if the open circle wasnt open, would this still be considered a function? is this detemined by

[pulsar willow]

As long as it passes the vertical line test, it would be a function

if the circle was closed, it would no longer pass the vertical line test

[deep spear]

If the open circle was closed

I know thats what you meant but to make it easier to read

[pulsar willow]

i think it was quite obvious, there is no other circle that isn't closed

[deep spear]

The kind of person who is posting vertical line test question doesnt strike me as the smartest

Which is why you have to simplify down to the very core

[pulsar willow]

They might just be wondering cuz it's the edge of a line and it can be seen as an edge case

[deep spear]

Im not shitting on anyone i was the same a couple months ago

[valid oxide]

i was referring to the only open circle sorry

[deep spear]

I might have made this question somepoint in time here

[pulsar willow]

I think this is the type of question I would ask, random semi-edge case ones

[valid oxide]

is it bc it would collide with the closed circle on the same x axis as it?

[pulsar willow]

there would be 2 y values for the same x

if it was closed

[valid oxide]

and thats not valid in terms of being a function right

[pulsar willow]

Yea

[valid oxide]

there are two functions in that graph right

[pulsar willow]

I think it should be taken as a piecewise function

[valid oxide]

which is a function that jumps basically

[pulsar willow]

Piecewise functions can be continuous

[valid oxide]

but only if one end is closed and the othe ris open if the ends are right ontop of each other

[pulsar willow]

there's probably non-continuous non-piecewise functions

it should not been seen as a rule

[valid oxide]

thanks

[pulsar willow]

Yea if they overlap then it's not a function