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    Find the equation of the lines that pass through point (-1,3) and are tangent to the curve y = x^3 - 2x + 1

    So far:

    f(x) =  x^3 - 2x + 1

    f'(x) =  3x^2 - 2, f'(a) = 3a^2 - 2

    Therefore, tangent is:

    y -  (a^3 - 2a + 1) = (3a^2 -2)(x+a)

    Is my next step to expand the right hand side and move the left hand side to the right?

    I don't get how I'm going to find what a is since there's a^3 and a^2...I was thinking factorise 'a' then that will leave me with a quadratic to factorise but it doesn't...so confused!

    Thanks
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    I think you have generated a family of lines that are tangent to your curve, parameterised by a, the x-value of the point at which they are tangent.

    At this stage, I'd plug in x=-1, y=3 to restrict the family to those that pass through (-1, 3) and then solve for a.
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    (Original post by gottastudy)
    Therefore, tangent is:

    y -  (a^3 - 2a + 1) = (3a^2 -2)(x+a)

    Is my next step to expand the right hand side and move the left hand side to the right?
    I think you have an error in that it should be (x-a) on the right.

    I'd next substitute the point (-1,3) into your equation and solve for a. It does work out.
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    I've got a = -3/2

    Is that wrong?
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    (Original post by gottastudy)
    I've got a = -3/2

    Is that wrong?
    That's one answer.
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    (Original post by ghostwalker)
    That's one answer.
    Is the other 0? Thanks!
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    (Original post by gottastudy)
    Is the other 0? Thanks!

    Yes.
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    (Original post by ghostwalker)
    Yes.
    So now I have the 2 values of a, do I substitute back into y -  (a^3 - 2a + 1) = (3a^2 -2)(x+a) ?
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    (Original post by gottastudy)
    So now I have the 2 values of a, do I substitute back into y -  (a^3 - 2a + 1) = (3a^2 -2)(x+a) ?
    Since that's the equation of the tangent, then yes; though see my previous comment regarding the sign.
 
 
 
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