# Thread: equation of a tangent line

1. ## equation of a tangent line

I am asked to find the equation of a line with a slope of -1 that is tangent to the curve
$y=\frac{1}{x-1}$
I know I am doing something wrong here, but I just can't figure out what.
first, I have the equation for the slope, y=-1x+k
next, I set the y-values to equal eachother
$y=\frac{1}{x-1}=-1x+k$

now, here's where I think I begin to mess this up. I am trying to rearrange this into a quadratic equation.
$1=-x^2+x+kx-k \\ 0=-x^2+x+kx-k-1$
a=-1 b=(1+k) c=(-k-1)
Then I use $b^2-4ac$

$1+k^2-4(-1)(-k-1) =0 \\ 1+k^2+4(-k-1)=0 \\ 1+k^2-4k-4=0$
now I'm stuck and don't know where to go from here.

2. ## Re: equation of a tangent line

Originally Posted by whit221
a=-1 b=(1+k) c=(-k-1)
Then I use $b^2-4ac$

$1+k^2-4(-1)(-k-1) =0$
$b^2=(1+k)^2=1+2k+k^2\ne1+k^2$.

Originally Posted by whit221
$1+k^2+4(-k-1)=0 \\ 1+k^2-4k-4=0$
now I'm stuck and don't know where to go from here.
With the adjustment above, you are correct. Just solve this equation in $k$.

3. ## Re: equation of a tangent line

Thank you very much! Now it all makes sense!