1. Sep 19, 2006

### mushroom

hi guys this is my first time posting on here hope it's in the right section
well heres my problem i have a ploted an inverse relationship ( hyperbole ) on my graph now my teacher wants me to make the graph into a linear relationship, he also gave us an equation which is y=k/x, now how would i go about doing this? i will provide more information if this is not enough...

thanks

2. Sep 20, 2006

### andrevdh

Welcome mushroom,

The relationship between the variables of a linear graph, say $u,v$, is of the form

$$v = grad\ u + v_o$$

where $grad$ is the gradient of the graph and $v_o$ is the value of the dependent variable when the independent variable is zero. That is $v$ when $u=0$.

A special case arises when $v_o = 0$ then whe have

$$v = grad\ u$$

a directly proportional relationship with a straight line graph through the origin.

By plotting your $y$ values against $u = \frac{1}{x}$ values the relationship take on the same form:

$$y = grad\ u$$

so that we get a directly proportional relationship between $y$ and $\frac{1}{x}$.

Last edited: Sep 20, 2006
3. Sep 20, 2006

### HallsofIvy

Staff Emeritus
Andrevdh's point was, that if y= k/x so that plotting y against x gives a hyperbola, then plotting y against 1/x rather than x (so the X-coordinate of the point is 1/x rather than x) gives the straight line graph y= kX.

I would also recommend noting that log(y)= log(k/x)= log(k)- log(x). If you graph log(y) against log(x) you get a straight line. In fact it used to be (probably still is, but computers have changed so many things) possible to buy "log-log" graph paper which was ruled so that marking "3" on an axis you were actually marking "log(3)".

4. Sep 20, 2006