/sci/ - Science & Math

[Anonymous No.6065396]

wait i have a question
if you were to graph (x^2-4)/(x-2)
would there be a hole at x=2? or do you factor and remove the denominator before graphing

[Anonymous No.6065398]

that would be undefined because of 0/0

[Anonymous No.6065403]

>>6065398
atleast at 2 anyways. if you factor your still left with x-2 which is still a hole

[Anonymous No.6065407]

>>6065403
so i still pretend the restriction exists even if the new function is really just x+2?

[Anonymous No.6065415]

>>6065396
asymptotic

[Anonymous No.6065420]

>>6065407
Plug in two for x into the equation and then try and simplify. See what happens?

[Anonymous No.6065436]

>>6065396
the graph of y= (x^2-4)/(x-2)
is the line of equation y=x+2, with a hole at x=2 (since y is undefined for this value)

Usually you draw the line, and mark with a "O" the blank

[Anonymous No.6065508]

When x->2 the graph wraps around the universe and comes back from the other side

[Anonymous No.6065555]

>>6065407
Yes.
It's called a Mathematical singularity.
Don't ask why, it's one of those definition things that makes the rest of maths easier.

[Anonymous No.6065615]

The value at x=2 can be defined as the limit from both sides so it works fine. Graphing software might choke on it though

[Anonymous No.6065657]

>>6065555
>don't ask why

[Anonymous No.6066050]

>>6065396

Yeah the hole is something called a "removable discontinuity". It's the same as x + 2 everywhere but at x = 2. If you then define the graph's value as 4 at x = 2, nothing breaks and everything else is the same, so that's how you "remove" the discontinuity.

[Anonymous No.6066082]

>>6065420
magic!!!

[Anonymous No.6066158]

Dammit, google.

[Anonymous No.6066353]

Note the function is continuous on its (implied) domain. This is because the implied domain doesn't contain 2.

[Anonymous No.6067435]

1. Plug in x=2.
2. Simplify 0/0 = 1
3. Plot the graph