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5. Let f be the function given by f(x) = x3- 7x + 6.

a. Find the zeros of f
b. Write an equation of the line tangent to the graph of f at x = -1
c. Find the x coordinate of the point where the tangent line is parallel to the secant line on the interval [1, 3].

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  1. Well, first off, if x=1 we have a zero.
    So divide by (x-1) and get
    x^2+x-6
    factor that
    (x-2)(x+3)
    so the zeros are at x = 1, 2, -3

    At x - -1
    y = -1 + 7 + 6 = 12
    y' = 3 x^2 - 7 which is -4 at x = -1
    so this line goes through (-1,12) and has slope m = -4
    You can do the rest

    at x = 1, y = 0
    at x = 3, y = 27-21+6 = 12
    line slope m = 12/2 = 6
    where is slope of tangent line = 6?
    3x^2 -7 = 6
    3 x^2 = 13
    x^2 = 13/3
    x = sqrt (13/3)

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