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College Algebra Fifth Edition James Stewart  Lothar Redlin  Saleem Watson

College Algebra Fifth Edition James Stewart  Lothar Redlin  Saleem Watson. Polynomial and Rational Functions. 4. Quadratic Functions and Models. 4.1. Introduction.

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College Algebra Fifth Edition James Stewart  Lothar Redlin  Saleem Watson

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  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) = anxn + an–1xn–1 + … +a1x + a0 • We have already studied polynomial functions of degree 0 and 1. These are functions of the formP(x) = a0 and P(x) = a1x + a0, 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) = ax2 + 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) = x2, 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) = x2by the transformations given in Section 3.5.

  9. Standard Form of a Quadratic Function • A quadratic function f(x) = ax2 + bx + c can be expressed in the standard formf(x) = a(x –h)2 + kby 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) = 2x2 – 12x + 23. • Express f in standard form. • Sketch the graph of f.

  12. Example (a) E.g. 1—Standard Form • Since the coefficient of x2 is not 1, we must factor this coefficient from the terms involving x before we complete the square.

  13. Example (b) E.g. 1—Standard Form • The standard form tells us that we get the graph of f by: • Taking the parabola y =x2. • Shifting it to the right 3 units. • Stretching it by a factor of 2. • Moving it upward 5 units. • The vertex of the parabola is at (3, 5) and the parabola opens upward.

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

  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 foccurs at x = h.

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

  20. E.g. 2—Minimum Value of a Quadratic Function • Consider the quadratic functionf(x) = 5x2 – 30x + 49 • Express f in standard form. • Sketch the graph of f. • Find the minimum value of f.

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

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

  23. Example (c) E.g. 2—Minimum Value • Since the coefficient of x2 is positive, f has a minimum value. • The minimum value is f(3) = 4.

  24. E.g. 3—Maximum Value of a Quadratic Function • Consider the quadratic functionf(x) = –x2 + x + 2 • Express f in standard form. • Sketch the graph of f. • Find the maximum value of f.

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

  26. Example (b) 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.

  27. Example (b) 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.

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

  29. Example (c) E.g. 3—Maximum Value • Since the coefficient of x2 is negative, f has a maximum value. • This is f(1/2) = 9/4.

  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 – b2/(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) = ax2 + bx + c occurs at:x = –b/2a • If a > 0, then the minimum valueis f(–b/2a). • If a < 0, then the maximum valueis f(–b/2a).

  34. E.g. 4—Finding Maximum & Minimum Values • Find the maximum or minimum value of each quadratic function. • f(x) = x2 + 4x • g(x) = –2x2 + 4x – 5

  35. Example (a) E.g. 4—Finding Max. & Min. Values • f(x) = x2 + 4x is a quadratic function with a = 1 and b = 4. • Thus, the maximum or minimum value occurs at:

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

  37. Example (b) E.g. 4—Finding Max. & Min. Values • g(x) = –2x2 + 4x – 5 is a quadratic function with a = –2 and b = 4. • Thus, the maximum or minimum value occurs at:

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

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

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

  48. Example (a) 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

  49. Example (a) E.g. 6—Max. Rev. from Ticket Sales • Next, we express attendance in terms of x.

  50. Example (a) 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 – 1000x2

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