Equation of a Tangent Line - PreCalc

[KatelynO]

Find the equation of the line with slope -1 that is tangent to the curve y = 1/(x-1).


The equation of a line with slope -1 is y = -x + k
The curve is y = 1/(x-1)



Set the y-values to each other: 1/(x-1) = -x + k
Rearrange and set equal to 0:
1 = -x² + x + k
x² - x + 1 – k = 0
Find the discriminant:
b²- 4ac = 0
(-1)² - 4(1)(1-k) = 0
1 - 4(1-k) = 0
1 – 4 + 4k = 0
k = 3/4
Therefore the equation of the tangent line is y = -x + 3/4 but I know this is wrong because after looking at the graph the lines do not intercept at this point, what am I doing wrong? Any help is appreciated.

 

[Dick]

(x-1)*(-x+k) is not equal to -x^2+x+k. Multiply it out!

 

[Mark44]

Find the equation of the line with slope -1 that is tangent to the curve y = 1/(x-1).


The equation of a line with slope -1 is y = -x + k
The curve is y = 1/(x-1)



Set the y-values to each other: 1/(x-1) = -x + k
Rearrange and set equal to 0:
1 = -x² + x + k

I get a different equation; namely,
1 = -x² +x + kx - k

I don't think this will help you, though, since I believe your strategy is flawed. The graph of y = 1/(x - 1) is a hyperbola, similar to the graph of y = 1/x translated to the right by one unit. The graph of the line, y = -x + k, will intersect either zero, one, or two points for each value of k. If you can figure out two values of k for which there is exactly one solution in x, each value of k will give you a line that is tangent to the hyperbola.

If what I said is difficult to follow, draw a graph of the hyperbola.

x² - x + 1 – k = 0
Find the discriminant:
b²- 4ac = 0
(-1)² - 4(1)(1-k) = 0
1 - 4(1-k) = 0
1 – 4 + 4k = 0
k = 3/4
Therefore the equation of the tangent line is y = -x + 3/4 but I know this is wrong because after looking at the graph the lines do not intercept at this point, what am I doing wrong? Any help is appreciated.

 

[Apphysicist]

Actually, I think his method nearly works out if he corrects the FOILing error. In a quick piece of work, I came up with values for K that are of opposite sign of what they should be, but that could be an error on my part for not devoting too much time to that method.

Since it's a PreCalc class, I wouldn't expect him to be too familiar differentiation beyond maybe power-rule, but using the chain-rule on the function and setting the derivative equal to -1, then solving for x, then solving for y, and replacing that information into y_1=-x_1+k_1 and y_2=-x_2+k_2 (1 and 2 being subscript for the two cases), you can reliably find values for k that fit the condition.

 

[Dick]

Sure. It works fine. If you have the correct equation for k, solving it gives you the same two lines you would get using calculus methods.

 

[KatelynO]

Actually, I think his method nearly works out if he corrects the FOILing error. In a quick piece of work, I came up with values for K that are of opposite sign of what they should be, but that could be an error on my part for not devoting too much time to that method.

Since it's a PreCalc class, I wouldn't expect him to be too familiar differentiation beyond maybe power-rule, but using the chain-rule on the function and setting the derivative equal to -1, then solving for x, then solving for y, and replacing that information into y_1=-x_1+k_1 and y_2=-x_2+k_2 (1 and 2 being subscript for the two cases), you can reliably find values for k that fit the condition.

I think you're on to something I mean that method is the only way I know how to solve this problem and I must have messed up my FOILing like you said, but to be honest I don't know how to correct it, could you or someone please let me know? I wasn't sure what to do with the 1/(x-1).
Thank you

 

[Mark44]

You have 1/(x-1) = -x + k, which is fine.

Multiply both sides of this equation by x - 1, which gives you the equation 1 = (x - 1)(-x + k).

See if you can expand the right side to get something different than you got before.

Please don't expect us to do this work for you...

 

[Mark44]

Actually, I think his method nearly works out if he corrects the FOILing error.

Since it's a PreCalc class, I wouldn't expect him to be too familiar differentiation beyond maybe power-rule

I would guess that KatelynO is a "she."

 

[KatelynO]

You have 1/(x-1) = -x + k, which is fine.

Multiply both sides of this equation by x - 1, which gives you the equation 1 = (x - 1)(-x + k).

See if you can expand the right side to get something different than you got before.

Please don't expect us to do this work for you...

I don't expect anyone to do it for me, I just didn't remember the method to convert the fraction believe me I know the only way to learn is to do it yourself, I just got stuck on this part of the problem. Thank you though, I appreciate the help.

 

[KatelynO]

You have 1/(x-1) = -x + k, which is fine.

Multiply both sides of this equation by x - 1, which gives you the equation 1 = (x - 1)(-x + k).

See if you can expand the right side to get something different than you got before.

Please don't expect us to do this work for you...

Alright so I redid the equation and expanded it to
1 = -x² + kx + x – k and then set it to zero like I was taught to do.
0 = -x² + x + kx – k – 1
But now I have too many values to find the discriminant in which there should only be a, b and c as I'm sure you already know, thank you for bearing with me, I just really want to understand this.

And yes I am a girl thank you for pointing that out :)

 

[Dick]

Write that as (-1)*x^2+(k+1)*x+(-k-1)=0. Compare that with ax^2+bx+c=0. Can you identity a, b and c now?

 

[KatelynO]

Write that as (-1)*x^2+(k+1)*x+(-k-1)=0. Compare that with ax^2+bx+c=0. Can you identity a, b and c now?

Oh my goodness! That works! That's it ahhhh I'm so happy thank you! :)
One more thing though so that I know for the future, how did you rearrange the (k+1)*x from the original x + kx?? Thank you again!

 

[Dick]

Very welcome. You seem to be kind of tentative about using the distributive law. ab+ac=a(b+c). It's true. You can trust me. I wrote x+kx=1*x+k*x and factored out the x.

 

[KatelynO]

Very welcome. You seem to be kind of tentative about using the distributive law. ab+ac=a(b+c). It's true. You can trust me. I wrote x+kx=1*x+k*x and factored out the x.

Hmm I'll have to study the Distributive Law more then I guess I'm not familiar enough with it and I know it's very important thank you again, I really appreciate the help!