What is the discontinuity and zero of the function f(x) = the quantity of 2 x squared plus 5 x minus 12, all over x plus 4?

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What is the discontinuity and zero of the function f(x) = the quantity of 2 x squared plus 5 x minus 12, all over x plus 4?

Algebra
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A function is said to be discontinuous at a point if we can't draw the graph without lifting our pen at that point.
in a function like this the discontinuity will occur where the denominator = 0 \[f(x) = \frac{ 5x ^{2}+5x-12 }{ x+4 }\] If you set f(x) = 0 then you are left with just the quadratic to solve
CORRECTION - should be 2x^2 above BUT 2x^2+5x-12 = (x+4)(2x-3) so the equation becomes y=2x-3 which has no discontinuity. Have I interpreted the equation correctly?

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Isnt this still a discontinuous function?
if the equation is \[f(x) = \frac{ (2x)^{2}+5x-12 }{ (x+4) }\] THEN my comments above are true because(2x)^2 = 4x^2 @sweetburger It looks like it should be - but because the denominator cancels with the factor in the numerator it becomes a continulas straight line
Yea I realized when I graphed it that you are indeed correct.
Wait actually according to my graphing calculator there is a large gap in the graph
ill screenshot it one second
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the question asks for the 'zero' but in most cases there are 2 zeroes I think I may have misinterpreted the way the question is written - or it is written incorrectly.
welp i was doing 5x^2 instead of 2x^2 my bad
soz - that was my initial typo but there are 2 zeroes in other values for ax^2
Yea, I see the 2 zeroe values with the correct equation. I understand where your answer came from now.
no - my point is that if the equation is as I wrote then there is ONE zero (as it is a straight line - but NO discontinuity If the equation is other, then there are 2 zeroes , and a discontinuity @ x=-4
I mean I see where your coming from as the x+4 cancel out and your left with a straight line with no discontinuities represented by y=2x+3. At least i think this is what your were saying.
yes - that's it I still think there is an error with the question, or with my interpretation. But the asker is offline so I'll wait foe his reply.
Yea the question seems to be misworded in some way or the asker typed it in wrong.

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