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

  • Thread starter gibguitar
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  • #1
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Homework Statement



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

Homework Equations



y=1/x-1

The Attempt at a Solution



Slope of -1 means y=-1x+k

So...
-1x+k = 1/x-1

I don't know how to rearrange this into a quadratic equation so that I can solve this.
I know I use b^2-4ac=0 and substituting for a, b and c.... I'm just stuck on rearranging the equation into the form: ax^2+bx+c=0
 

Answers and Replies

  • #2
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Homework Statement



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

Homework Equations



y=1/x-1

The Attempt at a Solution



Slope of -1 means y=-1x+k

So...
-1x+k = 1/x-1

I don't know how to rearrange this into a quadratic equation so that I can solve this.
I know I use b^2-4ac=0 and substituting for a, b and c.... I'm just stuck on rearranging the equation into the form: ax^2+bx+c=0
You're missing a very important piece: how to find the slope of the tangent line to y = 1/x - 1.

I'm going to guess that you are in a class that has discussed how to find the tangent to a curve.
 
  • #3
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OTOH, maybe this actually is a precalculus-type problem. What does it mean to say that a line is tangent to a curve?

For your equation, -x + k = 1/x - 1, what about multiplying both sides by x?
 
  • #4
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OTOH, maybe this actually is a precalculus-type problem. What does it mean to say that a line is tangent to a curve?

For your equation, -x + k = 1/x - 1, what about multiplying both sides by x?

(-x+k)(x) = -x^2+k(x)
(1/x-1)(x) = ?

You're right, but I'm confused as to how to multiply the second part? x/x^2-x ?
 
  • #5
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Here's my second stab at it...

-1x+k = 1/x-1

Rearranged: -x^2+kx-1

Using the discriminant of the quadratic formula:

(b^2-4ac)

a=-1 b=k^2 c=-1

k^2-4(-1)(-1)=0
Solving:
k = -2

Therefore the equation is y=-1x-2
For the line with a slope of -1 that is tangent to the curve of 1/x-1
Can anybody verify if I got this right?
 
  • #6
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Here's my second stab at it...

-1x+k = 1/x-1

Rearranged: -x^2+kx-1
Where did the = go? Also, you have an error.
Using the discriminant of the quadratic formula:

(b^2-4ac)

a=-1 b=k^2 c=-1

k^2-4(-1)(-1)
Solving:
k = -2

Therefore the equation is y=-1x-2
For the line with a slope of -1 that is tangent to the curve of 1/x-1
Can anybody verify if I got this right?
 
  • #7
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(-x+k)(x) = -x^2+k(x)
(1/x-1)(x) = ?

You're right, but I'm confused as to how to multiply the second part? x/x^2-x ?
What is x * (1/x)?
What is x * (-1)?
 
  • #8
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From post #3:
What does it mean to say that a line is tangent to a curve?
 
  • #9
Ray Vickson
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Homework Statement



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

Homework Equations



y=1/x-1

The Attempt at a Solution



Slope of -1 means y=-1x+k

So...
-1x+k = 1/x-1

I don't know how to rearrange this into a quadratic equation so that I can solve this.
I know I use b^2-4ac=0 and substituting for a, b and c.... I'm just stuck on rearranging the equation into the form: ax^2+bx+c=0

What you wrote means [itex] (1/x) - 1.[/itex] Is that what you meant, or did you really mean [itex] 1/(x-1)[/itex]? If you meant the latter, you need to use brackets, but if you meant the former then what you wrote is OK.

RGV
 
  • #10
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What you wrote means [itex] (1/x) - 1.[/itex] Is that what you meant, or did you really mean [itex] 1/(x-1)[/itex]? If you meant the latter, you need to use brackets, but if you meant the former then what you wrote is OK.
I think he meant (1/x) - 1, but I'm not certain of it.
 

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