THREE 129 c) Show that the time, T, that the dancer is in the air is related to H, the maimum height of the jump, by the equation H = 4T A football player kicks a ball at an angle of37 above the ground with an initial speed of 20 meters/second. The height in meters, h, as a function of the horizontal distance traveled, d, is given by: h = 0.75d d 2. a) Graph the path the ball follows. b) When the ball hits the ground, how far is it from the spot where the football player kicked it? c) What is the maimum height the ball reaches during its flight? d) What is the horizontal distance the ball has traveled when it reaches its maimum height? 4 CHAPTER SUMMARY General Formulas for Quadratic Functions Standard form: Factored form: Verte form: y = a 2 +b+c,a 0 y = a r) s) y = a h) 2 +k Graphs of Quadratic Functions Graphs are parabolas Verte h,k) Ais of symmetry, = h Effect of parameter a Opens upward concave up) if a > 0, minimum at h,k) Opens downward concave down) if a < 0, maimum ath,k) Factored form displays zeros at = r and = s Solving Quadratic Equations Factoring Quadratic formula Completing the square = b± b 2 4ac 2a REVIEW EXERCISES AND PROBLEMS FOR CHAPTER THREE Eercises Are the functions in Eercises 1 3 quadratic? If so, write the function in standard form. 1. f) = 27 ) LP) = P +1)1 P) 3. gm) = mm 2 2m)+3 14 m3 3 ) + 3m In Eercises 4 9, find the zeros if any) of the function algebraically. 4. y = y = y = y = y = y = Show that the function y = has no real zeros. 11. Find the verte and ais of symmetry of the graph of w) = In Eercises 12 16, find a possible formula for the parabola with the given conditions. 12. The verte is 1, 2) and the y-intercept isy = The verte is 4, 2) and the y-intercept isy = The verte is 7, 3) and it passes through the point 3, 7). 15. The parabola goes through the origin and its verte is 1, 1). 16. The -intercepts are at = 1 and = 2 and 2, 16) is on the function s graph. 4 Adapted from R. Halliday, D. Resnick, and K. Krane, Physics New York: Wiley, 1992), p. 58.

16 130 Chapter Three QUADRATIC FUNCTIONS In Eercises 17 20, find a formula for the parabola. 19. y 20. y 2, 36) 17. 0,4) y 18. 4,7) y 2 4, 0) 5,0) 5,5) 3,3) 5 3, 5) Problems For each parabola in Problems 21 24, state the coordinates of the verte, the ais of symmetry, the y-intercept, and whether the curve is concave up or concave down. 21. y = 2 3/4) 2 2/3 22. y = 1/2+6) y = 0.6) y = In Problems 25 28, write each function in factored form or verte form and then state its verte and zeros. 25. y = y = y = y = /3) For between = 2 and = 4, determine the concavity of the graph off) = 1) 2 +2 by calculating average rates of change over intervals of length A tomato is thrown vertically into the air at time t = 0. Its height, dt) in feet), above the ground at time t in seconds) is given by dt) = 16t 2 +48t. a) Graphdt). b) Find t when dt) = 0. What is happening to the tomato the first timedt) = 0? The second time? c) When does the tomato reach its maimum height? d) What is the maimum height that the tomato reaches? 31. When slam-dunking, a basketball player seems to hang in the air at the height of his jump. The height ht), in feet above the ground, of a basketball player at time t, in seconds since the start of a jump, is given by ht) = 16t 2 +16Tt, where T is the total time in seconds that it takes to complete the jump. For a jump that takes 1 second to complete, how much of this time does the basketball player spend at the top 25% of the trajectory? [Hint: Find the maimum height reached. Then find the times at which the height is 75% of this maimum.] STRENGTHEN YOUR UNDERSTANDING Are the statements in Problems 1 15 true or false? Give an eplanation for your answer. 1. The quadratic function f) = + 2) is in factored form. 2. If f) = +1)+2), then the zeros of f are 1 and A quadratic function whose graph is concave up has a maimum. 4. All quadratic equations have the form f) = a If the height above the ground of an object at time t is given byst) = at 2 +bt+c, thens0) tells us when the object hits the ground. 6. To find the zeros of f) = a 2 + b + c, solve the equation a 2 +b+c = 0 for. 7. Every quadratic equation has two real solutions. 8. There is only one quadratic function with zeros at = 2 and = A quadratic function has eactly two zeros. 10. The graph of every quadratic function is a parabola. 11. The maimum or minimum point of a parabola is called its verte. 12. If a parabola is concave up its verte is a maimum point. 13. If the equation of a parabola is written as y = a h) 2 +k, then the verte is located at the point h,k). 14. If the equation of a parabola is written as y = a h) 2 +k, then the ais of symmetry is found at = h. 15. If the equation of a parabola is y = a 2 + b + c and a < 0, then the parabola opens downward.

17 SKILLS REFRESHER FOR CHAPTER 3: QUADRATIC EQUATIONS SKILLS FOR FACTORING Epanding an Epression The distributive property for real numbers a,b, and c tells us that and ab+c) = ab+ac, b+c)a = ba+ca. We use the distributive property and the rules of eponents to multiply algebraic epressions involving parentheses. This process is sometimes referred to as epanding the epression. 131 Eample 1 Multiply the following epressions and simplify. a) ) 6 3 b) 2t) 2 5 ) t a) ) 6 3 = 3 2) )+ 3 2) ) = b) 2t) 2 5 ) t = 2t) 2 t) 5 t = 4t 2) t 1/2) 5t 1/2 = 4t 5/2 5t 1/2. If there are two terms in each factor, then there are four terms in the product: a+b)c+d) = ac+d)+bc+d) = ac+ad+bc+bd. The following special cases of the above product occur frequently. Learning to recognize their forms aids in factoring. a+b)a b) = a 2 b 2 a+b) 2 = a 2 +2ab+b 2 a b) 2 = a 2 2ab+b 2 Eample 2 Epand the following and simplify by gathering like terms. a) ) 4) b) 2 r +2)4 r 3) c) 3 12 ) 2 a) ) 4) = 5 2) )+ 5 2) 4)+2))+2) 4) = b) 2 r+2)4 r 3) = 2)4) r) 2 +2) 3) r)+2)4) r)+2) 3) = 8r+2 r 6. c) 3 1 ) 2 ) 1 2 = ) ) 2 2 = Factoring To write an epanded epression in factored form, we un-multiply the epression. Some techniques for factoring are given in this section. We can check factoring by multiplying the factors. Removing a Common Factor It is sometimes useful to factor out the same factor from each of the terms in an epression. This is basically the distributive law in reverse: ab+ac = ab+c).

18 132 SKILLS REFRESHER FOR CHAPTER THREE One special case is removing a factor of 1, which gives a b = a+b) Another special case is a b) = b a) Eample 3 Factor the following: 2 a) 3 2 y y b) 2p+1)p3 3p2p+1) c) s2 t 8w st2 16w a) y y = 2 3 y+2). b) 2p+1)p 3 3p2p+1) = p 3 3p)2p+1) = pp 2 3)2p+1). Note that the epression 2p+1) was one of the factors common to both terms.) c) s2 t 8w st2 16w = st s+ t ). 8w 2 Grouping Terms Even though all the terms may not have a common factor, we can sometimes factor by first grouping the terms and then removing a common factor. Eample 4 Factor 2 h +h. 2 h +h = 2 h ) h) = h) h) = h) 1). Factoring Quadratics One way to factor quadratics is to mentally multiply out the possibilities. Eample 5 Factor t 2 4t 12. If the quadratic factors, it is of the form t 2 4t 12 = t+?)t+?). We are looking for two numbers whose product is 12 and whose sum is 4. By trying combinations, we find t 2 4t 12 = t 6)t+2). Eample 6 Factor 4 2M 6M 2. By a similar method to the previous eample, we find 4 2M 6M 2 = 2 3M)2+2M). Perfect Squares and the Difference of Squares Recognition of the special products + y) 2, y) 2 and + y) y) in epanded form is useful in factoring. Reversing the results given on page 131, we have a 2 +2ab+b 2 = a+b) 2, a 2 2ab+b 2 = a b) 2, a 2 b 2 = a b)a+b).

19 SKILLS FOR FACTORING 133 When we see squared terms in an epression to be factored, it is often useful to look for one of these forms. The difference of squares identity the third one listed previously) is especially useful. Eample 7 Factor: a) 16y 2 24y +9 b) 25S 2 R 4 T 6 c) 2 2)+162 ) a) 16y 2 24y +9 = 4y 3) 2. b) 25S 2 R 4 T 6 = 5SR 2) 2 T 3 ) 2 = 5SR 2 T 3) 5SR 2 +T 3). c) 2 2)+162 ) = 2 2) 16 2) = 2) 2 16 ) = 2) 4)+4). Solving Quadratic Equations Eample 8 Give eact and approimate solutions to 2 = 3. The eact solutions are = ± 3; approimate ones are ±1.73, or ±1.732, or ± since 3 = ). Notice that the equation 2 = 3 has only two eact solutions, but many possible approimate solutions, depending on how much accuracy is required. Solving by Factoring Some equations can be put into factored form such that the product of the factors is zero. Then we solve by using the fact that ifa b = 0, then either a or b or both) is zero. Eample 9 Solve +1)+3) = 15. Although it is true that if a b = 0, then a = 0 or b = 0, it is not true that a b = 15 means that a = 15 or b = 15, or that a and b are 3 and 5. To solve this equation, we epand the left-hand side and rearrange so that the right-hand side is equal to zero: Then, factoring gives Thus = 2 and = 6 are the solutions = 15, = 0. 2)+6) = 0. Eample 10 Solve 2+3) 2 = 5+3). You might be tempted to divide both sides by + 3). However, if you do this you will overlook one of the solutions. Instead, write 2+3) 2 5+3) = 0 +3)2+3) 5) = 0 +3)2+6 5) = 0 +3)2+1) = 0. Thus, = 1/2 and = 3 are solutions. Note that if we had divided by + 3) at the start, we would have lost the solution = 3, which was obtained by setting+3 = 0.

20 134 SKILLS REFRESHER FOR CHAPTER THREE Solving with the Quadratic Formula Instead of factoring, we can solve the equation a 2 +b+c = 0 by using the quadratic formula: = b± b 2 4ac. 2a The quadratic formula is derived by completing the square for y = a 2 +b+c. See page 136. Eample 11 Solve 11+2 = 2. The equation is = 0. The epression on the left does not factor using integers, so we use = )11) 2 1) = )11) 2 1) = = = = = = = 1 2 3, = The eact solutions are = and = The decimal approimations to these numbers = = and = = are approimate solutions to this equation. The approimate solutions could also be found directly from a graph or calculator. Eercises for Skills for Factoring For Eercises 1 15, epand and simplify ) 4+5+4)3 4) ) 2. 4y +6) y) y) ) 6. 3z2 9z) 7. 10r5r + 6rs) )+23 8) 9. 5z 2) 3 2) )+3) 11. 2)+6) )2 3) 13. y +1)z +3) y 5)8w +7) 15. 5z 3) 2) Multiply and write the epressions in Problems without parentheses. Gather like terms ) 25 ) 17. 5) )) 19. Pp 3q) ) ) u u 1 +2 u) 2 u For Eercises 23 67, factor completely if possible y z t w u 4 4u u 7 +12u y r 4 s 2 21rst

21 SKILLS FOR FACTORING ac+ad+bc+bd y +3z +6yz a 2 2 b πr 2 +2πrh 49. B 2 10B c c y a 4 a t+3) y a 3 2a 2 +3a b 3 3b 2 9b c 2 d 2 25c 2 9d h h rr s) 2s r) 60. y 2 3y e 3 +2e t 2 e 5t +3te 5t +2e 5t 63. P1+r) 2 +P1+r) 2 r z dk +2dm 3ek 6em 66. πr 2 2πr +3r gs 12hs+10gm 15hm Solve the equations in Eercises y 2 5y 6 = s 2 +3s 15 = = = y 1 = t 2 +96t+12 = g 3 4g = 3g = p 3 +p 2 18p 9 = N 2 2N 3 = 2NN 3) t3 = t = = = 16t 2 +96t n = 5n 4 +16n 83. 5a 3 +50a 2 = 4a y 2 +4y 2 = z z 2 3z = = ) L 2 = q +1 1 q 1 = r2 +24 = = = 1 2 v = 7π 3+4) 2) 93. 5) 1) = 0 In Eercises 94 97, solve for the indicated variable. l 94. T = 2π g, forl. 95. Ab 5 = C, for b = 7, for. 2 5m+4m = 0, for m Solve the systems of equations in Eercises { y = 2 2 y = 3 { 2 +y 2 = 36 y = 3 { y = 3 1 y = e { y = 1/ 99. y = 4 { y = y 2 = Let l be the line of slope 3 passing through the origin. Find the points of intersection of the line l and the parabola whose equation is y = 2. Sketch the line and the parabola, and label the points of intersection. Determine the points of intersection for Problems y 2 = 25 y 105. y = 1 y y = 2 y = 15 2

22 136 SKILLS REFRESHER FOR CHAPTER THREE COMPLETING THE SQUARE An eample of changing the form of an epression is the conversion of a 2 +b+c into the form a h) 2 +k. We make this conversion by completing the square, a method for producing a perfect square within a quadratic epression. A perfect square is an epression of the form: +n) 2 = 2 +2n+n 2. In order to complete the square in an epression, we must find that number n, which is half the coefficient of. Before giving a general procedure, let s work through an eample. Eample 1 Complete the square to rewrite in the form a h) 2 +k. Step 1: We divide the coefficient of by 2, giving 1 10) = 5. Step 2: We square the result of step 1, giving 5) 2 = 25. Step 3: Now add and subtract the 25 after the -term: = = ) }{{} Perfect square Step 4: Notice that we have created a perfect square, The net step is to factor the perfect square and combine the constant terms, 25+4, giving the final result: Thus, a = +1, h = +5, and k = = 5) Visualizing the Process of Completing the Square We can visualize how to find the constant that needs to be added to 2 + b in order to obtain a perfect square by thinking of 2 + b as the area of a rectangle. For eample, the rectangle in Figure 3.15 has area + b) = 2 + b. Now imagine cutting the rectangle into pieces as in Figure 3.16 and trying to rearrange them to make a square, as in Figure The corner piece, whose area isb/2) 2, is missing. By adding this piece to our epression, we complete the square: 2 +b+b/2) 2 = +b/2) 2. +b/2 b/2 +b/2 +b Figure 3.15: Rectangle with sides and +b b/2 b/2 Figure 3.16: Cut off 2 strips of width b/2 b/2 b/2 Figure 3.17: Rearrange to see a square with missing corner of areab/2) 2 The procedure we followed can be summarized as follows:

23 COMPLETING THE SQUARE 137 To complete the square in the epression 2 +b+c, divide the coefficient ofby 2, giving b/2. Then add and subtract b/2) 2 = b 2 /4 and factor the perfect square: 2 +b+c = + b ) 2 b c. To complete the square in the epression a 2 +b+c, factor out a first. The net eample has a coefficient a with a 1. After factoring out the coefficient, we follow the same steps as in Eample 1. Eample 2 Complete the square in the formula h) = We first factor out 5: h) = ). Now we complete the square in the epression Step 1: Divide the coefficient of by 2, giving 3. Step 2: Square the result:3 2 = 9. Step 3: Add the result after the term, then subtract it: h) = ). }{{} Perfect square Step 4: Factor the perfect square and simplify the rest: h) = 5 +3) 2 11 ). Now that we have completed the square, we can multiply by the 5: h) = 5+3) Deriving the Quadratic Formula We derive a general formula to find the zeros of q) = a 2 +b+c, with a 0, by completing the square. To find the zeros, setq) = 0: a 2 +b+c = 0. Before we complete the square, we factor out the coefficient of 2 : a 2 + b a + c ) = 0. a Since a 0, we can divide both sides by a: 2 + b a + c a = 0.

24 138 SKILLS REFRESHER FOR CHAPTER THREE To complete the square, we add and then subtract b/a)/2) 2 = b 2 /4a 2 ): 2 + b b2 + } a {{ 4a 2 b2 } 4a 2 + c a = 0. Perfect square We factor the perfect square and simplify the constant term, giving: + b ) 2 2a b 2 ) 4ac 4a 2 + b ) 2 = b2 4ac 2a 4a 2 + b 2a = ± b2 4ac 4a 2 = ± b2 4ac 2a = b 2a ± b2 4ac 2a ) = 0 since b2 4a 2 + c a = b2 4a 2 + 4ac 4a 2 = b 2 4ac 4a 2 = b± b 2 4ac. 2a adding b 2 4ac 4a 2 to both sides taking the square root subtracting b/2a Eercises for Skills for Completing the Square For Eercises 1 8, complete the square for each epression w 2 +7w 3. 2r 2 +20r 4. 3t 2 +24t a 2 2a 4 6. n 2 +4n r 2 +9r g 2 +8g +5 In Eercises 9 12, rewrite in the forma h) 2 +k In Eercises 13 22, complete the square to find the verte of the parabola. 13. y = y = y = y = y = y = y = y = y = y = In Eercises 23 29, solve by completing the square. 23. g 2 = 2g p 2 2p = d 2 d = r 2 +4r 5 = s 2 = 1 10s 28. 7r 2 3r 6 = p 2 +9p = 1 In Eercises 30 35, solve by using the quadratic formula. 30. n 2 4n 12 = y 2 +5y = k 2 +11k = w 2 +w = z 2 +4z = q 2 +6q 3 = 0 In Eercises 36 46, solve using factoring, completing the square, or the quadratic formula. 36. r 2 2r = s 2 +3s = z 3 +2z 2 = 3z u 2 +4 = 30u 40. v 2 4v 9 = y 2 = 6y p = 14p 43. 2w = 6w 2 +8w = m 2 +70m+22 = = 2