# 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
$
y=\frac{1}{x-1}
$

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

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

set it to equal 0:
$
-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.

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

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

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

Occurs @ x = 2

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

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

$1 = 1(2) + b$

Equation is: $y_1 = -x - 1$

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

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

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

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

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

$y=-x-1$

And a similar approach for $x=2$