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April 3, 2014 April 3, 2014 Presentation Transcript

  • Holt Algebra 1 9-3 Graphing Quadratic FunctionsApril 3, 2014 Today:  2 STAR Tests Completed  Warm-Up  Getting to Know the Quadratic Function: (how the a, b, & c values change the parabola)  Class Work
  • Holt Algebra 1 9-3 Graphing Quadratic Functions x = 0 x = 1 (0, 2) 1. y = 4x2 – 7 2. y = x2 – 3x + 1 Find the axis of symmetry. 3. y = –2x2 + 4x + 3 (2, -12)5. y = x2 + 4x + 5 6. y = -2x2 + 2x – 8 Find the vertex and state whether the graph opens up or down. Warm-Up
  • Holt Algebra 1 9-3 Graphing Quadratic Functions For a quadratic function in the form y = ax2 + bx + c, when x = 0, y = c. The y-intercept of a quadratic function is c Finding the Y intercept Find the vertex and the y-intercept 1. y = x2 – 2 y = x2 – 4x + 4 y = -2x2 – 6x - 3
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Effects of the a, b, & c values With your graph paper, graph the function: y = x2 This is called the parent function. All other quadratic functions are simply transformations of the parent. For the parent function f(x) = x2: • The axis of symmetry is x = 0, or the y-axis. • The vertex is (0, 0) • The function has only one zero, 0.
  • Holt Algebra 1 9-3 Graphing Quadratic FunctionsEffects of the a, b, & c values
  • Holt Algebra 1 9-3 Graphing Quadratic Functions The value of a in a quadratic function determines not only the direction a parabola opens, but also the width of the parabola. Effects of the a, b, & c values
  • Holt Algebra 1 9-3 Graphing Quadratic FunctionsEffects of the a, b, & c values Example 1A: Comparing Widths of Parabolas Order the functions from narrowest graph to widest. f(x) = 3x2, g(x) = 0.5x2, h(x) = 1.5x2 f(x) = 3x2 h(x) = 1.5x2 g(x) = 0.5x2 The function with the narrowest graph has the greatest |a|.
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Effects of the a, b, & c values
  • Holt Algebra 1 9-3 Graphing Quadratic FunctionsEffects of the a, b, & c values The value of c in a quadratic function determines not only the value of the y-intercept but also a vertical translation of the graph of f(x) = ax2 up or down the y-axis.
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Comparing Graphs of Quadratic Functions Compare the graph of the function with the graph of f(x) = x2 opens downward and the graph of f(x) = x2 opens upward. • The graph of is wider than the graph of • The graph of f(x) = x2.
  • Holt Algebra 1 9-3 Graphing Quadratic FunctionsCompare the graph of each the graph of f(x) = x2. g(x) = –x2 – 4 • The graph of g(x) = –x2 – 4 opens downward and the graph of f(x) = x2 opens upward. The vertex of g(x) = –x2 – 4 f(x) = x2 is (0, 0). is translated 4 units down to (0, –3). • The vertex of • The axis of symmetry is the same.
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Example 1: Graphing a Quadratic Function Graph y = 3x2 – 6x + 1. Step 1 Find the axis of symmetry. = 1 The axis of symmetry is x = 1. Simplify. Use x = . Substitute 3 for a and –6 for b. Step 2 Find the vertex. y = 3x2 – 6x + 1 = 3(1)2 – 6(1) + 1 = 3 – 6 + 1 = –2 The vertex is (1, –2). The x-coordinate of the vertex is 1. Substitute 1 for x. Simplify. The y-coordinate is –2.
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Example 1 Continued Step 3 Find the y-intercept. y = 3x2 – 6x + 1 y = 3x2 – 6x + 1 The y-intercept is 1; the graph passes through (0, 1). Identify c.
  • Holt Algebra 1 9-3 Graphing Quadratic Functions 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. Let x = –1. y = 3(–1)2 – 6(–1) + 1 = 3 + 6 + 1 = 10 Let x = –2. y = 3(–2)2 – 6(–2) + 1 = 12 + 12 + 1 = 25 Substitute x-coordinates. Simplify. Two other points are (–1, 10) and (–2, 25). Example 1 Continued
  • Holt Algebra 1 9-3 Graphing Quadratic Functions 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 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve. Example 1 Continued x = 1(–2, 25) (–1, 10) (0, 1) (1, –2) x = 1 (–1, 10) (0, 1) (1, –2) (–2, 25)
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Because a parabola is symmetrical, each point is the same number of units away from the axis of symmetry as its reflected point. Helpful Hint
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Example 2 Graph the quadratic function. y = 2x2 + 6x + 2 Step 1 Find the axis of symmetry. Simplify. Use x = . Substitute 2 for a and 6 for b. The axis of symmetry is x .
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Step 2 Find the vertex. y = 2x2 + 6x + 2 Simplify. Example 2 Continued = 4 – 9 + 2 = –2 The x-coordinate of the vertex is . Substitute for x. The y-coordinate is . The vertex is .
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Step 3 Find the y-intercept. y = 2x2 + 6x + 2 y = 2x2 + 6x + 2 The y-intercept is 2; the graph passes through (0, 2). Identify c. Example 2 Continued
  • Holt Algebra 1 9-3 Graphing Quadratic Functions 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 y = 2(–1)2 + 6(–1) + 1 = 2 – 6 + 2 = –2 Let x = 1 y = 2(1)2 + 6(1) + 2 = 2 + 6 + 2 = 10 Substitute x-coordinates. Simplify. Two other points are (–1, –2) and (1, 10). Example 2 Continued Since the axis of symmetry is x = –1 , choose x values greater than –1 .
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points. Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve. y = 2x2 + 6x + 2 Example 2 Continued (–1, –2) (1, 10) (–1, –2) (1, 10)
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Class Work: Graphing Quadratic Functions
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Example 3 Graph the quadratic function. y + 6x = x2 + 9 Step 1 Find the axis of symmetry. Simplify. Use x = . Substitute 1 for a and –6 for b. The axis of symmetry is x = 3. = 3 y = x2 – 6x + 9 Rewrite in standard form.
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Step 2 Find the vertex. Simplify. Example 3 Continued = 9 – 18 + 9 = 0 The vertex is (3, 0). The x-coordinate of the vertex is 3. Substitute 3 for x. The y-coordinate is 0. . y = x2 – 6x + 9 y = 32 – 6(3) + 9
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Step 3 Find the y-intercept. y = x2 – 6x + 9 y = x2 – 6x + 9 The y-intercept is 9; the graph passes through (0, 9). Identify c. Example 3 Continued
  • Holt Algebra 1 9-3 Graphing Quadratic Functions 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 y = 1(2)2 – 6(2) + 9 = 4 – 12 + 9 = 1 Let x = 1 y = 1(1)2 – 6(1) + 9 = 1 – 6 + 9 = 4 Substitute x-coordinates. Simplify. Two other points are (2, 1) and (1, 4). Example 3 Continued
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points. Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve. y = x2 – 6x + 9 Example 3 Continued x = 3 (3, 0) (0, 9) (2, 1) (1, 4) (0, 9) (1, 4) (2, 1) x = 3 (3, 0)
  • Holt Algebra 1 9-3 Graphing Quadratic Functions 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.
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Example 2 Continued 1 Understand the Problem The answer includes three parts: the maximum height, the time to reach the maximum height, and the time to reach the ground. • The function f(x) = –16x2 + 32x models the height of the basketball after x seconds. List the important information:
  • Holt Algebra 1 9-3 Graphing Quadratic Functions 2 Make a Plan 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. Example 2 Continued
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Solve3 Step 1 Find the axis of symmetry. Use x = . Substitute –16 for a and 32 for b. Simplify. The axis of symmetry is x = 1. Example 2 Continued
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Step 2 Find the vertex. f(x) = –16x2 + 32x = –16(1)2 + 32(1) = –16(1) + 32 = –16 + 32 = 16 The vertex is (1, 16). The x-coordinate of the vertex is 1. Substitute 1 for x. Simplify. The y-coordinate is 16. Example 2 Continued
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Step 3 Find the y-intercept. Identify c.f(x) = –16x2 + 32x + 0 The y-intercept is 0; the graph passes through (0, 0). Example 2 Continued
  • Holt Algebra 1 9-3 Graphing Quadratic Functions 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. (0, 0) (1, 16) (2, 0) Example 2 Continued
  • Holt Algebra 1 9-3 Graphing Quadratic Functions 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. Example 2 Continued (0, 0) (1, 16) (2, 0)
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Look Back4 Check by substitution (1, 16) and (2, 0) into the function. 16 = 16 0 = 0 Example 2 Continued 16 = –16(1)2 + 32(1) ? 16 = –16 + 32 ? 0 = –16(2)2 + 32(0) ? 0 = –64 + 64 ?
  • Holt Algebra 1 9-3 Graphing Quadratic Functions The vertex is the highest or lowest point on a parabola. Therefore, in the example, it gives the maximum height of the basketball. Remember!
  • Holt Algebra 1 9-3 Graphing Quadratic Functions 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 + 24x, 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.
  • Holt Algebra 1 9-3 Graphing Quadratic Functions 1 Understand the Problem The answer includes three parts: the maximum height, the time to reach the maximum height, and the time to reach the pool. Check It Out! Example 2 Continued List the important information: • The function f(x) = –16x2 + 24x models the height of the dive after x seconds.
  • Holt Algebra 1 9-3 Graphing Quadratic Functions 2 Make a Plan 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. Check It Out! Example 2 Continued
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Solve3 Step 1 Find the axis of symmetry. Use x = . Substitute –16 for a and 24 for b. Simplify. The axis of symmetry is x = 0.75. Check It Out! Example 2 Continued
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Step 2 Find the vertex. f(x) = –16x2 + 24x = –16(0.75)2 + 24(0.75) = –16(0.5625) + 18 = –9 + 18 = 9 The vertex is (0.75, 9). Simplify. The y-coordinate is 9. The x-coordinate of the vertex is 0.75. Substitute 0.75 for x. Check It Out! Example 2 Continued
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Step 3 Find the y-intercept. Identify c.f(x) = –16x2 + 24x + 0 The y-intercept is 0; the graph passes through (0, 0). Check It Out! Example 2 Continued
  • Holt Algebra 1 9-3 Graphing Quadratic Functions 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 = 0.75, choose an x-value that is less than 0.75. Let x = 0.5 f(x) = –16(0.5)2 + 24(0.5) = –4 + 12 = 8 Another point is (0.5, 8). Substitute 0.5 for x. Simplify. Check It Out! Example 2 Continued
  • Holt Algebra 1 9-3 Graphing Quadratic Functions 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. (1.5, 0) (0.75, 9) (0, 0) (0.5, 8) (1, 8) Check It Out! Example 2 Continued
  • Holt Algebra 1 9-3 Graphing Quadratic Functions The vertex is (0.75, 9). So at 0.75 seconds, Molly's dive has reached its maximum height of 9 feet. The graph shows the zeros of the function are 0 and 1.5. At 0 seconds the dive has not begun, and at 1.5 seconds she reaches the pool. Molly reaches the pool in 1.5 seconds. (1.5, 0) (0.75, 9) (0, 0) (0.5, 8) (1, 8) Check It Out! Example 2 Continued
  • Holt Algebra 1 9-3 Graphing Quadratic Functions Look Back4 Check by substitution (0.75, 9) and (1.5, 0) into the function. 9 = 9  Check It Out! Example 2 Continued 0 = 0  9 = –16(0.75)2 + 24(0.75) ? 9 = –9 + 18 ? 0 = –16(1.5)2 + 24(1.5) ? 0 = –36 + 36 ?
  • Holt Algebra 1 9-3 Graphing Quadratic Functions 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
  • Holt Algebra 1 9-3 Graphing Quadratic Functions