# Thread: Tangents to a curve

1. ## Tangents to a curve

This is a University level homework question that I am having some difficulty with....

Find the equation of the line with the slope -1 that is the tangent to the curve
$\displaystyle y=\frac{1}{x-1}$

This is what I believe I should do:
use the slope to get the equation
$\displaystyle y=-1x+k$

Then set the y-values to equal eachother
$\displaystyle \frac{1}{x-1}=-1x+k$

set it to equal 0:
$\displaystyle -1x^{2}+kx-k=0$

Am I correct so far and where do I go from here?

Thank You

2. You're working too hard. Find y'. That's the slope of any line tangent to y at that point.

$\displaystyle y' = \frac{-1}{(x-1)^2}$

Line has a slope of -1, set y' to -1 and solve for x

$\displaystyle -1 = \frac{-1}{(x-1)^2}$

Occurs @ x = 2

Plug into original equation to get the point and you get (2, 1)

Use $\displaystyle y_1 = -1x + b$, plug in your point

$\displaystyle 1 = 1(2) + b$

Equation is: $\displaystyle y_1 = -x - 1$

3. If you want a tangent line that has slope of 1, then you must have some $\displaystyle x$ value where $\displaystyle \frac{dy}{dx}=-1$

$\displaystyle \frac{dy}{dx}=-\frac{1}{(x-1)^{2}}=-1$

Simply algebra will give you two values where that is true; $\displaystyle x=0,x=2$

The equation of the line when $\displaystyle x=0$:

$\displaystyle y-\frac{1}{0-1}=-1(x-0)$

$\displaystyle y=-x-1$

And a similar approach for $\displaystyle x=2$