Quadratic Functions and Motion

Quadratic Equations

Standard Form of Quadratic Equations

• The standard form of a quadratic equation is given by:

  • $f(x) = ax^2 + bx + c$

• Parabola Direction:

  • If $a > 0$, the parabola opens upward.

  • If $a < 0$, the parabola opens downward.

• Stretch/Compression:

  • The absolute value of the coefficient $a$ determines the width of the parabola:

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

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

• Y-Intercept:

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

• Vertex:

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

  • $x = -\frac{b}{2a}$

Vertex Form of Quadratic Equations

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

  • $f(x) = a(x - h)^2 + k$

• Vertex:

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

• Axis of Symmetry:

  • The axis of symmetry corresponds to the vertical line given by $x = h$.

• Parabola Direction: Same as in standard form:

  • If $a > 0$, opens upward.

  • If $a < 0$, opens downward.

Average Rate of Change of a Function

Example Calculation

• To find the average rate of change of the function:

  • $f(x) = x^2 + 4$
    over the interval $[1, 5]$.

  • Steps:

  1. Compute the function values at the endpoints of the interval:

     • $f(5) = (5)^2 + 4 = 25 + 4 = 29$

     • $f(1) = (1)^2 + 4 = 1 + 4 = 5$

  2. Apply the average rate of change formula:

     • $\text{Average Rate of Change} = \frac{f(5) - f(1)}{5 - 1}$

     • $= \frac{29 - 5}{4}$

     • $= \frac{24}{4} = 6$

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

Vertical Motion Word Problems

General Formula for Vertical Motion

• The standard formula for vertical motion is:

  • $h(t) = -16t^2 + V<em>0t + h</em>0$

  • Where:

  • $V_0$ = initial velocity

  • $h_0$ = initial height

Example Problem 1

• A football is kicked from the ground with an initial vertical velocity of 48 ft/s. Determine the time until it hits the ground:

  • Using the formula:

  • $h(t) = -16t^2 + 48t + 0$

  • Set equation to zero to find when it hits the ground:

  • $0 = -16t(t - 3)$

  • Solving yields:

  • $0 = -16t <br>ightarrow t = 0 ext{ sec}$

  • $0 = t - 3 <br>ightarrow t = 3 ext{ sec}$

  • Time until it hits the ground: 3 seconds.

Example Problem 2

• In a shot put event, an athlete releases the shot put from a height of 5 ft with an initial vertical velocity of 38 ft/s. Calculate its height after 2 seconds:

  • Applying the formula:

  • $h(t) = -16t^2 + 38t + 5$

  • Substitute $t = 2$:

  • $h(2) = -16(2)^2 + 38(2) + 5$

  • $= -64 + 76 + 5 = 17 ext{ ft}$

  • Height after 2 seconds: 17 ft.

Finding Maximum/Minimum Values of a Quadratic Function

From Standard Form

Steps:

1. Determine the Direction of the Parabola:

   • Examine the leading coefficient $a$ of the quadratic equation:

     • If $a > 0$, the parabola opens upward (minimum vertex).

     • If $a < 0$, the parabola opens downward (maximum vertex).

2. Find the Vertex:

   • Use the formula to find the x-coordinate of the vertex:

     • $x = -\frac{b}{2a}$

3. Calculate the Maximum or Minimum Value:

   • Determine the corresponding y-coordinate by plugging the x-value of the vertex back into the original function.

   • This y-value represents the minimum or maximum value of the function.

From Vertex Form

Steps:

1. Identify the 'k' Value:

   • The 'k' value is the constant term outside the parentheses in the vertex form equation.

     • For instance, in $y = 2(x - 3)^2 + 5$, the 'k' value is 5.

2. Determining Minimum or Maximum:

   • Check the coefficient $a$ in front of the parenthesis:

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

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