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    The tangent at point P on the curve y= (3x-5)(x+2) is parallel to the line y = 7x+1. Find the co-ordinates of point P.

    I've differentiated the first equation to get 6x+1 and made this equal to 7, then solved to find x, which is 1. Now what?
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    (Original post by gomc)
    The tangent at point P on the curve y= (3x-5)(x+2) is parallel to the line y = 7x+1. Find the co-ordinates of point P.

    I've differentiated the first equation to get 6x+1 and made this equal to 7, then solved to find x, which is 1. Now what?
    assuming your working are correct put x=1 into the equation of the curve to get the corresponding y coodinate
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    (Original post by gomc)
    The tangent at point P on the curve y= (3x-5)(x+2) is parallel to the line y = 7x+1. Find the co-ordinates of point P.

    I've differentiated the first equation to get 6x+1 and made this equal to 7, then solved to find x, which is 1. Now what?
    So you have x=1, what value does your curve take when you sub x=1 into it?

    Your final answer will be P(1, \, \mathrm{ something }\, )
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    (Original post by TeeEm)
    assuming your working are correct put x=1 into the equation of the curve to get the corresponding y coodinate
    Beaten again.
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    (Original post by gomc)
    The tangent at point P on the curve y= (3x-5)(x+2) is parallel to the line y = 7x+1. Find the co-ordinates of point P.

    I've differentiated the first equation to get 6x+1 and made this equal to 7, then solved to find x, which is 1. Now what?

    The gradient of the tangent to the curve at point P is equal to the gradient of the line y = 7x + 1.

    You need to put the differential of the first equation equal to the gradient of the line and solve for x.
    Does it make sense why you would do that?
    Never mind, I can't read, apparently. Ignore me! :P
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    but when i plot the two graphs they dont cross at (1,-6)?? no where near
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    (Original post by Zacken)
    Beaten again.
    is all this LaTeX typing ...
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    (Original post by gomc)
    but when i plot the two graphs they dont cross at (1,-6)?? no where near
    Because the line y = 7x + 1 is parallel to the tangent at P, it's not the actual tangent and hence will not cross the line.

    The tangent to the curve at P is given by y + 6 = 7(x-1) \iff y = 7x - 13 \neq 7x+1.
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    (Original post by Zacken)
    Because the line y = 7x + 1 is parallel to the tangent at P, it's not the actual tangent and hence will not cross the line.

    The tangent to the curve at P is given by y + 6 = 7(x-1) \iff y = 7x - 13 \neq 7x+1.
    god how did i miss that im so tired lol thanks though
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    (Original post by gomc)
    god how did i miss that im so tired lol thanks though
    Welcome, happens to the best of us!
 
 
 
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