1. College Algebra Fifth Edition James Stewart Lothar Redlin Saleem Watson

2. Polynomial and Rational Functions 4

3. Quadratic Functions and Models 4.1

4. Introduction A polynomial function is a function that is defined by a polynomial expression. So a polynomial function of degree n is a function of the form: P(x) = a n x n + a n–1 x n–1 + … +a 1 x + a 0 We have already studied polynomial functions of degree 0 and 1. These are functions of the form P(x) = a 0 and P(x) = a 1 x + a 0, respectively, whose graphs are lines.

5. Quadratic Functions In this section, we study polynomial functions of degree 2. These are called quadratic functions. We see in this section how quadratic functions model many real-world phenomena.

6. Quadratic Functions A quadratic function is a polynomial function of degree 2. So a quadratic function is a function of the form: f(x) = ax 2 + bx + c(a ≠ 0)

7. Graphing Quadratic Functions Using the Standard Form

8. Quadratic Function If we take a = 1 and b = c = 0, we get the simple quadratic function f(x) = x 2, whose graph is the parabola that we drew in Example 1 of Section 3.2. In fact, the graph of any quadratic function is a parabola. It can be obtained from the graph of f(x) = x 2 by the transformations given in Section 3.5.

9. Standard Form of a Quadratic Function A quadratic function f(x) = ax 2 + bx + c can be expressed in the standard form f(x) = a(x – h) 2 + k by completing the square.

10. Standard Form of a Quadratic Function The graph of f is a parabola with vertex (h, k). The parabola opens upward if a > 0 or downward if a < 0.

11. E.g. 1—Standard Form of Quadratic Function Let f(x) = 2x 2 – 12x + 23. (a)Express f in standard form. (b)Sketch the graph of f.

12. E.g. 1—Standard Form Since the coefficient of x 2 is not 1, we must factor this coefficient from the terms involving x before we complete the square. Example (a)

13. E.g. 1—Standard Form The standard form tells us that we get the graph of f by: 1.Taking the parabola y = x 2. 2.Shifting it to the right 3 units. 3.Stretching it by a factor of 2. 4.Moving it upward 5 units. The vertex of the parabola is at (3, 5) and the parabola opens upward. Example (b)

14. E.g. 1—Standard Form We sketch the graph after noting that the y-intercept is f(0) = 23. Example (b)

15. Maximum and Minimum Values of Quadratic Functions

16. Maximum & Minimum Values of Quadratic Functions If a quadratic function has vertex (h, k), then the function has: A minimum value at the vertex if it opens upward. A maximum value at the vertex if it opens downward.

17. Maximum & Minimum Values of Quadratic Functions For example, the function graphed here has minimum value 5 when x = 3—since the vertex (3, 5) is the lowest point on the graph.

18. Maximum or Minimum Value of a Quadratic Function Let f be a quadratic function with standard form f(x) = a(x – h) 2 + k The maximum or minimum value of f occurs at x = h.

19. Maximum or Minimum Value of a Quadratic Function If a > 0, the minimum value of f is f(h) = k. If a < 0, the maximum value of f is f(h) = k.

20. E.g. 2—Minimum Value of a Quadratic Function Consider the quadratic function f(x) = 5x 2 – 30x + 49 (a)Express f in standard form. (b)Sketch the graph of f. (c)Find the minimum value of f.

21. E.g. 2—Minimum Value To express this quadratic function in standard form, we complete the square. Example (a)

22. E.g. 2—Minimum Value The graph is a parabola that has its vertex at (3, 4) and opens upward. Example (b)

23. E.g. 2—Minimum Value Since the coefficient of x 2 is positive, f has a minimum value. The minimum value is f(3) = 4. Example (c)

24. E.g. 3—Maximum Value of a Quadratic Function Consider the quadratic function f(x) = –x 2 + x + 2 (a)Express f in standard form. (b)Sketch the graph of f. (c)Find the maximum value of f.

25. E.g. 3—Maximum Value To express this quadratic function in standard form, we complete the square. Example (a)

26. E.g. 3—Maximum Value From the standard form, we see that the graph is a parabola that opens downward and has vertex (1/2, 9/4). As an aid to sketching the graph, we find the intercepts. The y-intercept is f(0) = 2. Example (b)

27. E.g. 3—Maximum Value To find the x-intercepts, we set f(x) = 0 and factor the resulting equation. Thus, the x-intercepts are x = 2 and x = –1. Example (b)

28. E.g. 3—Maximum Value The graph of f is sketched here. Example (b)

29. E.g. 3—Maximum Value Since the coefficient of x 2 is negative, f has a maximum value. This is f(1/2) = 9/4. Example (c)

30. Finding Maximum or Minimum Value Expressing a quadratic function in standard form helps us sketch its graph as well as find its maximum or minimum value. If we are interested only in finding the maximum or minimum value, then a formula is available for doing so.

31. Finding Maximum or Minimum Value The formula is obtained by completing the square for the general quadratic function:

32. Finding Maximum or Minimum Value This equation is in standard form with h = –b/(2a) and k = c – b 2 /(4a) Since the maximum or minimum value occurs at x = h, we have the following result.

33. Maximum or Minimum Value of a Quadratic Function The maximum or minimum value of a quadratic function f(x) = ax 2 + bx + c occurs at: x = –b/2a If a > 0, then the minimum value is f(–b/2a). If a < 0, then the maximum value is f(–b/2a).

34. E.g. 4—Finding Maximum & Minimum Values Find the maximum or minimum value of each quadratic function. (a) f(x) = x 2 + 4x (b) g(x) = –2x 2 + 4x – 5

35. E.g. 4—Finding Max. & Min. Values f(x) = x 2 + 4x is a quadratic function with a = 1 and b = 4. Thus, the maximum or minimum value occurs at: Example (a)

36. E.g. 4—Finding Max. & Min. Values Since a > 0, the function has the minimum value f(–2) = (–2) 2 + 4(–2) = –4 Example (a)

37. E.g. 4—Finding Max. & Min. Values g(x) = –2x 2 + 4x – 5 is a quadratic function with a = –2 and b = 4. Thus, the maximum or minimum value occurs at: Example (b)

38. E.g. 4—Finding Max. & Min. Values Since a < 0, the function has the maximum value f(1) = –2(1) 2 + 4(1) – 5 = –3 Example (b)

39. Modeling with Quadratic Functions We study some examples of real-world phenomena that are modeled by quadratic functions. In the next example, we find the maximum value of a quadratic function that models the gas mileage for a car.

40. E.g. 5—Maximum Gas Mileage for a Car Most cars get their best gas mileage when traveling at a relatively modest speed. The gas mileage M for a certain new car is modeled by the function where: s is the speed in mi/h. M is measured in mi/gal.

41. E.g. 5—Maximum Gas Mileage for a Car What is the car’s best gas mileage? At what speed is it attained?

42. E.g. 5—Maximum Gas Mileage for a Car The function M is a quadratic function with a = –1/28 and b = 3. Thus, its maximum value occurs when The maximum is:

43. E.g. 5—Maximum Gas Mileage for a Car So, the car’s best gas mileage is 32 mi/gal, when it is traveling at 42 mi/h.

44. E.g. 6—Maximizing Revenue from Ticket Sales A hockey team plays in an arena with a seating capacity of 15,000 spectators. With the ticket price set at $14, average attendance at recent games has been 9,500. A market survey indicates that, for each dollar the ticket price is lowered, the average attendance increases by 1,000.

45. E.g. 6—Maximizing Revenue from Ticket Sales (a)Find a function that models the revenue in terms of ticket price. (b) Find the price that maximizes revenue from ticket sales. (c) What ticket price is so high that no one attends and so no revenue is generated?

46. E.g. 6—Max. Rev. from Ticket Sales The model we want is: A function that gives the revenue for any ticket price. Example (a)

47. E.g. 6—Max. Rev. from Ticket Sales 1. Express the model in words. We know that: Revenue = Ticket price x Attendance Example (a)

48. E.g. 6—Max. Rev. from Ticket Sales 2. Choose the variable. There are two varying quantities—ticket price and attendance. Since the function we want depends on price, we let: x = ticket price Example (a)

49. E.g. 6—Max. Rev. from Ticket Sales Next, we express attendance in terms of x. Example (a) In WordsIn Algebra Ticket pricex Amount ticket price is lowered 14 – x Increase in attendance1000(14 – x) Attendance9500 + 1000(14 – x) = 23,500 – 10,000x

50. E.g. 6—Max. Rev. from Ticket Sales 3. Set up the model. The model is the function R that gives the revenue for a given ticket price x. Revenue = Ticket price x Attendance R(x) = x(23,500 – 1000x) R(x) = 23,500x – 1000x 2 Example (a)

51. E.g. 6—Max. Rev. from Ticket Sales 4. Use the model. We use the model to answer the questions in parts (b) and (c).

52. E.g. 6—Max. Rev. from Ticket Sales Since R(x) = 23,500x – 1000x 2 is a quadratic function with a = –1000 and b = 23,500, the maximum occurs at: Example (b)

53. E.g. 6—Max. Rev. from Ticket Sales So a ticket price of $11.75 gives the maximum revenue. At this price, the revenue is: R(11.75) = 23,500(11.75) – 1000(11.75) 2 = $138,062.50 Example (b)

54. E.g. 6—Max. Rev. from Ticket Sales We want to find the ticket price x for which: R(x) = 23,500x – 1000x 2 = 0 We can solve this quadratic equation algebraically or graphically. Example (c)

55. E.g. 6—Max. Rev. from Ticket Sales From this graph, we see that: R(x) = 0 when x = 0 or x = 23.5 So, according to our model, the revenue would drop to zero if the ticket price is $23.50 or higher. Of course, revenue is also zero if the ticket price is zero! Example (c)