# equation of the line that is tangent to the curve

• Jul 23rd 2011, 01:51 AM
raymac62
equation of the line that is tangent to the curve
Find the equation of the line with a slope -1 that is the tangent to the curve:

$\displaystyle y= \frac {1}{x-1}$

This is the process I am using to find the solution but I am getting stuck.
$\displaystyle -1x+k= \frac {1}{x-1}$
• Jul 23rd 2011, 02:23 AM
Siron
Re: equation of the line that is tangent to the curve
What is k in your 'solution'? ...

What you have to do is taking the first derivative of f(x) and then let f'(x)=-1.
• Jul 23rd 2011, 05:22 AM
skeeter
Re: equation of the line that is tangent to the curve
Quote:

Originally Posted by raymac62
Find the equation of the line with a slope -1 that is the tangent to the curve:

$\displaystyle y= \frac {1}{x-1}$

This is the process I am using to find the solution but I am getting stuck.
$\displaystyle -1x+k= \frac {1}{x-1}$

your equation has two unknowns, $\displaystyle x$, and the y-intercept of the tangent line, $\displaystyle k$ . you need another equation ...

$\displaystyle \frac{d}{dx}\left(\frac{1}{x-1}\right) = -1$

I have to ask since you posted this in the PreCalculus forum, do you know what this second equation means?

... if so, are you free to use the basic derivative rules, or are you required to find the derivative from first principles (the limit definition)?
• Jul 23rd 2011, 05:58 AM
earboth
Re: equation of the line that is tangent to the curve
Quote:

Originally Posted by raymac62
Find the equation of the line with a slope -1 that is the tangent to the curve:

$\displaystyle y= \frac {1}{x-1}$

This is the process I am using to find the solution but I am getting stuck.
$\displaystyle -1x+k= \frac {1}{x-1}$

Actually you are calculating the points of intersection. You'll get:

$\displaystyle x = \frac{k+1}{2} \pm \sqrt{\frac{k^2-2k-3}{4}}$

The line is a tangent if there exists only one point of intersection: That's the tangent point. This will happen if the discriminant equals zero:

$\displaystyle \frac{k^2-2k-3}4=0~\implies~k=3~\vee~k=-1$
• Jul 23rd 2011, 03:23 PM
raymac62
Re: equation of the line that is tangent to the curve
Quote:

Originally Posted by earboth
Actually you are calculating the points of intersection. You'll get:

$\displaystyle x = \frac{k+1}{2} \pm \sqrt{\frac{k^2-2k-3}{4}}$

The line is a tangent if there exists only one point of intersection: That's the tangent point. This will happen if the discriminant equals zero:

$\displaystyle \frac{k^2-2k-3}4=0~\implies~k=3~\vee~k=-1$

Exactly, thank you very much.
• Jul 23rd 2011, 04:04 PM
raymac62
Re: equation of the line that is tangent to the curve
I do have another question though. How did you get from: $\displaystyle -1x+k = \frac {1}{x-1}$ to $\displaystyle x = \frac{k+1}{2} \pm \sqrt{\frac{k^2-2k-3}{4}}$

I do understand that the solution is the values of k where the discriminant is equal to 0.

I am having difficulties understanding the algebraic process in order to determine the discriminant.

I am doing this:
$\displaystyle -1x+k = \frac {1}{x-1} ;$

$\displaystyle (x-1)(-x)+(x-1)(k) = 1;$

$\displaystyle -x^2+x+kx-k-1=0$

I get here and I am like, "OK, using the quadratic formula what terms in this equation are $\displaystyle b$ and $\displaystyle c$ so that I know how to set up the discriminant?" based on the commonly presented:

$\displaystyle b^2-4ac$ (just to clarify what i am saying)

It is clear to me that $\displaystyle b$ must be $\displaystyle k$ based on what you have presented. But how does one exactly know that?

Algebraically, what steps are you taking?

Sincerely,

Raymond MacNeil
• Jul 23rd 2011, 11:41 PM
earboth
Re: equation of the line that is tangent to the curve
Quote:

Originally Posted by raymac62
I do have another question though. How did you get from: $\displaystyle -1x+k = \frac {1}{x-1}$ to $\displaystyle x = \frac{k+1}{2} \pm \sqrt{\frac{k^2-2k-3}{4}}$

I do understand that the solution is the values of k where the discriminant is equal to 0.

I am having difficulties understanding the algebraic process in order to determine the discriminant.

I am doing this:
$\displaystyle -1x+k = \frac {1}{x-1} ;$

$\displaystyle (x-1)(-x)+(x-1)(k) = 1;$

$\displaystyle -x^2+x+kx-k-1=0$

I get here and I am like, "OK, using the quadratic formula what terms in this equation are $\displaystyle b$ and $\displaystyle c$ so that I know how to set up the discriminant?" based on the commonly presented:

$\displaystyle b^2-4ac$ (just to clarify what i am saying)

It is clear to me that $\displaystyle b$ must be $\displaystyle k$ based on what you have presented. But how does one exactly know that?

Algebraically, what steps are you taking?

l

I'm taking this line:

$\displaystyle -x^2+x+kx-k-1=0 ~\implies~x^2-(k+1)x+(k+1)=0$

So you have:
a = 1
b = -(k + 1)
c = (k +1)

Now apply the quadratic formula:

$\displaystyle x = \dfrac{(k+1)\pm \sqrt{(k+1)^2-4(k+1)}}{2}$

Expand the brackets in the discriminant and you'll get the term I posted in my previous post.
• Jul 24th 2011, 02:09 PM
raymac62
Re: equation of the line that is tangent to the curve
Factoring! OK. Thank you so much. Though I need to clarify some more things.

All the best,

Raymond
• Jul 24th 2011, 04:17 PM
raymac62
Re: equation of the line that is tangent to the curve
Quote:

Originally Posted by earboth
I'm taking this line:

$\displaystyle -x^2+x+kx-k-1=0 ~\implies~x^2-(k+1)x+(k+1)=0$

So you have:
a = 1
b = -(k + 1)
c = (k +1)

Now apply the quadratic formula:

$\displaystyle x = \dfrac{(k+1)\pm \sqrt{(k+1)^2-4(k+1)}}{2}$

Expand the brackets in the discriminant and you'll get the term I posted in my previous post.

Actually, there are still some things I need to clarify if you don't mind. Why do you have:

$\displaystyle -x^2+x+kx-k-1=0 ~\implies~x^2-(k+1)x+(k+1)=0$ ?

Should it not be: $\displaystyle -x^2+x+kx-k-1=0 ~\implies~x^2+(k+1)x-(k+1)=0$?

Would $\displaystyle x^2-(k+1)x+(k+1)$ not give $\displaystyle x^2-kx-x+k+1$ ?

**EDIT: OK algebraically I see how it makes sense, but how did you know your were suppose to make that modification to correctly solve for K?**

And, in your original post, why were you able to divide the discriminant by 4? I thought the discriminant was simply: $\displaystyle b^2-4ac=0$
• Jul 24th 2011, 10:48 PM
earboth
Re: equation of the line that is tangent to the curve
Quote:

Originally Posted by raymac62
Actually, there are still some things I need to clarify if you don't mind. Why do you have:

$\displaystyle -x^2+x+kx-k-1=0 ~\implies~x^2-(k+1)x+(k+1)=0$ ?

Should it not be: $\displaystyle -x^2+x+kx-k-1=0 ~\implies~x^2+(k+1)x-(k+1)=0$? <-- no. See remark at [A]

Would $\displaystyle x^2-(k+1)x+(k+1)$ not give $\displaystyle x^2-kx-x+k+1$ ? <--- yes. See remark at [A]

**EDIT: OK algebraically I see how it makes sense, but how did you know your were suppose to make that modification to correctly solve for K?**

And, in your original post, why were you able to divide the discriminant by 4? <--- See remark at [.B.] I thought the discriminant was simply: $\displaystyle b^2-4ac=0$

[A] I've made the experience that for me a leading factor a = -1 causes a lot of mistakes. So I usually divide through the equation by the leading factor such that a = 1. Then I don't have to think about it in the following calculations any more:

$\displaystyle -x^2+x+kx-k-1=0 | \div (-1)~\implies~x^2-x-kx+k+1=0~\implies~x^2-(k+1)x+(k+1)=0$

[B] That's a question of definition: I've learned that the discriminant is the term under the root-sign (in Germany such a term is called radicand) if it determines different solutions of an equation (dicriminare is a Latin word which means to separate, to put apart)

$\displaystyle \sqrt{\dfrac{k^2-2k-3}4} = \dfrac{\sqrt{k^2-2k-3}}2$

So in my opinion the discriminant could be:

$\displaystyle \sqrt{\dfrac{k^2-2k-3}4}$ ... or ... $\displaystyle \sqrt{k^2-2k-3}$