8-2 Quadratic Functions (Part # 1)

Example 1

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Example 2 - Vertex Formula

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a) y = 2x^2+4x 
    1. we get the a, b, c in the formula y = ax^2 + bx + c
      • a = 2
      • b = 4
      • c = 0
    2. then we plug in a, b into the formula
      • x = -b / (2a)
      • x = -4 / (2*2) = -4/4 = -1
B) y = -x^2 + 4x - 5 
    1. we get the a, b, c in the formula y = ax^2 + bx + c
      • a = -1
      • b = 4
      • c = -5
    2. then we plug in a, b into the formula
      • x = -b / (2a)
      • x = -4 / (2*-1) = -4/-2 = 2

Example 3 + up/down test

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up/down test
Example 3
  1. a) y= x^2 + 3x + 4 
    • opens upwards because a is Positive
  2. b) y = -3x^2 + 5x
    •  opens downward because a is Negative
  3. c) y = 2x - x^2 + 6
    •  opens downward because a is Negative


Example 4 Graph f(x) = x^2 -2x - 8

Steps to Graph ax^2 + bx + c

Graph f(x) = x^2 -2x - 8 

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for now, I'm just gonna type my work and figure out what to do next

  1. find the line of symmetry -
    1. a = 1, b = -2, c = -8
  2. use this to find the vertex
    1. x = (b/2a)
    2. x = -(-2) / 2(1) = 1
  3. since we know that the along the x axis at 1 will be the vertex we replace x with 1 in the original formula
    1. x=1  
    2. y = x^2 -2x - 8
    3. y = 1^2 + -2 * 1 - 8 = 1- 2 - 8 = -9 
    4. y = -9
  4. the vertex is (1, -9)
  5. since the vertex is -1,-9 we know that x=1 is the axis of symmetry
  6. finding the y-intercept is the easiest to start with because we just replace x with 0
  7. x = 0 | y = x^2 -2x - 8
  8. y = 0 - 8 = -8
  9. y-intercept = (0,-8)
  10. so so To find the x-intercepts, you can set y equal to zero and solve for x:

  11. y = 0 | x = (-b ± sqrt(b^2 - 4ac)) / 2a

    1. x = (-(-2) ± sqrt((-2)^2 - 4(1)(-8))) / 2(1)

      x = (2 ± sqrt(4 + 32)) / 2

      x = (2 ± sqrt(36)) / 2

      x = (2 ± 6) / 2

      x = 8 / 2 or x = -4/ 2

      x = 4 or x = -2 

      sooooo (-2,0) & (4,0)
  12. so since we know 3 y axis points on the graph and the axis of symmetry we can get another point without doing much work 
    1. symmetry line = x = 1, 
    2. calc'd x-intercept 0,-8
      1. the symmetry line is 1 and the known point is  0 since 1-0 = 1 we can add that to the x coordinate of y and keep the same y coordinate to get the mirrored point making another point on the graph (2,-8)
    3.  since we need one more point for the graph we can choose say x=3, | x^2 -2x - 8
      1. y = 3^2 -3*2 - 8 = -5
        1. soooo the new point is (3,-5) if we mirror that along 1,-9 we get  (-1, -5 ) because 3 is 2 more than 1, and 2 less than 1 is -1. we also keep the same y coordinate
    4. so all points are:
      1. (1, -9)
      2. (0,-8)
      3. (2, -8)
      4. (3,-5)
      5. (-1,-5)

Example 5: Graph y = x^2 + 2x + 3

Find the vertex and the axis of symmetry. Sketch these in.
• Find the x-intercept by plugging in 0 for y.
• Find the y-intercept by plugging in 0 for x.
• Reflect your points across the axis of symmetry and connect your dots with a smooth
U-shaped (not V-shaped) curve.

fix the fallowing

  1. find the line of symmetry -
    1. x = (b/2a)
    2. x = -(2) / 2(1) =1
  2. use this to find the vertex
  3. since we know that the along the x axis at -1 will be the vertex we replace x with 1 in the original formula
    1. x=1  
    2. y = x^2 -2x - 8
    3. y = 1^2 + -2 * 1 - 8 = 1- 2 - 8 = -9 
    4. y = -9
  4. the vertex is (1, -9)
  5. since the vertex is -1,-9 we know that x=1 is the axis of symmetry
  6. finding the y-intercept is the easiest to start with because we just replace x with 0
  7. x = 0 | y = x^2 -2x - 8
  8. y = 0 - 8 = -8
  9. y-intercept = (0,-8)
  10. so so To find the x-intercepts, you can set y equal to zero and solve for x:

  11. y = 0 | x = (-b ± sqrt(b^2 - 4ac)) / 2a

    1. x = (-(-2) ± sqrt((-2)^2 - 4(1)(-8))) / 2(1)

      x = (2 ± sqrt(4 + 32)) / 2

      x = (2 ± sqrt(36)) / 2

      x = (2 ± 6) / 2

      x = 8 / 2 or x = -4/ 2

      x = 4 or x = -2 

      sooooo (-2,0) & (4,0)
  12. so since we know 3 y axis points on the graph and the axis of symmetry we can get another point without doing much work 
    1. symmetry line = x = 1, 
    2. calc'd x-intercept 0,-8
      1. the symmetry line is 1 and the known point is  0 since 1-0 = 1 we can add that to the x coordinate of y and keep the same y coordinate to get the mirrored point making another point on the graph (2,-8)
    3.  since we need one more point for the graph we can choose say x=3, | x^2 -2x - 8
      1. y = 3^2 -3*2 - 8 = -5
        1. soooo the new point is (3,-5) if we mirror that along 1,-9 we get  (-1, -5 ) because 3 is 2 more than 1, and 2 less than 1 is -1. we also keep the same y coordinate
    4. so all points are:
      1. (1, -9)
      2. (0,-8)
      3. (2, -8)
      4. (3,-5)
      5. (-1,-5)

Example 6: Graph y = 2x^2 - 8x

• Find the vertex and the axis of symmetry. Sketch these in.
• Find the x-intercept by plugging in 0 for y.
• Find the y-intercept by plugging in 0 for x.
• Reflect your points across the axis of symmetry and connect your dots with a smooth
U-shaped (not V-shaped) curve.

Example 7: h = 16t^2 + 72t + 520

Suppose a particular “star” is projected from a firework at a starting height of 520 feet with an initial upward velocity of 72 ft/sec.

The equation:

h = 16t^2 + 72t + 520

gives the star’s height h in feet at time t in seconds.


a) How long will it take for the star to reach its maximum height?


b) What is the maximum height?