# sketching reciprocal function (1 Viewer)

#### mathsbrain

##### Member
Hi,

1) Just wondering what happens when you reciprocate a vertical asymptote.
For example if you are given the graph y=1/x and asked to reciprocate it, will the origin be an open circle or closed circle.

2) If you are NOT given the equation of y=1/x but given something that looks similar, how can you tell if you are going to get a straight line or a curve after reciprocation?

Thanks!

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#### blyatman

##### Well-Known Member
1) Generally, it'll be a closed circle. It behaves similar to say, y = x/x, which is equal to 1 as long as x =/= 0. So in a similar fashion, the reciprocal of 1/x is 1/(1/x), which is equal to x as long as x =/= 0.

Here's a relatable post: https://math.stackexchange.com/questions/213928/what-is-cot-pi-2

2) Doesn't really matter, as the examiner will only care about the general behaviour of the curve (i.e. whether it's increasing or decreasing). Personally, I'd plot them as curves since we shouldn't assume the original graph to be a hyperbola.

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• BLIT2014

#### mathsbrain

##### Member
Thanks, but just a little confused regarding point 1), you mentioned x=/=0, so why is that a closed circle then?

1) Generally, it'll be a closed circle. It behaves similar to say, y = x/x, which is equal to 1 as long as x =/= 0. So in a similar fashion, the reciprocal of 1/x is 1/(1/x), which is equal to x as long as x =/= 0.

Here's a relatable post: https://math.stackexchange.com/questions/213928/what-is-cot-pi-2

2) Doesn't really matter, as the examiner will only care about the general behaviour of the curve (i.e. whether it's increasing or decreasing). Personally, I'd plot them as curves since we shouldn't assume the original graph to be a hyperbola.

#### blyatman

##### Well-Known Member
Sorry, typo, open* circle.