Quadratic Equations and Motion

Standard Form of Quadratic Equations

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

    • f(x)=ax2+bx+cf(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 aa determines the width of the parabola:

    • If a > 1, the parabola is narrower.

    • If |a| < 1, it is wider.

  • Y-Intercept:

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

  • Vertex:

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

    • x=−b2ax = -\frac{b}{2a}

Vertex Form of Quadratic Equations

  • The vertex form of a quadratic equation is:

    • f(x)=a(x−h)2+kf(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 is described by the vertical line:

    • x=hx = h.

  • Parabola Direction in Vertex Form:

    • Consistent with standard form, if a > 0, the parabola opens upward; if a < 0, it opens downward.

Average Rate of Change of a Function

  • Example:

    • Find the average rate of change of the function f(x)=x2+4f(x) = x^2 + 4 on the interval [1,5][1, 5].

    • Steps:

    1. Evaluate the function at the endpoints of the interval:

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

      • f(1)=(1)2+4=1+4=5f(1) = (1)^2 + 4 = 1 + 4 = 5

    2. Apply the formula for the average rate of change:

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

      • =29−54=244=6= \frac{29 - 5}{4} = \frac{24}{4} = 6

    • Result:

    • The average rate of change is 6.

Vertical Motion Problems

  • Standard Formula for Vertical Motion:

    • h(t)=−16t2+V<em>0t+h</em>0h(t) = -16t^2 + V<em>0 t + h</em>0

    • Where:

      • V0V_0 = initial velocity

      • h0h_0 = initial height

  • Example 1: A football kicked from the ground with an initial vertical velocity of 48 ft/s:

    • To find the time until it hits the ground:

    • h(t)=−16t2+48t+0h(t) = -16t^2 + 48t + 0

    • Factorization: h(t)=−16t(t−3)h(t) = -16t(t - 3)

    • Set heights to zero:

      • 0=−16t(t−3)0 = -16t(t - 3)

      • Solutions:

      • t=0 extsect = 0 \, ext{sec} (initial time)

      • t=3 extsect = 3 \, ext{sec} (time until hitting ground)

    • Example 2: Shot put thrown with an initial vertical velocity of 38 ft/s from a height of 5 ft:

    • Find height after 2 seconds:

    • h(t)=−16t2+38t+5h(t) = -16t^2 + 38t + 5

    • Substituting t=2t = 2:

      • h(2)=−16(22)+38(2)+5h(2) = -16(2^2) + 38(2) + 5

      • =−16(4)+76+5= -16(4) + 76 + 5

      • =−64+76+5=17 extft= -64 + 76 + 5 = 17 \, ext{ft}

Finding Maximum/Minimum Values of Parabolas

  • From Standard Form:

    • Step 1: Determine the direction of the parabola based on the leading coefficient aa:

    • 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.

    • Step 2: Find the vertex:

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

      • x=−b2ax = -\frac{b}{2a}

    • Step 3: Calculate maximum or minimum value:

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

  • From Vertex Form:

    • Step 1: Identify the value of 'k':

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

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

    • Step 2: Determine if it's a minimum or maximum:

    • Check the coefficient 'a':

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

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