Tangent question

**Slipery** · February 10th 2009, 07:49 AM

Hey guys,

Here is another question I am struggling with:

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

(5 marks: 1 mark each for a graph, for the general equation of the line, for getting the quadratic, for solving for k, for the solution)

So far I have this. I am taught in my book to solve by setting the y-values equal to each other.

**earboth** · February 10th 2009, 07:59 AM

Originally Posted by Slipery

Hey guys,

Here is another question I am struggling with:

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

(5 marks: 1 mark each for a graph, for the general equation of the line, for getting the quadratic, for solving for k, for the solution)

So far I have this. I am taught in my book to solve by setting the y-values equal to each other....

The last line of your work, which is correct, is:

$1=-x^2+kx+x-k~\implies~x^2-(k+1)x+(k+1) = 0$

This is a quadratic equation in x. Use the formula to solve a quadratic equation:

$x = \dfrac{(k+1) \pm \sqrt{(k+1)^2-4 \cdot (k+1)}}{2}$

You only get one point of intersection (= tangent point) if the radicand equals zero. Solve for k.

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

I'll leave the rest for you.

**Slipery** · February 10th 2009, 08:02 AM

Hmmm, can you clarify what you did in the first line there? Why you changed all of the signs and got rid of an x and added the 1 to the k's? I could just figure out the question from your help, but I also like to know what happened so that I can actually learn it and not just replicate it. I understand the rest except for that part

**earboth** · February 10th 2009, 08:20 AM

Originally Posted by Slipery

Hmmm, can you clarify what you did in the first line there? Why you changed all of the signs and got rid of an x and added the 1 to the k's? I could just figure out the question from your help, but I also like to know what happened so that I can actually learn it and not just replicate it. I understand the rest except for that part

I moved all terms to the LHS:

$1=-x^2+kx+x-k~\implies~x^2{\bold{\color{blue}-kx-x}}+{\bold{\color{red}k+1 }}= 0$

Then I factored out -x at the blue terms and collected the constants in one bracket (this step isn't necessary here, but I wanted to make clear this part)

$x^2{\bold{\color{blue}-(k+1)x}}+{\bold{\color{red}(k+1)}} = 0$

Additional remark: Where the (k+1) comes from:

If you want to factor out a value from a sum you have to divide each summand by the factor. The result of the divison is put into bracket:

$kx+x = x\left(\dfrac{kx}x + \dfrac xx\right) = x(k+1)$

**Slipery** · February 10th 2009, 08:35 AM

Alright, thank you very much.

However

I can't find the right answer?

In the book with this kind of question, they end up with a nice easy answer. For example, from the formula y= x^2 + 5x +6 -x, they come up with k= -1/4.

When I proceed with what you left me off with. I come up with:
k^2-2k+5 = 0

Now, I am not sure if I did that right, because I was a little confused as to the reason you rearanged the formula in the last line you gave me. To me, it gave the same answer as if I worked it out in left side of that statement.

Anyways, I am wondering why I am left with another quadratic equation when I should have a single answer? Argh, why couldn't they make it easy like the example they give me?

I am supposed to use the quadratic formula again on that? But then k is no longer in my equation? And I am supposed to get a single answer, not roots.. So I am confused.

**earboth** · February 10th 2009, 09:00 AM

Originally Posted by Slipery

Alright, thank you very much.

However

I can't find the right answer?

In the book with this kind of question, they end up with a nice easy answer. For example, from the formula y= x^2 + 5x +6 -x, they come up with k= -1/4.

When I proceed with what you left me off with. I come up with:
k^2-2k+5 = 0

Now, I am not sure if I did that right, because I was a little confused as to the reason you rearanged the formula in the last line you gave me. To me, it gave the same answer as if I worked it out in left side of that statement.

Anyways, I am wondering why I am left with another quadratic equation when I should have a single answer? Argh, why couldn't they make it easy like the example they give me?

I am supposed to use the quadratic formula again on that? But then k is no longer in my equation? And I am supposed to get a single answer, not roots.. So I am confused.

I re-arranged the last line to give you a hint how to solve the equation without much work:

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

This is a product of 2 factors which is only zero if one factor is zero too:

$k+1=0~\vee~k+1-4=0~\implies~k=-1~\vee~k=3$

Now plug in this value to get the x-coordinate of the tangent point: (there are 2 tangent points!):
T_1(0, -1), and T_2(2, 1)

**Slipery** · February 10th 2009, 09:09 AM

Ahhh.

I never would have figured this out without your help. Thank you so much

Thread: Tangent question

LinkBack

Thread Tools

Display

Tangent question

Similar Math Help Forum Discussions

Tangent question.

tangent question

tangent question

tangent circle question

Tangent question :)

Search Tags