Where is (x+2)/((x^2-2x-8) discontinuous

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Where is (x+2)/((x^2-2x-8) discontinuous

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Oh. Where a line is discontinuous is basically where a point on the graph of the function cannot exist. This is indicated by a open circle on the graph. |dw:1437150289650:dw|
So to first do this, factor the denominator of \[(x+2)/((x^2-2x-8) \]
(x+2)(x-4)

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Other answers:

Yes! Just another side note, is there by chance answer choices?
–2 –2, 4 4 f(x) is continuous everywhere
So the function cannot exist if the denominator equals 0. So this means...?
It has a discontinuity?
Basically you have to find the "zeroes" of the denominator, So x - 4 = 0, and x+2 = 0
SO if you solve for x, what would they be?
Hello?
Sorry internet trouble
So it would be 4 and -2
yes. thats correct. sorry i was helping another person

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