|x| because just Google it if I'm wrong
When a line and a curve are tangent to each other, they are joined together by 1 point (x,y). So, you equate both equations either in terms of x or y. For this problem, let's equate in terms of y.
y = k - x
y = x² + 3x + 1
k - x = x² + 3x + 1
k = x² + 3x + x + 1
k = x² + 4x + 1
This would be the value of k. Since we are not given the common point, it can only be expressed in terms of x or y.
The function
... f(x) = (x+2)/(x-1) = 1 + 3/(x-1)
is symmetrical about the line y=x, hence is its own inverse.
We can evaluate the desired derivative directly.
... f'(x) -3/(x-1)²
so f'(2) is
... f'(2) = -3/(2-1)²
