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Math Help - Secant Line

  1. #1
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    Lightbulb Secant Line

    I would just like to make sure I'm doing this right..thanks
    let f(x)=x^2 x=-2, and x=1.9
    f(-1.9)^2=-3.61
    f(-2)^2=-4
    (-3.61)-(-4)
    (-1.9)-(-4)
    =.39
    2.1
    =0.185714285
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  2. #2
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    Quote Originally Posted by lisa1984wilson View Post
    I would just like to make sure I'm doing this right..thanks
    let f(x)=x^2 x=-2, and x=1.9 --------------------------> [is it x = 1.9 or x = -1.9 ? ]
    f(-1.9)^2=-3.61
    f(-2)^2=-4
    (-3.61)-(-4)
    (-1.9)-(-4)
    =.39
    2.1
    =0.185714285
    Is it x = 1.9 or x = -1.9 ?
    I assume that it is x = -1.9

    f(-1.9) = (-1.9)^2 = (-1.9)\times (-1.9) = +3.61

    f(-2)=(-2)^2=+4

    slope = \frac{3.61-4}{-1.9-(-2)}=\frac{-0.39}{-1.9+2}=\frac{-0.39}{0.1}=-3.9
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  3. #3
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    Lightbulb Tangent

    I need to compute the slope of the line that is tangent to the graph when x=-2 and compare with the slope of the secant in the first part! Thanks
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  4. #4
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    Quote Originally Posted by lisa1984wilson View Post
    I need to compute the slope of the line that is tangent to the graph when x=-2 and compare with the slope of the secant in the first part! Thanks
    the slope of tangent line to the curve f(x) = x^2 at x= -2 is -4

    You already have found the slope of secant in first part.
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