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If f(x) = 1/x, then this function is an hyperbola. As x -> (tends to) zero from the positive side, f(x) -> +infinity. As x -> 0 from the negative side, f(x) -> negative infinity. As x -> negative infinity, f(x) will tend to zero from below, as x -> +infinity, f(x) will tend to zero from above.
If y = 1/x, then the graph will be reflected in the line y = x, and also in the line y = -x.

2006-11-24 18:56:07 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

4 answers

If you look at a picture (aka a graph) of the function, it will all become clear...

http://upload.wikimedia.org/wikipedia/en/3/3e/1-over-x.png

2006-11-24 19:04:50 · answer #1 · answered by Anonymous · 0 0

plot the graph , by taking some points ( x = -5,-2,-1,-0.5,1,2,5 ) and calculate the y = 1/x , so for instance x = -5 then y = -1/5

then yuou can answer all the questions yourself.

2006-11-25 03:51:15 · answer #2 · answered by gjmb1960 7 · 0 0

It is describing the curve you will get if you graph the function
y = 1 / x..


Doug

2006-11-25 03:07:13 · answer #3 · answered by doug_donaghue 7 · 0 0

I dont know what your question exactly is, I think that you have a problem to understand the geometry of this

This is more or less what you get:

This is the straight line y=x. If forms 45o with the x axis. The straigh line y=-x forms -45o and its perpendicular to the one I draw below

.................o..........x
.................o.......x
.................o....x
.................o.x
ooooooooooooooooooo
..............x..o
...........x....o
........x........o
.....x...........o

And this is the hiperbola graph:

.................o.k........x
.................o..k.....x
.................o....kx
.................o.x... k..k
ooooooooooooooooooo
...k...k....x..o
...........k....o
........x.....k.o
.....x.........ko

Hope you see the symmetry here now.

Ana

2006-11-26 09:13:18 · answer #4 · answered by Ilusion 4 · 0 0

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