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

In summary, to find the equation of a tangent line with a slope of -1 to the curve y=1/x-1, you can use the derivative of the function or visually identify the point of tangency. The slope of a tangent line represents the rate of change of the function at the point of tangency and can be undefined in certain cases. To check the accuracy of the calculated equation, you can graph the original function and the tangent line or plug in the coordinates of a known point on the line.
  • #1
gibguitar
9
0

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
 
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  • #2
gibguitar said:

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
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
Mark44 said:
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
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
gibguitar said:
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.
gibguitar said:
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
gibguitar said:
(-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
From post #3:
What does it mean to say that a line is tangent to a curve?
 
  • #9
gibguitar said:

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
Ray Vickson said:
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.
 

1. How do I find the equation of the line with slope -1 that is tangent to the curve y=1/x-1?

To find the equation of a tangent line, you will need to use the derivative of the function. In this case, the derivative of y=1/x-1 is y'=-1/x^2. The slope of the tangent line will be equal to the value of the derivative at the point of tangency. So, for a slope of -1, we can set -1=-1/x^2 and solve for x. The resulting x-value is the point of tangency. Then, using the point-slope form of a line, the equation of the tangent line can be written as y-1=(-1)(x-x0), where x0 is the x-value of the point of tangency.

2. Can I find the equation of the tangent line without using derivatives?

Yes, it is possible to find the equation of a tangent line without using derivatives. One method is to graph the function and visually identify the point of tangency. Then, using the point-slope form of a line, the equation of the tangent line can be written as y-y0=m(x-x0), where m is the slope of the tangent line and (x0, y0) is the point of tangency.

3. What is the significance of the slope of a tangent line?

The slope of a tangent line represents the rate of change of the function at the point of tangency. It also represents the instantaneous rate of change, as it is calculated at a specific point rather than over an interval. The slope of a tangent line can also be used to find the equation of the tangent line and to determine other properties of the curve.

4. Can the slope of a tangent line ever be undefined?

Yes, the slope of a tangent line can be undefined in certain cases. This occurs when the derivative of the function is undefined at the point of tangency. In other words, the function is not differentiable at that point. This can happen at sharp turns or cusps in the curve, or at vertical asymptotes.

5. How can I check if my calculated equation of the tangent line is correct?

One way to check if your calculated equation of the tangent line is correct is by graphing the original function and the tangent line on the same coordinate plane. The tangent line should pass through the point of tangency and have the correct slope. You can also plug in the coordinates of a known point on the line (other than the point of tangency) into the equation to see if it satisfies the equation.

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