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

Mathematics
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at x=2 function is not defind.
so the function is discontinuous at that point.....
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

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thnk you... but how did you solve it?
Aeron mentioned one of the values, but there is also another one!! Make sure you don't miss that trouble maker! :3

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