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Math Help - Find Co-ords of parallel tangent to quad function

  1. #1
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    Find Co-ords of parallel tangent to quad function

    Looking for the co-ordinates of the tangent parallel to this when x=0.6

    2x^3-6x^2+10x+9

    so far got - x=0.6 y=13.272 gradient = 4.96 (of original function)

    how do i find the parallel tangent and then its co-ordinates.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    You mean you're looking for the equation of the tangent of the curve?
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  3. #3
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    A 2nd tangent parallel to one at (0.6,13.272)
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    Quote Originally Posted by stophe73 View Post
    Looking for the co-ordinates of the tangent parallel to this when x=0.6

    2x^3-6x^2+10x+9

    so far got - x=0.6 y=13.272 gradient = 4.96 (of original function)

    how do i find the parallel tangent and then its co-ordinates.
    Quote Originally Posted by stophe73 View Post
    A 2nd tangent parallel to one at (0.6,13.272)
    I don't get this either. You are looking for "the co-ordinates of the tangent parallel to this" y =
    2x^3-6x^2+10x+9 at the point (0.6, 13.272)? We can certainly get the tangent line to this curve at x = 0.6. Is this what you are looking for? What do you mean by "2nd tangent?" This is badly worded.

    -Dan
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  5. #5
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    this function(graph is s shaped 2 knees? turns?) has a tangent to it at co ords 0.6,13.272 with gradient 4.96.
    I'm looking for another tangent with a parallel gradient , but different co-ordinates.
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  6. #6
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    I can only interpret this as meaning that you are looking for a value of x, other than .6 that has the same slope.

    With f(x)= 2x^3-6x^2+10x+9, f'(x)= 6x^2-12x+ 10. When x= .6, that is
    f'(.6)= 6(.6)^2- 12(.6)+ 10= 4.96

    so now you are looking for another value of x that satisfies f'(x)= 6x^2- 12x+10= 4.96
    That is the same as solving 6x^2- 12x+ 5.04 and that is especially easy to solve because you already know that x= .6 is a solution: divide 6x^2- 12x+ 5.04 by x- .6.
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  7. #7
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    The question asks for the x,y on the same 'line' where the same gradient / slope occurs. i.e the co-ordinates where a parallel tangent exists.

    So how would I carry out it out for the y, co-ordinate? ('original' y by calculation 13.272)
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  8. #8
    MHF Contributor Unknown008's Avatar
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    From the post of HallsofIvy, you should see that you will get another value of x where a tangent will have the same gradient as the tangent at x = 0.6.

    Once you get the other x coordinate, you can find the other coordinate.
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  9. #9
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    You mean with y=mx+c , how do i find c for the "2nd tangent" , sorry about the confusing grammar.
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  10. #10
    MHF Contributor Unknown008's Avatar
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    No, I mean you look for the other value of x which satisfies the equation given by HallsofIvy:

    6x^2 -12x +5.04 = 0

    One solution is 0.6, as given by your question. There is another one. Find it, then use that x to get the y-coordinate.
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  11. #11
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    I see...but how do i find the y co-ord and the intercept?
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  12. #12
    MHF Contributor Unknown008's Avatar
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    You do just like you did for the first one, you use the value of that new x and plug it in the original equation.

    Then, you take the point and the gradient of the tangent to get directly the equation of the line.
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