Finding Tangent Lines to f(x)=x/(x-1) Passing through (-1,5)

[Bashyboy]

Find equations of the tangent lines to the graph of $f(x)=\frac{x}{x-1}$ that pass through the point $(-1, 5)$.

Well, first I took the derivative, and afterwards, I made the connection that the derivative was a slope at any instant on the graph. By this, I inferred that $f'(x) = m$. I knew that the given point was not on the graph, so I had to find a point on the graph that had the same slope at that point, and also had to pass through the point $(-1, 5)$. So, I calculated the slope to be $m = \frac{y - 5}{x + 1}$; and so, I set this equal to the derivative, and by doing this I had a system of two equations--those being the derivative and the original function.

But what followed was great dismay: I found that when I substituted in for y and solved for x, I procured imaginary solutions. I rather thought I was going on a course of brilliance. What did I do wrong?

 

[SammyS]

Find equations of the tangent lines to the graph of $f(x)=\frac{x}{x-1}$ that pass through the point $(-1, 5)$.

Well, first I took the derivative, and afterwards, I made the connection that the derivative was a slope at any instant on the graph. By this, I inferred that $f'(x) = m$. I knew that the given point was not on the graph, so I had to find a point on the graph that had the same slope at that point, and also had to pass through the point $(-1, 5)$. So, I calculated the slope to be $m = \frac{y - 5}{x + 1}$; and so, I set this equal to the derivative, and by doing this I had a system of two equations--those being the derivative and the original function.

But what followed was great dismay: I found that when I substituted in for y and solved for x, I procured complex solutions. I rather thought I was going on a course of brilliance. What did I do wrong?

Take your equation for the slope, $\displaystyle m = \frac{y - 5}{x + 1}$ and substitute $\displaystyle \frac{x}{x-1}$ in for y, and whatever expression you have for $f'(x)$ in for m . Solve for x to find what location(s) on the graph give the second point for each line.

 

[HallsofIvy]

You don't say what you got for the derivative of the curve so it's impossible to tell what, if anything, you did wrong. When I do essentially what you said, I get a quadratic equation for the x coordinate of the point of tangency but it has real roots.

 

[Bashyboy]

You don't say what you got for the derivative of the curve so it's impossible to tell what, if anything, you did wrong. When I do essentially what you said, I get a quadratic equation for the x coordinate of the point of tangency but it has real roots.

For my derivative, I get $f'(x) = \frac{-1}{(x-1)^2}$

 

[SammyS]

For my derivative, I get $f'(x) = \frac{-1}{(x-1)^2}$

Good !

 

[Bashyboy]

Well, I repeated the process of substituting the the equation $m = \frac{y - 5}{x + 1}$ in for $f'(x)$, which gave $\frac{y - 5}{x + 1} = \frac{-1}{(x - 1)^2}$. Solving for y gave me $y = \frac{-(x + 1)}{(x - 1)^2} + 5$. Finally I substituted this into my original function, $\frac{-(x + 1)}{(x - 1)^2} + 5 = \frac{x}{x - 1}$, and solved for x, which gave me $x = \pm2$. These aren't the correct solutions, though; when I went to graph the function and the two tangents lines, one of them clearly was not a tangent line.

I attached the graph with the supposed tangent lines.

 

[e^(i Pi)+1=0]

1. Take your derivative and plug in the x-coordinate you're looking for, this will give you the slope of the tangent (m) at that point x.

2. Use the point-slope formula to find the tangent line. y-y₁=m(x-x₁)

That's all there is to it. I don't know what the deal is with all that substitution and whatnot above, it seems like a lot more work than necessary. Maybe I'm misreading something.

 

[SammyS]

Well, I repeated the process of substituting the the equation $m = \frac{y - 5}{x + 1}$ in for $f'(x)$, which gave $\frac{y - 5}{x + 1} = \frac{-1}{(x - 1)^2}$. Solving for y gave me $y = \frac{-(x + 1)}{(x - 1)^2} + 5$. Finally I substituted this into my original function, $\frac{-(x + 1)}{(x - 1)^2} + 5 = \frac{x}{x - 1}$, and solved for x, which gave me $x = \pm2$. These aren't the correct solutions, though; when I went to graph the function and the two tangents lines, one of them clearly was not a tangent line.

I attached the graph with the supposed tangent lines.

Check your solutions to $\displaystyle \frac{-(x + 1)}{(x - 1)^2} + 5 = \frac{x}{x - 1}$ again. x = 2 is correct. x = -2 is not.

 

[Bashyboy]

1. Take your derivative and plug in the x-coordinate you're looking for, this will give you the slope of the tangent (m) at that point x.

2. Use the point-slope formula to find the tangent line. y-y₁=m(x-x₁)

That's all there is to it. I don't know what the deal is with all that substitution and whatnot above, it seems like a lot more work than necessary. Maybe I'm misreading something.

The reason for those seemingly unnecessary steps is that the given coordinate is not a point on the graph of the original function. If you plug -1 into the original function, you'll get (-1, 1/2)

Check your solutions to $\displaystyle \frac{-(x + 1)}{(x - 1)^2} + 5 = \frac{x}{x - 1}$ again. x = 2 is correct. x = -2 is not.

I got 1/2, would that happened to be the other correct solution?

 

[Infinitum]

I got 1/2, would that happened to be the other correct solution?

Yep. 1/2 and 2 are correct

 

[Bashyboy]

That was a very entertaining problem. Thank you everyone for your insight you so generously provided.

 

[e^(i Pi)+1=0]

I see now, I did misread the problem.

 

[SammyS]

Here's a graph from WolframAlpha:

 

[Bashyboy]

Here's a graph from WolframAlpha:

Thank you for the graph, but how were you able to graph multiple equations on one coordinate plane? I've only ever been able to do one at a time.

 

[SammyS]

Thank you for the graph, but how were you able to graph multiple equations on one coordinate plane? I've only ever been able to do one at a time.

Here's a link to WolframAlpha with the parameters I used.
http://www.wolframalpha.com/input/?i=plot+x%2F%28x-1%29%2C-x%2B4%2C-4%28x-1%2F2%29-1

If you don't include the "plot" instruction, but just enter y=x/(x-1),y=-x+4,y=-4(x-1/2)-1, you get this: http://www.wolframalpha.com/input/?i=y%3Dx%2F%28x-1%29%2Cy%3D-x%2B4%2Cy%3D-4%28x-1%2F2%29-1

 

[Bashyboy]

Here's a link to WolframAlpha with the parameters I used.
http://www.wolframalpha.com/input/?i=plot+x%2F%28x-1%29%2C-x%2B4%2C-4%28x-1%2F2%29-1

If you don't include the "plot" instruction, but just enter y=x/(x-1),y=-x+4,y=-4(x-1/2)-1, you get this: http://www.wolframalpha.com/input/?i=y%3Dx%2F%28x-1%29%2Cy%3D-x%2B4%2Cy%3D-4%28x-1%2F2%29-1

Thank you very much.