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Math Help - Tangent question

  1. #1
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    Tangent question

    Hey guys,

    Here is another question I am struggling with:

    Find the equation of line with the slope -1 that is tangent to the curve 1/(x-1).

    (5 marks: 1 mark each for a graph, for the general equation of the line, for getting the quadratic, for solving for k, for the solution)

    So far I have this. I am taught in my book to solve by setting the y-values equal to each other.



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  2. #2
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    Quote Originally Posted by Slipery View Post
    Hey guys,

    Here is another question I am struggling with:

    Find the equation of line with the slope -1 that is tangent to the curve 1/(x-1).

    (5 marks: 1 mark each for a graph, for the general equation of the line, for getting the quadratic, for solving for k, for the solution)

    So far I have this. I am taught in my book to solve by setting the y-values equal to each other....
    The last line of your work, which is correct, is:

    1=-x^2+kx+x-k~\implies~x^2-(k+1)x+(k+1) = 0

    This is a quadratic equation in x. Use the formula to solve a quadratic equation:

    x = \dfrac{(k+1) \pm \sqrt{(k+1)^2-4 \cdot (k+1)}}{2}

    You only get one point of intersection (= tangent point) if the radicand equals zero. Solve for k.

    (k+1)^2-4(k+1)=0~\implies~(k+1)((k+1)-4)=0

    I'll leave the rest for you.
    Last edited by earboth; February 10th 2009 at 09:01 AM.
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  3. #3
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    Hmmm, can you clarify what you did in the first line there? Why you changed all of the signs and got rid of an x and added the 1 to the k's? I could just figure out the question from your help, but I also like to know what happened so that I can actually learn it and not just replicate it. I understand the rest except for that part
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  4. #4
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    Quote Originally Posted by Slipery View Post
    Hmmm, can you clarify what you did in the first line there? Why you changed all of the signs and got rid of an x and added the 1 to the k's? I could just figure out the question from your help, but I also like to know what happened so that I can actually learn it and not just replicate it. I understand the rest except for that part
    I moved all terms to the LHS:

    1=-x^2+kx+x-k~\implies~x^2{\bold{\color{blue}-kx-x}}+{\bold{\color{red}k+1 }}= 0

    Then I factored out -x at the blue terms and collected the constants in one bracket (this step isn't necessary here, but I wanted to make clear this part)

    x^2{\bold{\color{blue}-(k+1)x}}+{\bold{\color{red}(k+1)}} = 0

    Additional remark: Where the (k+1) comes from:

    If you want to factor out a value from a sum you have to divide each summand by the factor. The result of the divison is put into bracket:

    kx+x = x\left(\dfrac{kx}x + \dfrac xx\right) = x(k+1)
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  5. #5
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    Alright, thank you very much.

    However I can't find the right answer?

    In the book with this kind of question, they end up with a nice easy answer. For example, from the formula y= x^2 + 5x +6 -x, they come up with k= -1/4.

    When I proceed with what you left me off with. I come up with:
    k^2-2k+5 = 0

    Now, I am not sure if I did that right, because I was a little confused as to the reason you rearanged the formula in the last line you gave me. To me, it gave the same answer as if I worked it out in left side of that statement.

    Anyways, I am wondering why I am left with another quadratic equation when I should have a single answer? Argh, why couldn't they make it easy like the example they give me?

    I am supposed to use the quadratic formula again on that? But then k is no longer in my equation? And I am supposed to get a single answer, not roots.. So I am confused.
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  6. #6
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    Quote Originally Posted by Slipery View Post
    Alright, thank you very much.

    However I can't find the right answer?

    In the book with this kind of question, they end up with a nice easy answer. For example, from the formula y= x^2 + 5x +6 -x, they come up with k= -1/4.

    When I proceed with what you left me off with. I come up with:
    k^2-2k+5 = 0

    Now, I am not sure if I did that right, because I was a little confused as to the reason you rearanged the formula in the last line you gave me. To me, it gave the same answer as if I worked it out in left side of that statement.

    Anyways, I am wondering why I am left with another quadratic equation when I should have a single answer? Argh, why couldn't they make it easy like the example they give me?

    I am supposed to use the quadratic formula again on that? But then k is no longer in my equation? And I am supposed to get a single answer, not roots.. So I am confused.
    I re-arranged the last line to give you a hint how to solve the equation without much work:

    (k+1)((k+1)-4)=0

    This is a product of 2 factors which is only zero if one factor is zero too:

    k+1=0~\vee~k+1-4=0~\implies~k=-1~\vee~k=3

    Now plug in this value to get the x-coordinate of the tangent point: (there are 2 tangent points!):
    T_1(0, -1), and T_2(2, 1)
    Attached Thumbnails Attached Thumbnails Tangent question-zwei_tangenten.png  
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  7. #7
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    Ahhh.

    I never would have figured this out without your help. Thank you so much
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