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Math Help - quadradic function

  1. #1
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    quadradic function

    1a)

    The bending moment M for a beam at distance X metres from the support is given by the quadratic equation M = 2x squared - 13x + 15, using the quadratic formula calculate the values of X where the bending moment M=0.

    1b)

    Determine the turning point on the curve and find whether it is a maximum or a minimum.

    1c)

    Draw a graph of S=3T squared + 10T - 8 between T= -5 and T=3 and henc solve this equation 3T squared + 10T - 8 = 0.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by andy18 View Post
    1a)

    The bending moment M for a beam at distance X metres from the support is given by the quadratic equation M = 2x squared - 13x + 15, using the quadratic formula calculate the values of X where the bending moment M=0.

    1b)

    Determine the turning point on the curve and find whether it is a maximum or a minimum.
    Standard form of a quadratic function:

    ax^2+bx+c=0

    M=f(x)=0

    2x^2-13x+15=0

    Coefficient of x^2 is positive so you have a minimal value.

    Vertex of the parabola is (\frac{-b}{2a}, f(\frac{-b}{2a}))

    \frac{-b}{2a}=\frac{13}{4}

    f(\frac{13}{4})=2(\frac{12}{4})^2-13(\frac{13}{4})+15 = \frac{-48}{8}

    Turning point is the vertex (\frac{13}{4}, \frac{-49}{8})

    Calculate x by factoring:
    (2x-3)(x-5)=0
    x=\frac{3}{2} \; \; or \; \;x=5

    or use the quadratic formula: x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
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  3. #3
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    Quote Originally Posted by andy18 View Post
    1a)

    The bending moment M for a beam at distance X metres from the support is given by the quadratic equation M = 2x squared - 13x + 15, using the quadratic formula calculate the values of X where the bending moment M=0.

    1b)

    Determine the turning point on the curve and find whether it is a maximum or a minimum.
    I am assuming this is not a Calculus based problem.

    M = 2x^2 - 13x + 15 = 0

    So
    x = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(2)(15)}}{2 \cdot 2}

    x = \frac{13} \pm \sqrt{49}}{4}

    x = \frac{13 \pm 7}{4}

    x = \frac{5, \frac{3}{2}

    For the turning point, the axis of symmetry of the parabola is at
    x = -\frac{-13}{2 \cdot 2} = \frac{13}{4}

    So
    M = 2 \left ( \frac{13}{4} \right ) ^2 - 13 \left ( \frac{13}{4} \right ) + 15 = -\frac{49}{8}

    Since the leading coefficient (2) is positive this will be a minimum point.

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by andy18 View Post
    1c)

    Draw a graph of S=3T squared + 10T - 8 between T= -5 and T=3 and henc solve this equation 3T squared + 10T - 8 = 0.
    Here's the graph. What do you get for the answer?

    -Dan
    Attached Thumbnails Attached Thumbnails quadradic function-graph.jpg  
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