# Solving the equation of a line, tangent to a curve

Having some trouble with this..

Need to find equation of a line with a slope of -1 that is tangent to the curve y=1/(x-1).
So, rearranging slope formula as y=-1x+k and setting the equations equal,
y=-1x+k=1/(x-1)
y=(-1x+k)(x-1)=1

Here is where my multiplication is either totally wrong or I am not sure how to fit into the format ax^2+bx+c=0, to use quadratic formula to find root.

I have
0=-1x^2+x+kx-k-1
What do I do with the latter half of that equation??

Tried inputting a=-1, b=1 and c=(-k-1) but ended up with x=0/4 so obviously missing something.
Any help appreciated.

Cheers

Your method is good for finding a point where the line and curve intersect, but there is an infinite number of points where a straight line could cross the curve. In order to find the point where the line forms a tangent, you need the line and curve to intersect at the point where the gradient of the curve is also -1.

Having some trouble with this..

Need to find equation of a line with a slope of -1 that is tangent to the curve y=1/(x-1).
So, rearranging slope formula as y=-1x+k and setting the equations equal,
y=-1x+k=1/(x-1)
y=(-1x+k)(x-1)=1

Here is where my multiplication is either totally wrong or I am not sure how to fit into the format ax^2+bx+c=0, to use quadratic formula to find root.

I have
0=-1x^2+x+kx-k-1
What do I do with the latter half of that equation??

Tried inputting a=-1, b=1 and c=(-k-1) but ended up with x=0/4 so obviously missing something.
Any help appreciated.

Cheers

This is wrong, you won't find tangent lines to the curve with that equation, you'll only find lines that cross the curve, whether they're tangent or not. To find tangent lines you need to derive y=1/(x-1) which is y' = -1/(x-1)^2 and that'll be the slope of your line (the derivative function is essentially the slope of the original function's tangent line at any point x). So now you'll only to find the x's for which the slope is -1: -1 = -1/(x-1)^2 => x = 2 or x = 0

Thanks mates.

Unfortunately I am not supposed to be using derivatives. The only examples I have involve setting the equations equal and solving out to ax^2+bx+c=0, and using the discriminant of the quadratic formula (listed as b^2-4ac =0).
Can someone try to tackle this question for me? Im lost.

SammyS
Staff Emeritus
Homework Helper
Gold Member
Unfortunately I am not supposed to be using derivatives. The only examples I have involve setting the equations equal and solving out to ax^2+bx+c=0, and using the discriminant of the quadratic formula (listed as b^2-4ac =0).
Can someone try to tackle this question for me? I'm lost.
OK.

With the above in mind, let's look at what you first tried.
Need to find equation of a line with a slope of -1 that is tangent to the curve y=1/(x-1).
So, rearranging slope formula as y=-1x+k and setting the equations equal,
y=-1x+k=1/(x-1)
y=(-1x+k)(x-1)=1

You don't need the y in either equation, and it's downright incorrect in the second one.

Here is where my multiplication is either totally wrong or I am not sure how to fit into the format ax^2+bx+c=0, to use quadratic formula to find root.

I have
0=-1x^2+x+kx-k-1
What do I do with the latter half of that equation??
You're OK so far. (A little spacing will help readability .)

0 = -1x^2 + x + kx - k - 1

Factor x out of the 2nd & 3rd terms and get rid of that coefficient of -1 on the x2.

Yes, you need to get the right hand side into the form Ax2 + Bx + C .

Then the quadratic formula should only give one solution, because (for this particular function, y = 1/(x - 1) ) the tangent line will intersect the function in only one place. What does that mean in terms of the discriminant ?

Sorry for the readability. I dont understand how you could factor the x's out in the middle of the equation. Should I just divide both sides by -1 to get rid of that?

SammyS
Staff Emeritus