Quadratic Equations and Motion

Standard Form of a Quadratic Equation

• Definition: The standard form of a quadratic equation is defined as:

  • f(x) = ax^2 + bx + c

Characteristics of Standard Form

• Parabola Direction:

  • If a > 0, the parabola opens upward.

  • If a < 0, the parabola opens downward.

• Stretch/Compression:

  • The absolute value of a determines the width of the parabola.

  • If |a| > 1, the parabola is narrower.

  • If |a| < 1, the parabola is wider.

• Y-intercept:

  • The constant c represents the y-intercept of the graph.

• Vertex:

  • The x-coordinate of the vertex is calculated using the formula:

  • x = -\frac{b}{2a}

Vertex Form of a Quadratic Equation

• Definition: The vertex form of a quadratic equation is expressed as:

  • f(x) = a(x - h)^2 + k

Characteristics of Vertex Form

• Vertex:

  • The vertex of the parabola is located at the point (h, k).

• Axis of Symmetry:

  • The vertical line of symmetry is defined by x = h.

• Parabola Direction:

  • Similar to standard form:

  • If a > 0, the parabola opens upward.

  • If a < 0, the parabola opens downward.

Average Rate of Change from a Function

• Example Problem: Find the average rate of change of the function f(x) = x^2 + 4 on the interval [1, 5].

Steps to Solve

1. Find the Values at Endpoints of the Interval:

   • Calculate f(5):

     • f(5) = (5)^2 + 4 = 25 + 4 = 29

   • Calculate f(1):

     • f(1) = (1)^2 + 4 = 1 + 4 = 5

2. Apply the Average Rate of Change Formula:

   • Average Rate of Change = \frac{f(b) - f(a)}{b - a}

   • Substituting the calculated values:

     • Average Rate of Change = \frac{29 - 5}{5 - 1} = \frac{24}{4} = 6

3. Result: The average rate of change is 6.

Vertical Motion Word Problems

• Standard Formula for Vertical Motion:

  • h(t) = -16t^2 + V0t + h0

  • where V0 is the initial vertical velocity and h0 is the initial height.

Practice Problem 1

• Problem Statement: A football is kicked from the ground with an initial vertical velocity of 48 ft/s. How long before it hits the ground?

  • Given:

  • h_0 = 0 (kicked from the ground)

  • h(t) = -16t^2 + 48t + 0

Steps to Solve

1. Set up the equation:

   • 0 = -16t^2 + 48t

   • Factor out t:

     • 0 = -16t(t - 3)

   • Set each factor to zero:

     • 0 = -16t
       ightarrow t = 0

     • 0 = t - 3
       ightarrow t = 3

Solution: The football will take 3 seconds to hit the ground.

Practice Problem 2

• Problem Statement: In a shot put event, an athlete throws the shot put with an initial vertical velocity of 38 ft/s and releases it from a height of 5 ft. How high above the ground is it after 2 seconds?

  • Given:

  • h(t) = -16t^2 + 38t + 5

Steps to Solve

1. Plug in t = 2:

   • h(2) = -16(2)^2 + 38(2) + 5

   • Perform the calculations:

   • h(2) = -16(4) + 76 + 5

     • h(2) = -64 + 76 + 5 = 17 ft

Solution: The height above the ground after 2 seconds is 17 ft.

Finding Maximum/Minimum Value of a Parabola from Standard Form

1. Determine the Direction of the Parabola:

   • Examine the leading coefficient a in the quadratic equation f(x) = ax^2 + bx + c:

     • If a > 0, the parabola opens upward, and the vertex is a minimum.

     • If a < 0, the parabola opens downward, and the vertex is a maximum.

2. Find the Vertex:

   • The x-coordinate of the vertex is computed as:

     • x = -\frac{b}{2a}

3. Calculate the Maximum or Minimum Value:

   • Insert the x-value of the vertex into the original function to find the corresponding y-coordinate. This y-value represents the minimum or maximum value of the function.

Finding Maximum/Minimum Value from Vertex Form

1. Identify the 'k' Value:

   • The 'k' value is the constant added or subtracted outside the parentheses in the vertex form equation.

   • Example: In y = 2(x - 3)^2 + 5, the 'k' value is 5.

2. Determine if it's a Minimum or Maximum:

   • Check the coefficient 'a' in front of the parentheses:

     • If a > 0, the parabola opens upwards, and 'k' is the minimum value.

     • If a < 0, the parabola opens downwards, and 'k' is the maximum value.