Find the equation of the tangent line(S) to the graph of
f(x) = (x)/(x-1) that pass through the point (-1, 5)
Then graph the function and the lines.

I found out one of the lines but I don't understand how there is multiple tangent lines??

f(x) = (x)/(x-1)
f'(x) = (1)/(x-1) - (x)/(x-1)^2 or -1/(x-1)^2

-.5 - .25 = -.75

y - 5 = -3/4 (x+1)

y = -3/4x - 23/4

That's one of the tangent lines, correct?

Now what would be the other tangent line?

This post was edited by Trev on Jun 30 2013 08:39pm

i'm not sure what method you used to find your derivative, but I think that is the problem. I suggest using the quotient rule

if this doesn't help, i'll work the problem out step-by-step to show you the method

I have both versions, both quotient and multiple rule. (multiple rule = x * (x-1)^-1)

I have both versions up there tho! (Hence the or)

to find the equation of the tangent line, you need these things: a function and a coordinate.

Step 1: first take your function and first its derivative.
f(x)=(x)/(x-1)
this is a quotient function, so you would use the quotient rule

f'(x) = [(x-1)(1) - (x)(1)]/(x-1)^2

this simplifies to: -1/(x-1)^2 = f'(x)

Step 2: now use your given x-coordinate and plug it into your derivative to find the slope:
f(-1) = -1/(-1-1)^2 = -1/4 <--- slope

Step 3: now the easy part, some simple algebra. plug your values into the point-slope formula.

y-y1 = m(x-x1)
y - 5 = (-1/4)(x - (-1)) -----> y = (-1/4)x + (19/4)

This is strange
http://www.derivative-calculator.net/

Gives both and they end up with different results for both or did I some how mess it up and it should be -1/4 in my calculations? :/

Whoops, added the -1/4 to the -1/2. Okay so the -1/4 is the slope.

That's one equation, now what's the second equation? Lol.. I don't understand how there can be two equations.

This post was edited by Trev on Jun 30 2013 10:11pm

The second tangent line is y = 1 - 4x, except I have no idea how to get that.

The point (-1 ; 5) is not in the graph of f (x) = x / ( x - 1 ).

Maybe the problem was talking about 2 points, of X-coordonates : -1 and 5.
That would be the points :
A ( -1 ; 1/2 ) and B ( 5 ; 5/4 )

... with 2 tangent lines :

y1 = (-1/4) x + 1/4

y2 = (-1/16) x + 25/16

By definition, a tangent line on a point is unique.

y = 1 - 4x is the tangent line on the point ( 1/2 ; -1 ) ...

This post was edited by feanur on Jul 1 2013 03:33am

As an aside, I find its easier to use the product rule combined with the chain rule, rather than remembering the quotient rule. x/(x+2) = x*(x+2)^-1

d/dx x(x+2)^-1 = (x+2)^-1 - x(x+2)^-2 = 2*(x+2)^-2 = 2/(x+2)^2

You can have multiple tangents because the point (-1, 5) is not on the curve. So you could have several tangents intersecting at the point, although each of them would be tangent to the curve at only one, unique point (obviously). To help visualize this, think of a simple f(x) = x^2. How many tangents go through the point (0, -1)? Two, one with a negative slope and one with a positive one, each touching the curve on the symmetrically opposite sides of the y-axis.