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Math Help - equation of a tangent line

  1. #1
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    equation of a tangent line

    I have this question, Find equation of the line with a slope of -1 that is tangent to y = 1/x - 1. This is what i have so far, I know that y = mx + b and therefore y = -1x + k. and that to set the y values equal to each other to get; 1/x-1 = -1x + k. Then I get stuck on rearranging this equation so that it equals zero, I know schoolgirl error, once the equation is rearranged i can use the discriminant part of the quadratic equation to solve for k, (b^2 - 4ac), when this equals zero there is just one solution to the quadratic, right. But it is just that rearranging part that is getting me stuck, im not sure off it.
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  2. #2
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    Quote Originally Posted by missfuzzy View Post
    I have this question, Find equation of the line with a slope of -1 that is tangent to y = 1/x - 1.
    Is this y= (1/x)- 1 or y= 1/(x-1)?

    [tex] This is what i have so far, I know that y = mx + b and therefore y = -1x + k. and that to set the y values equal to each other to get; 1/x-1 = -1x + k. Then I get stuck on rearranging this equation so that it equals zero, I know schoolgirl error, once the equation is rearranged i can use the discriminant part of the quadratic equation to solve for k, (b^2 - 4ac), when this equals zero there is just one solution to the quadratic, right. But it is just that rearranging part that is getting me stuck, im not sure off it.[/QUOTE]
    Since you know the slope of the tangent line is -1, you know the derivative of y must be -1 at that point. For what x is the derivative -1?
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  3. #3
    Senior Member Twig's Avatar
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    Will the tangent line be unique if the original function is  y = \frac{1}{x}-1 ? Because \frac{dy}{dx}= -\frac{1}{x^{2}} \mbox{ which has two solutions for } \frac{dy}{dx} = -1
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