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

• dylanjames
In summary: Yes, that is correct.In summary, this person is having trouble finding a line that is tangent to a curve, and is looking for a slope of -1 that is also the gradient of the curve. They tried rearranging the slope formula and setting the equations equal, but it was incorrect. They then tried to factor the x's out of the equation and ended up with x = 0/4. Multiply by -1 or divide by -1.
dylanjames
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.

dylanjames said:
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? I am lost.

dylanjames said:
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.
dylanjames said:
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.

dylanjames said:
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 don't 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?

dylanjames said:
Sorry for the readability. I don't 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?
Multiply by -1 or divide by -1.

x + kx = (1 + k)⋅x , Right?

## 1. What is the equation of a line that is tangent to a curve?

The equation of a line tangent to a curve is given by the derivative of the curve at the point of tangency. This means that the slope of the tangent line is equal to the slope of the curve at that point, and the equation can be written in the form y = mx + b, where m is the slope and b is the y-intercept.

## 2. How do you find the point of tangency between a line and a curve?

The point of tangency between a line and a curve can be found by setting the derivative of the curve equal to the slope of the line. This will give you an equation with one variable, which can be solved to find the x-coordinate of the point. Once you have the x-coordinate, you can plug it into the original equation of the curve to find the corresponding y-coordinate.

## 3. Can there be more than one tangent line to a curve at a given point?

Yes, it is possible for there to be more than one tangent line to a curve at a given point. This can happen if the curve has a sharp turn or a cusp at that point. In this case, there would be two or more lines that have the same slope as the curve at that point, and therefore they would all be tangent to the curve.

## 4. How do you determine if a line is tangent to a curve?

A line is tangent to a curve if it has the same slope as the curve at the point of tangency. This can be determined by finding the derivative of the curve and comparing it to the slope of the line. If they are equal, then the line is tangent to the curve at that point.

## 5. Can you find the equation of a line tangent to a curve without using derivatives?

No, it is not possible to find the equation of a line tangent to a curve without using derivatives. The derivative represents the rate of change of the curve at a given point, and this is crucial in determining the slope of the tangent line. Without using derivatives, it would be impossible to accurately determine the slope of the tangent line and therefore find its equation.

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