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Chapter 8: Quadratic Equations and Functions

Chapter 8: Quadratic Equations and Functions. 8.1 Square Root Property and Completing the Square. Some quadratic equations can be solved by factoring and applying the zero product rule; however, the zero product rule can be used only if the equation is factorable.

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Chapter 8: Quadratic Equations and Functions

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  1. Chapter 8: Quadratic Equations and Functions

  2. 8.1 Square Root Property and Completing the Square

  3. Some quadratic equations can be solved by factoring and applying the zero product rule; however, the zero product rule can be used only if the equation is factorable. There are other methods that may be used to solve all quadratic equations, factorable and nonfactorable.

  4. The Square Root Property For any real number, k, if x2 = k, then The solution may also be written as , read “x equals plus or minus the square root of k.” Example

  5. Any equation ax2 + bx + c = 0 (a≠ 0) can be rewritten in the form (x + h )2 = k by using a process called completing the square. • For a trinomial ax2 + bx + c (a≠ 0), • ax2 is called the quadratic term, • bx is called the linear term, and • c is called the constant term.

  6. Follow these steps to solve a quadratic equation in the form ax2 + bx + c = 0 (a≠ 0) by completing the square and applying the square root property: • Divide both sides by a to make the leading coefficient 1. • Isolate the variable terms on one side of the equation. • Complete the square: Add the square of one-half the linear term coefficient to both sides of the equation. Then factor the resulting perfect square trinomial. • Apply the square root property and solve for x.

  7. Example Solve the quadratic equation by completing the square and applying the square root property. Lead coefficient is 1. Isolate variable terms. Complete the square. Factor. Apply the square root property. Solve for x.

  8. Exercise 1 Solve the equation: 5y2 + 17 = 3y2 – 239 A) • y = 8, y = –8 • No solution D)

  9. Exercise 2 Solve by completing the square. y2 + 12y – 1 = 0 A) B) C) D) No solution

  10. Exercise 3 Solve by completing the square. 3x2 – 12x + 21 = 0 A) B) C) D)

  11. 8.2 Quadratic Formula

  12. The Quadratic Formula For any quadratic equation of the form ax2 + bx + c = 0 (a≠ 0) the solutions are To solve a quadratic equation using the quadratic formula, first write the equation in the form ax2 + bx + c = 0 and identify the values of a, b, and c.

  13. Example Solve the quadratic equation by using the quadratic formula. The solutions are 3 and –6.

  14. Using the Discriminant to Determine the Number and Type of Solutions to a Quadratic Equation Consider the equation ax2 + bx + c = 0, where a, b, and c are rational numbers and a≠ 0. The expression b2 – 4ac is called the discriminant. • If b2 – 4ac > 0, there will be two real solutions. • If b2 – 4ac is a perfect square, the solutions are rational. • If b2 – 4ac is not a perfect square, the solutions are irrational. • If b2 – 4ac < 0, there will be two imaginary solutions. • If b2 – 4ac = 0, there will be one rational solution.

  15. Examples Use the discriminant to determine the number and type of solutions for each equation. Equation Discriminant Solution b2 – 4ac Type/Number • One rational • Two rational • Two imaginary

  16. The number of solutions tells us the number of x-intercepts. Example: f(x) = x2 – 4x + 3 b2 – 4ac = (–4)2 – 4(1)(3) = 4 There are two real solutions and two x-intercepts.

  17. Methods to Solve a Quadratic Equation • Factor and use the zero product rule. • Use the square root property. Complete the square, if necessary. • Use the quadratic formula.

  18. Exercise 4 Solve using the quadratic formula. 17y – 6 + 39y2 = 0  A) y = 0.2 + 2.9i, y = 0.2 – 2.9i B) y = 1/3, y = 2/5 C) y = –2/3, y = 3/13  D) y = 3/2, y = –13/3

  19. Exercise 5 Use the discriminant to determine the type and number of solutions. –3x2 + 2x + 4 = 0  A) Two rational solutions B) Two imaginary solutions C) One rational solution  D) Two irrational solutions

  20. Exercise 6 Find the x-intercepts of the function. h(x) = 16x2 – 64x + 64   A) (2, 0) and (–2, 0) B) (2, 0) C) (1/2, 0) and (–1/2, 0)  D) There are none.

  21. 8.3 Equations in Quadratic Form

  22. Some equations can be reduced to a quadratic equation and solved. • Some radical equations reduce to quadratic equations after both sides are squared. Check all potential solutions for extraneous solutions! • Some higher-order polynomials can be factored then solved like a quadratic.

  23. Some equations that are not quadratic can be manipulated to appear as equations in quadratic formby using substitution. Example To solve make the substitution u = 3m – 1. When using substitution, it is critical to reverse substitution to solve the equation in terms of the original variable.

  24. Exercise 7 Solve the equation. z4 – 625 = 0  A) z = 5 B) z = 5, z = –5 C) z = 5, z = –5, z = 5i, z = –5i D) z = 5, z = –5, z = 25i, z = –25i

  25. Exercise 8 Solve the equation. (t + 10)2 – (t + 10) – 12 = 0 A) t = 4, t = –3 B) t = –14, t = –7 C) t = 6, t = 13 D) t = –6, t = –13

  26. Exercise 9 Solve the equation. z2/3 + 2z1/3 – 15 = 0 A) z = 6, z = –6 B) z = 27, z = –125 C) z = 27, z = 125 D) z = 9, z = 25

  27. 8.4 Graphs of Quadratic Functions

  28. The graph of a quadratic function f(x) = ax2 + bx + c (a≠ 0) is a parabola. • If a > 0, the parabola opens up, and the vertex is the lowest point on the graph. • If a < 0, the parabola opens down, and the vertexis the highest point on the graph. • The axis of symmetry is the vertical line that passes through the vertex.

  29. Graphs of f(x) = x2 + k • If k > 0, then the graph of f(x) = x2 + k is the same as the graph of y = x2 shifted upk units. • If k < 0, then the graph of f(x) = x2 + k is the same as the graph of y = x2 shifted downk units.

  30. Graphs of f(x) = (x – h)2 • If h > 0, then the graph of f(x) = (x – h)2 is the same as the graph of y = x2 shifted h units to the right. • If h < 0, then the graph of f(x) = (x – h)2 is the same as the graph of y = x2 shifted h units to the left.

  31. Graphs of f(x) = ax2 • The graph of a quadratic function of the form f(x) = ax2 is a parabola that opens up when a > 0 and opens down when a < 0. • If a > 1, the graph of y = x2 is stretched vertically by a factor of a. • If 0 < a < 1, the graph of y = x2 is shrunk vertically by a factor of a.

  32. Graphs of f(x) = a(x – h)2 + k • The vertex is located at (h, k). • The axis of symmetry is the line x = h. • If a > 0, the parabola opens up and k is the minimum valueof the function. • If a < 0, the parabola opens down and k is the maximum valueof the function.

  33. Exercise 10 Which is the graph of g(x) = (x + 6)2? A) B) C) D)

  34. Exercise 11 Which is the graph of h(x) = –4x2? A) B) C) D)

  35. Exercise 12 Identify the maximum or minimum value of the function g(x) = –2(x – 8)2 + 4.   A) (8, 4); minimum B) (8, 4); maximum C) (–8, 4); maximum  D) (4, 8); minimum

  36. 8.5 Vertex of a Parabola and Applications

  37. The graph of a quadratic function is a parabola, and if the function is written in the form f(x) = a(x – h)2 + k (a≠ 0), then the vertex is at (h, k). vertex: (1, 7) axis of symmetry: x = 1 A quadratic function can be written in the form f(x) = a(x – h)2 + k (a≠ 0) by completing the square.

  38. Example Find the vertex of f(x) = x2 + 12x – 2. f(x) = x2 + 12x – 2 = 1(x2 + 12x) – 2 Factor the leading coefficient from the variable terms. = 1(x2 + 12x ) – 2 Complete the square on the expression in parentheses. = 1(x2 + 12x + 36 – 36) – 2 Add and subtract within parentheses. = 1(x2 + 12x + 36) – 36 – 2 Regroup. = (x + 6)2 – 38 Factor and simplify. The vertex is at (–6, –38).

  39. The Vertex Formula For f(x) = ax2 + bx + c (a≠ 0), the vertex is given by Using the vertex formula, the vertex off(x) = x2 + 12x – 2 is

  40. The location of the vertex and the direction that the parabola opens can be used to determine whether the function has any x-intercepts. For example, given f(x) = x2 + 12x – 2, the vertex (–6, –38) is below the x-axis. Furthermore, because a > 0, the parabola opens upward. Therefore, the parabola must cross the x-axis in two places, and so there are two x-intercepts.

  41. Finding the vertex of a parabola is useful in many applications because the vertex is either highest point or the lowest point on the graph. Given a quadratic function for a model, we may be able to find values such as • Maximum or minimum heights and the time it takes to reach these heights • Maximum or minimum profits and when these may have occurred

  42. Exercise 13 Write the quadratic function in the form f(x) = a(x – h)2 + k . f(x) = x2 + 18x – 8  A) f(x) = (x + 18)2 – 8 B) f(x) = (x – 18)2 + 8 C) f(x) = (x – 9)2 + 11/4 D) f(x) = (x + 9)2 – 89

  43. Exercise 14 Find the vertex of the parabolah(x) = –2x2 + 16x – 3. A) (16, –3) B) (4, 29) C) (0, –3) D) (–2, 13)

  44. Exercise 15 A model rocket is launched from a raised platform at a speed of 192 feet per second. Its height in feet is given by h(t) = –16t2 + 192t + 20 (t = seconds after launch). After how many seconds does the object reach its maximum height? A) 20 sec B) 6 sec C) 3 sec D) 8 sec

  45. D A B C D B C D B D B B D B B Exercise Answers

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