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1 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

2 . 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 6, ), ), 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

5 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) = 6 = + 6 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) = + 6 = ) + ) + 6 Factor the coefficient of from and.) ) = ) = ) + + ) + ) ) + 6 Group the perfect square trinomial.) = + ) 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!

6 . 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) = 6 = 0. Solving, we get = and =, so the -intercepts are, 0) and, 0). Setting = 0, we find g0) = 6, so the -intercept is 0, 6). Plotting these points gives us the graph below. We see that the range of g is, 5 ] , 5 ) 6 5 0, 6) 5 = 0, ), 0), 0) 5, ) f) = +, 0) =, 0) g) = 6 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.

7 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

8 . 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 96 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) = with the restriction carried over from the price-demand function) that The cost, in dollars, to produce PortaBo Game Sstems is given in Eample..5 as C) = 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) = , 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) = ) ) = Since the revenue function is valid when 0 66, P is also restricted to these values.. To find the -intercepts, we set P ) = 0 and solve = 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 ) ) and 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 ) = is in the general form of a quadratic function and appeal to Equation.. Substituting a =.5 and b = 70, we get = 70.5) = 70.

10 . 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, 56.67, ). 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.) 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 56.67, ). As above, represents the weekl sales of PortaBo sstems, so we can t sell game sstems. Comparing P 56) = 666 and P 57) = 666.5, we conclude that we will make a maimum profit of \$ 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 = 6.5. The price should be set at \$6.50. 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?

11 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) ) = 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 = 00 average alpaca. We conclude this section with the graph of a more complicated absolute value function. Eample..5. Graph f) = 6. Solution. Using the definition of absolute value, Definition., we have { f) = 6 ), if 6 < 0 6, if 6 0 The trouble is that we have et to develop an analtic techniques to solve nonlinear inequalities such as 6 < 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.

12 . Quadratic Functions 99 Nevertheless, we can attack this problem graphicall. To that end, we graph = g) = 6 using the intercepts and the verte. To find the -intercepts, we solve 6 = 0. Factoring gives ) + ) = 0 so = or =. Hence,, 0) and, 0) are -intercepts. The -intercept 0, 6) is found b setting = 0. To plot the verte, we find = b a = ) =, and = ) ) 6 = 5 = 6.5. Plotting, we get the parabola seen below on the left. To obtain points on the graph of = f) = 6, we can take points on the graph of g) = 6 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, 6) on = 6 would result in the point 0, 6 ) = 0, 6)) = 0, 6) on the graph of f) = 6. Proceeding in this manner for all points with -coordinates between and results in the graph seen below on the right = g) = 6 = f) = 6 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) = 6 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.

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