Question

y= (2x^2 + 5) / (x^2 - 2x) point of discontinuity

Answer 1

at x=2 function is not defind.

Answer 2

so the function is discontinuous at that point.....

Answer 3

We have discontinuities whenever the DENOMINATOR is zero. So let's set the denominator equal to 0, and solve for x!
Those values of X are the trouble makers! :O

\[\large (x^2-2x)=0\]Factoring an x out of each term gives us:
\[\large x(x-2)=0\]
Know what to do from here to solve for x? :D
Need to take advantage of the Zero Factor Property! :D

Answer 4

thnk you... but how did you solve it?

Answer 5

Aeron mentioned one of the values, but there is also another one!! Make sure you don't miss that trouble maker! :3