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. The point 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) to the left units and moves, ) to, ),, ) to, ), 0, 0) to, 0),, ) to, ) and, ) to 0, ). Net, we subtract from each of the -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
. 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 to each of the -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 and then add to that result. Geometricall, this is a vertical stretch b a factor of, followed b a 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, ), ), 7) 5, 7) 0, 0) f) = h) = f ) + = ) + The verte is, ) 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
90 Linear and Quadratic Functions which gives ) =. Etracting square roots gives = ±, so that when we add to each side, we get = ± ). Hence, our -intercepts are, 0.9, 0) and ) +, 0.7, 0). Although our graph doesn t show it, there is a -intercept which can be found b setting = 0. With h0) = 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) and h) into that form b epanding and collecting like terms. Doing so, we find g) = + ) = + + and h) = ) + = + 7. While these simplified formulas for g) and h) 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.. Standard and General Form of Quadratic Functions: Suppose f is a quadratic function. The general form of the quadratic function f is f) = a + b + c, where a, b and c are real numbers with a 0. The standard form of the quadratic function f is f) = a h) + k, where a, 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, where a, 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) to be, ). 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) to be, ). To see wh the formula in Theorem. produces the verte, consider the graph of the equation = a h) +k. When we substitute = h, we get = k, so h, k) is on the graph. If h, then h 0 so 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!
. 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, so the point 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. 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) = 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 You should use transformations to verif this!
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 = f0) =. 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) = = + 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) = + = ) + ) + Factor the coefficient of from and.) ) = ) + + + = ) + + ) + ) ) + Group the perfect square trinomial.) = + ) + 5 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!
. Quadratic Functions 9 From g) = + ) + 5, we get the verte to be, 5 ) and the ais of smmetr to be =. To get the -intercepts, we opt to set the given formula g) = = 0. Solving, we get = and =, so the -intercepts are, 0) and, 0). Setting = 0, we find g0) =, so the -intercept is 0, ). Plotting these points gives us the graph below. We see that the range of g is, 5 ]. 8 7, 5 ) 5 0, ) 5 = 0, ), 0), 0) 5, ) f) = +, 0) =, 0) g) = 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.
9 Linear and Quadratic Functions Equation.. Verte Formulas for Quadratic Functions: Suppose a, b, c, h and k are real numbers with a 0. If f) = a h) + k, the verte of the graph of = f) is the point h, k). If f) = a + b + c, the verte of the graph of = f) is the point 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
. 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 a = 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. If b ac < 0, the equation a + b + c = 0 has no real solutions. If b ac = 0, the equation a + b + c = 0 has eactl one real solution. 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.
9 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) =.5 + 50 with the restriction carried over from the price-demand function) that 0. The cost, in dollars, to produce PortaBo Game Sstems is given in Eample..5 as C) = 80 + 50 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) =.5 + 50, 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) =.5 + 50 ) 80 + 50) =.5 + 70 50. Since the revenue function is valid when 0, P is also restricted to these values.. To find the -intercepts, we set P ) = 0 and solve.5 + 70 50 = 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 ) ) 70 0 70 and 70+0 70. 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 ) =.5 + 70 50 is in the general form of a quadratic function and appeal to Equation.. Substituting a =.5 and b = 70, we get = 70.5) = 70.
. 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, 5.7,.7). 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.) 000 000 000 000 0 0 0 0 50 0 70 80 90 00 0 0. 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 ) is above the -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 5.7,.7). As above, represents the weekl sales of PortaBo sstems, so we can t sell 5.7 game sstems. Comparing P 5) = and P 57) =.5, we conclude that we will make a maimum profit of $.50 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 p57) =.557) + 50 =.5. The price should be set at $.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)stranged) 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?
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. Since w 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. We get A = wl = w00 w) = 00w w. We now have A as a function of w, Aw) = 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 Aw) = w +00w for 0 < w < 00. Since A is a quadratic function of w), we know that the graph of = Aw) 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 A50) = 50) + 0050) = 5000. 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 A50) = 5000, or 5000 square feet. If an average alpaca requires 5 square feet of pasture, Donnie can raise 5000 5 = 00 average alpaca. We conclude this section with the graph of a more complicated absolute value function. Eample..5. Graph f) =. Solution. Using the definition of absolute value, Definition., we have { f) = ), if < 0, if 0 The trouble is that we have et to develop an analtic techniques to solve nonlinear inequalities such as < 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.
. Quadratic Functions 99 Nevertheless, we can attack this problem graphicall. To that end, we graph = g) = using the intercepts and the verte. To find the -intercepts, we solve = 0. Factoring gives ) + ) = 0 so = or =. Hence,, 0) and, 0) are -intercepts. The -intercept 0, ) is found b setting = 0. To plot the verte, we find = b a = ) =, and = ) ) = 5 =.5. Plotting, we get the parabola seen below on the left. To obtain points on the graph of = f) =, we can take points on the graph of g) = 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, ) on = would result in the point 0, ) = 0, )) = 0, ) on the graph of f) =. Proceeding in this manner for all points with -coordinates between and results in the graph seen below on the right. 7 5 7 5 5 5 = g) = = f) = 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) = 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 reflection the portion of the line g) = which is below the -ais back above the -ais creating the characteristic shape.
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) =. f) = + 7 7. f) = + + 8. f) = + 5 + 9. 0 f) = 00 In Eercises 0 -, the cost and price-demand functions are given for different scenarios. For each scenario, Find the profit function P ). Find the number of items which need to be sold in order to maimize profit. Find the maimum profit. Find the price to charge per item in order to maimize profit. Find and interpret break-even points. 0. The cost, in dollars, to produce I d rather be a Sasquatch T-Shirts is C) = +, 0 and the price-demand function, in dollars per shirt, is p) = 0, 0 5.. 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, 0 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, 0 0.. The dail cost, in dollars, to produce Sasquatch Berr Pies is C) = +, 0 and 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 + 000, 0 and the price-demand function, in hundreds of dollars per scooter, is p) = 0, 0 70. 0 We have alread seen the graph of this function. It was used as an eample in Section. to show how the graphing calculator can be misleading.
. 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) = 0. + 5 and that the demand function for their cookies is p = 0.0. How man cookies should the bake in order to maimize their profit?. 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) = 0.007t + 0.5t +, 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 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) = 0 + 7 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 ht) = t +, 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 ht) = 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 ht) = t +.08t +. What is the hammer s maimum height? What is the hammer s total time in the air? Round our answers to two decimal places.
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 st) =.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?. 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 -, solve the quadratic equation for the indicated variable.. 0 = 0 for. = for. m = for. = for 5. = for. gt + v 0 t + s 0 = 0 for t 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.5 what shape a free hanging cable makes.
. 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 = 0 0 9 8 7 5. f) = + ) = -intercept, 0) -intercept 0, ) Domain:, ) Range:, 0] Increasing on, ] Decreasing on [, ) Verte, 0) is a maimum Ais of smmetr = 5 7 8. 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 = 5 7 8 9. f) = + ) + = + -intercepts, 0) and +, 0) -intercept 0, ) Domain:, ) Range:, ] Increasing on, ] Decreasing on [, ) Verte, ) is a maimum Ais of smmetr =
0 Linear and Quadratic Functions 5. f) = = ) -intercepts ) +, 0 and ), 0 -intercept 0, ) Domain:, ) Range: [, ) Increasing on [, ) Decreasing on, ] Verte, ) is a minimum Ais of smmetr =. f) = + 7 = ) 7 No -intercepts -intercept 0, 7) Domain:, ) Range:, 7 ] Increasing on, ] Decreasing on [, ) Verte, 7 ) is a maimum Ais of smmetr = 5 7 8 9 0 7. f) = + + = + ) + No -intercepts -intercept 0, ) Domain:, ) Range: [, ) Increasing on [, ) Decreasing on, ] Verte, ) is a minimum Ais of smmetr =
. Quadratic Functions 05 8. f) = + 5 + = ) 5 + 7 5 -intercepts ) 7 5+, 0 and ) 7, 0 -intercept 0, ) Domain:, ) Range:, 7 ] Increasing on, 5 ] Decreasing on [ 5, ) Verte 5, 7 ) is a maimum Ais of smmetr = 5 5 9. f) = 00 = ) 000 ) -intercepts and + 000 00 -intercept 0, ) Domain:, ) Range: [ 000 0000, ) Decreasing on, 00 ] 00 0000 ) 000 00 Increasing on [ 00, ) Verte 00, 000 ) 0000 is a minimum Ais of smmetr = 00 8 7 5 0. P ) = + 8, for 0 5. 7 T-shirts should be made and sold to maimize profit. The maimum profit is $7. The price per T-shirt should be set at $ to maimize profit. The break even points are = and =, so to make a profit, between and T-shirts need to be made and sold.. P ) = + 5 00, for 0 5 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. The maimum profit is $5. The price per bottle can be either $ to sell bottles) or $ to sell bottles.) Both will result in the maimum profit. 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.
0 Linear and Quadratic Functions. P ) = + 7 0, for 0 0 cups of lemonade need to be made and sold to maimize profit. The maimum profit is 9 or $.9. The price per cup should be set at 5 per cup to maimize profit. The break even points are = and = 0, so to make a profit, between and 0 cups of lemonade need to be made and sold.. P ) = 0.5 + 9, for 0 9 pies should be made and sold to maimize the dail profit. The maimum dail profit is $.50. The price per pie should be set at $7.50 to maimize profit. The break even points are = and =, so to make a profit, between and pies need to be made and sold dail.. P ) = + 0 000, for 0 70 0 scooters need to be made and sold to maimize profit. The maimum monthl profit is 800 hundred dollars, or $80,000. The price per scooter should be set at 80 hundred dollars, or $8000 per scooter. The break even points are = 0 and = 50, so to make a profit, between 0 and 50 scooters need to be made and sold monthl. 5. 95 cookies. The verte is approimatel) 9.0,.), which corresponds to a maimum fuel econom of. 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. 7. at PM 8 hours after AM.) 8. 5000 pens should be produced for a cost of $00. 9. 8 feet b feet; maimum area is 8 square feet. 0. 50 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. feet. The hammer is in the air approimatel. seconds.
. Quadratic Functions 07 5. a) The applied domain is [0, ). d) The height function is this case is st) =.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.0 seconds. e) The revised height function is st) =.9t + 5t + 5 which has zeros at t.0 and t.. We ignore the negative value and claim that the marble will hit the ground after. seconds. f) Shooting down means the initial velocit is negative so the height functions becomes st) =.9t 5t + 5.. Make the verte of the parabola 0, 0) so that the point on the top of the left-hand tower where the cable connects is 00, 00) and the point on the top of the right-hand tower is 00, 00). Then the parabola is given b p) = 9 000 + 0. Standing 50 feet to the right of the left-hand tower means ou re standing at = 50 and p 50) = 0.5. So the cable is 0.5 feet above the bridge deck there. 7. = 7 8., + ) 7 + 7,, ) 7 7 5 9. D) = ++) = 5 ++, D is minimized when = 5, so the point on = + closest to 0, 0) is 5, ) 5. = ± 0. = ± ). = m ± m +. = ± + 9 5. = ±. t = v 0 ± v 0 + gs 0 g