as the question says, are the graphs effected.... the basic equation is $\displaystyle \frac{1}{x}$ so if it'z like $\displaystyle 5 (\frac {1} {3(x)}) $ will the resulting graph be different than $\displaystyle \frac{1}{x}$

- Jan 25th 2008, 03:00 PMruscutie100Are asymptotes effected by vertical/horizontal expansion or compression??
as the question says, are the graphs effected.... the basic equation is $\displaystyle \frac{1}{x}$ so if it'z like $\displaystyle 5 (\frac {1} {3(x)}) $ will the resulting graph be different than $\displaystyle \frac{1}{x}$

- Jan 25th 2008, 03:47 PMchris_uknot affected
The asymotopes wnt be affected, in that example by compression and expansion as the asymotopes lie on the x and y axis, the general steepness of the graph will increase tho.

- Jan 25th 2008, 06:46 PMruscutie100
- Jan 26th 2008, 04:12 AMchris_uk
If you added a shift to it the expansion or compression would affect the asymotopes just as it would any other point

e.g.

if you had y=1/x +2 where the asymotopes are at y=2 and x=0

a vertical expansion of scale factor 2... y=2(1/x +2) can be written as y=2/x +4. Where the asymotopes are y=4 and x=0

note: y=1/x +1 is not equal to y=1/(x+1) it wasn't very clear wen i wrote it. - Jan 26th 2008, 01:35 PMmr fantastic
If you have $\displaystyle f(x) = \frac{1}{x}$, the vertical and horizontal asymptotes are obviously not changed by a dilation from either axis.

But consider $\displaystyle g(x) = \frac{1}{x - 1} + 2$, say. This can be got by applying appropriate translations to f(x). g(x) has a vertical asymptote x = 1 and a horizontal asymptote y = 2. (So applying translations to $\displaystyle f(x) = \frac{1}{x}$*will*change its asymptotes, if*that's*what you were asking).

But if you want to know whether dilations will have an affect on asymptotes*after*translations have been added to $\displaystyle f(x) = \frac{1}{x}$, then consider:

Dilation by factor a from horizontal axis: $\displaystyle g(x) \rightarrow a g(x)$. Clearly $\displaystyle g(x) \rightarrow h(x) = a g(x) = a \left( \frac{1}{x - 1} + 2 \right) = \frac{a}{x - 1} + 2a$.

h(x) still has a vertical asymptote x = 1 but the horizontal asymptote is y = 2a.

Dilation by factor 1/a from vertical axis: $\displaystyle g(x) \rightarrow g(ax)$. Clearly $\displaystyle g(x) \rightarrow k(x) = g(ax) = \frac{1}{ax - 1} + 2$.

k(x) now has a vertical asymptote x = 1/a and the horizontal asymptote is still y = 2.