Answer
$y=-x-1$
and
$y=-x+3$
Work Step by Step
Given the function $f(x)=\frac{1}{x-1}$, we are asked find the equation of all the straight lines with a slope of $-1$.
Since we are asked to find lines with a specific slope, the best start would be to find the derivative function to go about finding which x values have a slope of $-1$.
$f(x)={(x-1)}^{-1}$
$f'(x)=(-1)(x-1)^{-2}=\frac{-1}{(x-1)^2}$
We equate $f'(x)$ to $-1$
$\frac{-1}{(x-1)^2}=-1$
$(x-1)^2=1$
$x^2-2x+1=1$
$x(x-2)=0$
$x=0$ or $x=2$
It turns out that the function has two points at which the tangent line would have a slope of $-1$.
One of the coordinates with a slope of $-1$ is $(0,-1)$ (the y-value was found using $f(0)$). As such, the coordinate also readily gives up information of the y intercep. Therefore, the equation of the first tangent is $y=-x-1$
The second coordinate can be found using $f(2)=1$ which corresponds to a point $(2,1)$
The y intercept for this tangent is found as per the following:
$\frac{1-c}{2-0}=-1$
$c=3$
In the end, the equation of the second tangent is $y=-x+3$
