10.4

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  • 1. -3, -2 2. -4 3. no solution
  • 10.4

    1. 1. Vocabulary Quadratic Equation – an equation that can be written in the standard form ax 2 + bx + c = 0 where a ¹ 0 Zero(s) of a function – x value(s) for which y = 0* Zero(s) of a polynomial function and root(s) of apolynomial are the same!
    2. 2. Solve by Graphing (vs. Factoring)Recall Solve by Factoring: x 2 - 6x + 5 = 0 (x -1)(x - 5) = 0 x =1, x = 5Solve by Graphing:
    3. 3. Example 1 Solve a quadratic equation having two solutionsSolve x2 – 2x = 3 by graphing.SOLUTIONSTEP 1 Write the equation in standard form. x2 – 2x = 3 Write original equation. x2 – 2x – 3 = 0 Subtract 3 from each side.STEP 2 Graph the related function y = x2 – 2x – 3 . The x-intercepts are –1 and 3.
    4. 4. Example 1 Solve a quadratic equation having two solutionsANSWERThe solutions of the equation x2 – 2x = 3 are – 1 and 3.CHECK You can check –1 and 3 in the original equation. x2 – 2x = 3 x2 – 2x = 3 Write original equation. ? ( – 1)2 – 2 (– 1) = 3 ? ( 3)2 – 2( 3) = 3 Substitute for x. 3 = 3 3 = 3 Simplify. Each solution checks.
    5. 5. Example 2 Solve a quadratic equation having one solutionSolve – x2 + 2x = 1 by graphing.SOLUTIONSTEP 1 Write the equation in standard form. – x2 + 2x = 1 Write original equation. – x2 + 2x – 1 = 0 Subtract 1 from each side.STEP 2 Graph the related function y = – x2 + 2x – 1 . The x-intercept is 1.
    6. 6. Example 2 Solve a quadratic equation having one solutionANSWERThe solution of the equation – x2 + 2x = 1 is 1.
    7. 7. Example 3 Solve a quadratic equation having no solutionSolve x2 + 7 = 4x by graphing.SOLUTIONSTEP 1 Write the equation in standard form. x2 + 7 = 4x Write original equation. x2 – 4x + 7 = 0 Subtract 4x from each side.STEP 2 Graph the related function y = x2 – 4x + 7. The graph has no x-intercepts.
    8. 8. Example 3 Solve a quadratic equation having no solutionANSWERThe equation x2 + 7 = 4x has no solution.
    9. 9. Number of Solutions of a Quadratic Equation Two Solutions One Solution No SolutionA quadratic equation A quadratic equation A quadratic equationhas two solutions if has one solution if the has no real solution ifthe graph of its related graph of its related the graph of its relatedfunction has two x- function has one x- function has no x-intercepts. intercept. intercepts.
    10. 10. Example 4 Multiple Choice PracticeThe graph of the equationy = x2 + 6x – 7 is shown. Forwhat value or values of x is y = 0? x = –7 only x = 1 only x = – 7 and x = 1 x = –1 and x = 7SOLUTIONYou can see from the graph that the x-intercepts are –7and 1. So, y = 0 when x = –7 and x = 1.ANSWER The correct answer is C.
    11. 11. Example 5 Approximate the zeros of a quadratic functionApproximate the zeros of y = x2 + 4x + 1 to the nearesttenth.SOLUTIONSTEP 1 Graph the function y = x2 + 4x + 1. There are two x-intercepts: one between – 4 and –3 and another between –1 and 0.
    12. 12. Example 5 Approximate the zeros of a quadratic functionSTEP 2 Make a table of values for x-values between – 4 and – 3 and between – 1 and 0 using an increment of 0.1. Look for a change in the signs of the function values. x – 3.9 – 3.8 – 3.7 – 3.6 – 3.5 – 3.4 – 3.3 – 3.2 – 3.1 y 0.61 0.24 – 0.11 – 0.44 – 0.75 – 1.04 – 1.31 – 1.56 – 1.79 x – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 y – 1.79 – 1.56 – 1.31 – 1.04 – 0.75 – 0.44 – 0.11 0.24 0.61
    13. 13. Example 5 Approximate the zeros of a quadratic functionANSWERIn each table, the function value closest to 0 is – 0.11.So, the zeros of y = x2 + 4x + 1 are about – 3.7 andabout – 0.3.
    14. 14. Example 6 Solve a multi-step problemSPORTSAn athlete throws a shot put withan initial vertical velocity of 40 feetper second.a. Write an equation that models the height h (in feet) of the shot put as a function of the time t (in seconds) after it is thrown.b. Use the equation to find the time that the shot put is in the air.
    15. 15. Example 6 Solve a multi-step problemSOLUTIONa. Use the initial vertical velocity and height to write a vertical motion model. h = – 16t2 + vt + s Vertical motion model Substitute 40 for v and h = – 16t2 + 40t + 6.5 6.5 for s.b. The shot put lands when h = 0. To find the time t when h = 0, solve 0 = –16t 2 + 40t + 6.5 for t. To solve the equation, graph the related function h = –16t2 + 40t + 6.5 on a graphing calculator. Use the trace feature to find the t-intercepts.
    16. 16. Example 6 Solve a multi-step problemANSWERThere is only one positive t-intercept. The shot put is inthe air for about 2.6 seconds.
    17. 17. Relating Roots of Polynomials,Solutions of Equations, x-interceptsof Graphs, and Zeros of Functions The Roots of the Polynomial -x 2 +8x -12 are 2 & 6. The Solutions of the equation -x 2 +8x -12 = 0 are 2 & 6. The x-intercepts of the graph of y = -x +8x -12 occur 2 where y=0, so the x-intercepts are 2 and 6. The Zeros of the function y = -x +8x -12 are the 2 values of x for which y=0, so the zeros are 2 and 6.
    18. 18. 10.4 Warm-UpSolve the equation by graphing.1. x 2 + 5x + 6 = 02. x 2 +8x +16 = 03. x 2 - 2x + 3 = 0

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