SOLUTION: Find the equation of line with a slope -1 that is the tangent to the curve y=1/x-1 WITHOUT USING CALCULUS TO FIND THE DERIVATIVE. I am only in an advanced functions course an

To see what we are doing let's draw the graph
of y= 



Now I will draw a bunch of green lines which have slope -1.

You will notice that some of those green lines
intersect the graph twice, some don't intersect
it at all, and the two lines that are tangent
to the graph intersect it at just one point.


 
We want the equations of those two green lines
which intersect the graph at exactly one point only.

Let the lines that we want have equations of the form  y = mx + b,
with the slope = m = -1 so substituting -1 for m, the lines will 
have equations of the form:

y = -1x + b

or

y = -x + b

To find points of intersection of the line and the curve, 
we solve the system:



by substituting (-x+b) for y in the first equations:

-x+b = 

Multply both sides by (x-1)

(-x+b)(x-1) = (x-1)

-x² + x + bx - b = 1

Multiply by -1 to get the x² term positive:

x² - x - bx + b = -1

Get 0 on the right

x² - bx - x + b + 1 = 0

x² - (b + 1)x + (b + 1) = 0

We want there to be exactly one solution.

To guarantee that that quadratic equation
has exactly one real solution, we find the discriminant 
and set it equal to 0:

discriminant = B²-4AC

(Note: I used capital letters to avoid conflict of
notation with b and B):

where A=1, B=-(b+1),  C=b+1 

B²-4AC = [-(b+1)]² - 4(1)(b+1)] = 0

(b+1)² - 4(b+1) = 0

(b+1)[(b+1) - 4] = 0

(b+1)(b+1-4) = 0

(b+1)(b-3) = 0

b+1 = 0;  b-3 = 0
  b = -1;   b = 3

So the tangent lines are the lines
y = -x + b with these two values for b.

So they are 

y = -x - 1 
and
y = -x + 3

And here are their graphs:



Edwin