Calculus I

y = x^3 - 8x^2 + 6x -2 has a slope of 6 at two points, what are the coordinates?

I did the power rule to derive the equation: y'= 3x^2 - 8x + 6 , and then I tried to factor it to find the x intercepts but it doesn't work. But anyway, my point of confusion is where does the slope of 6 come in all of this? I know I may need the point slope formula, but can anyone give some tips of how to go about this problem?

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  1. You are told that the slope is 6, and you know that the derivative is the slope, so ...

    3x^2 - 15x + 6 = 6 , your y' has an error

    3x^2 - 15x = 0
    3x(x - 5) = 0
    x = 0 or x = 5

    if x = 0, y = -2
    if x = 5, y = 125 - 200 + 30 - 2 = -47

    So the two points are (0,-2) and (5,-47)

    It did not ask for the equation of those tangents, but the first one is easy to find, since it is a y-intercept, so
    y = 6x - 2
    I will check my work by plotting this line and the original curve.
    http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E3+-+8x%5E2+%2B+6x+-2,+y+%3D+6x-2

    looks line a nice tangent.

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  2. Thanks for your help! But wouldn't the derivative be y'=3x^2-16x+6 not 3x^2-15x+6?

    Then I'll have x(3x-16x)=0
    x=0
    3x-16x=0 ~> -15x=0 , x=0

    I'm confused even more now...

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  3. *** wait. I mean x(3x-16)=0, I'm not sure what's wrong with my eyes. But I will try to figure out how to do it now. Thank you again

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  4. Of course you are right, looks like I had a typo and carried it through,
    so you would end up with

    3x^2 - 16x = 0
    x(3x - 16) = 0
    x = 0, or x = 16/3

    the only change would be ...
    when x = 16/3
    y = (16/3)^3 - 8(16/3)^2 + 6(16/3)- 2 = -1238/27
    so the 2nd point is (16/3, -1238/27)

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