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  1. 9-3 Graphing Quadratic Functions Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 1 Holt Algebra 1

  2. Warm Up Find the axis of symmetry. 1. y = 4x2 – 7 2. y =x2 – 3x + 1 3. y = –2x2 +4x + 3 4.y = –2x2 + 3x – 1 Find the vertex. 5. y = x2 + 4x + 5 6. y = 3x2 + 2 7. y = 2x2 +2x – 8 x = 0 x = 1 (0, 2) (–2, 1)

  3. Objective Graph a quadratic function in the form y = ax2 + bx + c.

  4. Recall that a y-intercept is the y-coordinate of the point where a graph intersects the y-axis. The x-coordinate of this point is always 0. For a quadratic function written in the form y = ax2 + bx + c, when x = 0, y = c. So the y-intercept of a quadratic function is c.

  5. Use x = . Substitute 3 for a and –6 for b. Example 1: Graphing a Quadratic Function Graph y = 3x2– 6x + 1. Step 1 Find the axis of symmetry. = 1 Simplify. The axis of symmetry is x = 1. Step 2 Find the vertex. The x-coordinate of the vertex is 1. Substitute 1 for x. y = 3x2– 6x + 1 = 3(1)2– 6(1) + 1 = 3 – 6 + 1 Simplify. =–2 The y-coordinate is –2. The vertex is (1, –2).

  6. Example 1 Continued Step 3 Find the y-intercept. y = 3x2 – 6x + 1 y = 3x2 – 6x + 1 Identify c. The y-intercept is 1; the graph passes through (0, 1).

  7. Example 1 Continued Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y-intercept. Since the axis of symmetry is x = 1, choose x-values less than 1. Substitute x-coordinates. Let x = –1. Let x = –2. y = 3(–1)2 – 6(–1) + 1 y = 3(–2)2 – 6(–2) + 1 = 3 + 6 + 1 = 12 + 12 + 1 Simplify. =10 =25 Two other points are (–1, 10) and (–2, 25).

  8. (–2, 25) (–2, 25) x = 1 x = 1 (–1, 10) (–1, 10) (0, 1) (0, 1) (1, –2) (1, –2) Example 1 Continued Graph y = 3x2 – 6x + 1. Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points. Step 6Reflect the points across the axis of symmetry. Connect the points with a smooth curve.

  9. Helpful Hint Because a parabola is symmetrical, each point is the same number of units away from the axis of symmetry as its reflected point.

  10. Use x = . Substitute 2 for a and 6 for b. The axis of symmetry is x . Check It Out! Example 1a Graph the quadratic function. y = 2x2 + 6x + 2 Step 1 Find the axis of symmetry. Simplify.

  11. The x-coordinate of the vertex is . Substitute for x. = 4 – 9 + 2 The y-coordinate is . =–2 The vertex is . Check It Out! Example 1a Continued Step 2 Find the vertex. y = 2x2 + 6x + 2 Simplify.

  12. Check It Out! Example 1a Continued Step 3 Find the y-intercept. y = 2x2 + 6x + 2 y = 2x2 + 6x + 2 Identify c. The y-intercept is 2; the graph passes through (0, 2).

  13. Since the axis of symmetry is x = –1 , choose x values greater than –1 . Check It Out! Example 1a Continued Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y-intercept. Let x = –1 Let x = 1 Substitute x-coordinates. y = 2(–1)2 + 6(–1) + 2 y = 2(1)2 + 6(1) + 2 = 2 – 6 + 2 = 2 + 6 + 2 Simplify. =–2 =10 Two other points are (–1, –2) and (1, 10).

  14. (1, 10) (1, 10) (–1, –2) (–1, –2) Check It Out! Example 1a Continued y = 2x2 + 6x + 2 Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points. Step 6Reflect the points across the axis of symmetry. Connect the points with a smooth curve.

  15. Use x = . Substitute 1 for a and –6 for b. Check It Out! Example 1b Graph the quadratic function. y + 6x = x2 + 9 y = x2 – 6x + 9 Rewrite in standard form. Step 1 Find the axis of symmetry. =3 Simplify. The axis of symmetry is x = 3.

  16. Check It Out! Example 1b Continued Step 2 Find the vertex. y = x2 – 6x + 9 The x-coordinate of the vertex is 3. Substitute 3 for x. y = 32 – 6(3) + 9 = 9 – 18 + 9 Simplify. =0 The y-coordinate is 0. The vertex is (3, 0).

  17. Check It Out! Example 1b Continued Step 3 Find the y-intercept. y = x2 – 6x + 9 y = x2 – 6x + 9 Identify c. The y-intercept is 9; the graph passes through (0, 9).

  18. Check It Out! Example 1b Continued Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y- intercept. Since the axis of symmetry is x = 3, choose x-values less than 3. Let x = 2 Let x = 1 Substitute x-coordinates. y = 1(2)2 – 6(2) + 9 y = 1(1)2 – 6(1) + 9 = 4 – 12 + 9 = 1 – 6 + 9 Simplify. =1 =4 Two other points are (2, 1) and (1, 4).

  19. x = 3 (0, 9) x = 3 (0, 9) (1, 4) (1, 4) (2, 1) (2, 1) (3, 0) (3, 0) Check It Out! Example 1b Continued y = x2 – 6x + 9 Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points. Step 6Reflect the points across the axis of symmetry. Connect the points with a smooth curve.

  20. Example 2: Application The height in feet of a basketball that is thrown can be modeled by f(x) = –16x2 + 32x, where x is the time in seconds after it is thrown. Find the basketball’s maximum height and the time it takes the basketball to reach this height. Then find how long the basketball is in the air.

  21. 1 Understand the Problem Example 2 Continued The answer includes three parts: the maximum height, the time to reach the maximum height, and the time to reach the ground. List the important information: • The function f(x) = –16x2 + 32x models the height of the basketball after x seconds.

  22. Make a Plan 2 Example 2 Continued Find the vertex of the graph because the maximum height of the basketball and the time it takes to reach it are the coordinates of the vertex. The basketball will hit the ground when its height is 0, so find the zeros of the function. You can do this by graphing.

  23. 3 Solve Use x = . Substitute –16 for a and 32 for b. Example 2 Continued Step 1 Find the axis of symmetry. Simplify. The axis of symmetry is x = 1.

  24. Example 2 Continued Step 2 Find the vertex. f(x) = –16x2 + 32x The x-coordinate of the vertex is 1. Substitute 1 for x. = –16(1)2 + 32(1) = –16(1) + 32 = –16 + 32 Simplify. = 16 The y-coordinate is 16. The vertex is (1, 16).

  25. Example 2 Continued Step 3 Find the y-intercept. f(x) = –16x2 + 32x + 0 Identify c. The y-intercept is 0; the graph passes through (0, 0).

  26. (1, 16) (0, 0) (2, 0) Example 2 Continued Step 4 Graph the axis of symmetry, the vertex, and the point containing the y-intercept. Then reflect the point across the axis of symmetry. Connect the points with a smooth curve.

  27. (1, 16) (0, 0) (2, 0) Example 2 Continued The vertex is (1, 16). So at 1 second, the basketball has reached its maximum height of 16 feet. The graph shows the zeros of the function are 0 and 2. At 0 seconds the basketball has not yet been thrown, and at 2 seconds it reaches the ground. The basketball is in the air for 2 seconds.

  28. ? 16 = –16(1)2 + 32(1) ? 16 = –16 + 32  16 = 16 ? 0 = –16(2)2 + 32(2) ? 0 = –64 + 64  0 = 0 4 Example 2 Continued Look Back Check by substitution (1, 16) and (2, 0) into the function.

  29. Remember! The vertex is the highest or lowest point on a parabola. Therefore, in the example, it gives the maximum height of the basketball.

  30. Check It Out! Example 2 As Molly dives into her pool, her height in feet above the water can be modeled by the function f(x) = –16x2 + 16x + 12, where x is the time in seconds after she begins diving. Find the maximum height of her dive and the time it takes Molly to reach this height. Then find how long it takes her to reach the pool.

  31. 1 Understand the Problem Check It Out! Example 2 Continued The answer includes three parts: the maximum height, the time to reach the maximum height, and the time to reach the pool. List the important information: • The function f(x) = –16x2 + 16x + 12models the height of the dive after x seconds.

  32. Make a Plan 2 Check It Out! Example 2 Continued Find the vertex of the graph because the maximum height of the dive and the time it takes to reach it are the coordinates of the vertex. The diver will hit the water when its height is 0, so find the zeros of the function. You can do this by graphing.

  33. 3 Solve Use x = . Substitute –16 for a and 16 for b. Check It Out! Example 2 Continued Step 1 Find the axis of symmetry. Simplify. The axis of symmetry is x = 0.5.

  34. The x-coordinate of the vertex is 0.5. Substitute 0.5 for x. Check It Out! Example 2 Continued Step 2 Find the vertex. f(x) = –16x2 + 16x + 12 = –16(0.5)2 + 16(0.5)+12 = –16(0.25) + 8 +12 Simplify. = –4 + 20 = 16 The y-coordinate is 9. The vertex is (0.5, 16).

  35. Check It Out! Example 2 Continued Step 3 Find the y-intercept. f(x) = –16x2 + 16x + 12 Identify c. The y-intercept is 12; the graph passes through (0, 12).

  36. Check It Out! Example 2 Continued Step 4 Find another point on the same side of the axis of symmetry as the point containing the y-intercept. Since the axis of symmetry is x = 0.5, choose an x-value that is less than 0.5. Let x = 0.25 f(x) = –16(0.25)2 + 16(0.25) + 12 Substitute 0.25 for x. = –1 + 4 + 12 Simplify. = 15 Another point is (0.25, 15).

  37. Check It Out! Example 2 Continued Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and the other point. Then reflect the points across the axis of symmetry. Connect the points with a smooth curve.

  38. Check It Out! Example 2 Continued The vertex is (0.5, 16). So at 0.5 seconds, Molly's dive has reached its maximum height of 16 feet. The graph shows the zeros of the function are –0.5 and 1.5 seconds. At –0.5 seconds the dive has not begun, and at 1.5 seconds she reaches the pool. Molly reaches the pool in 1.5 seconds.

  39. ? 16 = –16(0.5)2 + 16(0.5) + 12 ? 16 = –4 + 8 +12   0 = 0 16 = 16 ? 0 = –16(1.5)2 + 16(1.5) +12 ? 0 = –36 + 24 + 12 4 Check It Out! Example 2 Continued Look Back Check by substituting (0.5, 16) and (1.5, 0) into the function.

  40. Lesson Quiz 1. Graph y = –2x2 – 8x + 4. 2. The height in feet of a fireworks shell can be modeled by h(t) = –16t2 + 224t, where t is the time in seconds after it is fired. Find the maximum height of the shell, the time it takes to reach its maximum height, and length of time the shell is in the air. 784 ft; 7 s; 14 s