# Slope and Equation of the Line

• Jul 9th 2013, 07:54 PM
Fermi
Slope and Equation of the Line

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

This is my work until I get stuck.

y=-1x+k
-1x+k=1/(x-1)
y=??

I get stuck from here. Can someone please solve this completely and be exact with the steps. Thanks!
• Jul 9th 2013, 08:28 PM
MarkFL
Re: Slope and Equation of the Line
You are on the right track...you have:

$k-x=\frac{1}{x-1}$

Multiply through by $x-1$, write the resulting quadratic in standard form, and then use the condition on the discriminant for one repeated root (do you see why?). You will find two values for $k$.
• Jul 10th 2013, 12:44 AM
Fermi
Re: Slope and Equation of the Line
Well that's what I was getting confused about. I ended up cross multiplying like you're suggesting. But then I don't know if the quadratic is correct, because it certainly looks wrong. And then the discriminant looks even worse.

(k-x)(x-1)-1=0
x^2+x(k+1)+(k-1)=0
I'm not certain how to get a value for k from the discriminant of this? I'm assuming the k doesn't equal 1.
• Jul 10th 2013, 12:50 AM
MarkFL
Re: Slope and Equation of the Line
Your first equation is correct. The quadratic in standard form that you should wind up with is:

$x^2-(k+1)x+(k+1)=0$

Now, when a quadratic has 1 root, what do we then know about the discriminant?
• Jul 11th 2013, 11:30 AM
Fermi
Re: Slope and Equation of the Line
So does K=-1? The discriminant would only cross the x axis once.
• Jul 11th 2013, 12:20 PM
MarkFL
Re: Slope and Equation of the Line
For the quadratic $ax^2+bx+c$ the discriminant is defined to be $b^2-4ac$. When the discriminant is zero, the there is only 1 repeated root for the quadratic. So, equating the discriminant in this case to zero, we find:

$(-(k+1))^2-4(1)(k+1)=0$

$(k+1)((k+1)-4)=0$

$(k+1)(k-3)=0$

Thus, we find two possible tangent lines:

$y=-x-1$

$y=-x+3$
• Jul 12th 2013, 02:14 PM
Fermi
Re: Slope and Equation of the Line
Thanks so much! I see what I was doing wrong.