# Thread: equation of a tangent line

1. ## 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.

2. Originally Posted by missfuzzy
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?

3. Will the tangent line be unique if the original function is $y = \frac{1}{x}-1$ ? Because $\frac{dy}{dx}= -\frac{1}{x^{2}} \mbox{ which has two solutions for } \frac{dy}{dx} = -1$