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f(x)=3-5x , (-1,8) find the slope of the tangent line to the graph of the function at the given point
how do you figure this out?

  • calculus - ,

    since your function is a straight line
    and it is in the form f(x) = mx + b
    the slope anywhere on the line is -5

    no work needed here.

  • calculus - ,

    To find the slope of the tangent line you have to take the first derivative of f(x). So f'(x)=-5. Or you can also see that from the formula y=mx+b, where m is slope and b is your y intercept, that -5 is the slope. Because the original function is a line, the slope is the same at every point.

  • calculus - ,

    To find the slope of the tangent line at a point means to find the derivative of f(x).

    f(x) = 3 - 5x
    f' = 0 - 5
    f' = -5

    Then, normally, you would plug in the given point, (-1, 8), to find the slope.

    But since f' = 5 with no variable x, for instance, there is nowhere to plug in the point.

    Therefore, the slope is everywhere.

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