# Determining slope of the tangent line.

1. Sep 20, 2009

### fghtffyrdmns

1. The problem statement, all variables and given/known data
I've never seen such a problem before: Find the two straight lines that are perpendicular to $$y=0.25x$$ and tanget to the curve $$f(x)=1/x$$

2. Relevant equations

$$y=0.25x$$ ; $$f(x)=1/x$$

3. The attempt at a solution

Using the power rule, I found the derivatives of: $$f(x)=-x^{-2}$$ and $$y=0.25$$

But I am not sure what I do now? Do I find the slope of the equations? Do I equate y=f(x)?

Last edited: Sep 20, 2009
2. Sep 20, 2009

### Gregg

If they are perpendicular to y=x/4 then the slope of the line is what?

Now, to find a tangent to y=1/x.

you need the equation for the straight line:

$$y-y_0 = (x-x_0)f'(x_0)$$

what x0 is required to for the desired slope?

3. Sep 20, 2009

### fghtffyrdmns

Is it -4?

4. Sep 20, 2009

### fghtffyrdmns

Any help anyone? :(

5. Sep 20, 2009

### Gregg

All you have to do is not get mixed up, we need a line that is tangent to 1/x. It also needs to be perpendicular to that line you described.So, to get a line that is normal to x/4, you do need a line that has a gradient or slope of -4.

$$y-y_0 = (x-x_0)m$$

First of all in order for the line to be tangent to 1/x it needs to parallel to it at a given point x0. The slope of f(x) is, as you say -1/x^2. The line is going to look like this:

$$y-y_0 = (x-x_0)f'(x_0)$$

and

$$f'(x_0)=-4$$

Get the x0s then find y0 and plug them into the line equation.

6. Sep 20, 2009

### fghtffyrdmns

"the x0s then find y0"

What are these? What are the equations that you come up with?

7. Sep 20, 2009

### Gregg

Well the general equation for the line is going to take the form y=mx+c, in order to get that equation you need the slope (m) and a point on the line, which will be (x0), y0. For example if I knew my slope was 3 and that a point on my curve was (1,2) then I can work out that the equation for line is:

$$y-2=3(x-1), y=3x-1$$

The same is with this problem with have here, we need to work out at what point the slope is -4 don't we. The derivative of the curve is -1/x^2.

$$\frac{1}{x_0^2} = 4 \Rightarrow x_0 = \pm \frac{1}{2}$$

So you have two. Now all you do is work out the corresponding y values and use the equation for a straight line to work it out.

8. Sep 20, 2009

### fghtffyrdmns

I figred out that the functions cross at 2 and -2. I can't use these points?

9. Sep 20, 2009

### Gregg

Just find out what value -1/x^2 = -4, that's x1. (there's two of them)

Then plug it into y=1/x to get the y1. (there's two of them)

Then plug it into

y-y1=-4(x-x1)

and you will get 2 lines.

10. Sep 20, 2009

### fghtffyrdmns

Okay, I got +/- 1/2 which gets me +/- 2

Now I'm confused which is x and which is y :/

11. Sep 20, 2009

### Bohrok

You have two x values; plug them into f(x) = 1/x and you'll get the y values of the coordinates of the points tangent to 1/x. The plug them into the line equation y = mx + b, solve for b, then get the two equations of the tangent lines.

12. Sep 20, 2009

### Gregg

x=-1/2
so what is y at this point, 1/(-1/2) = -2
So the corresponding y for -1/2 is -2. you can do the rest?

13. Sep 20, 2009

### fghtffyrdmns

Alright. I got y=-4x-4 and y=-4x+4

Are these correct?

14. Sep 20, 2009

### Bohrok

They are correct. You should try graphing on a calculator or by hand, or look for an online graphing calculator. 1/x should be easy to graph by hand.

15. Sep 20, 2009

### Gregg

Red is -4x-4
Yellow is -4x+4
Green is x/4
Blue is 1/x

Nice!

16. Sep 20, 2009

### fghtffyrdmns

Thank you guys :)!