## anonymous one year ago 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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1. amilapsn

A function is said to be discontinuous at a point if we can't draw the graph without lifting our pen at that point.

2. MrNood

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

3. MrNood

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?

4. sweetburger

Isnt this still a discontinuous function?

5. MrNood

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

6. sweetburger

Yea I realized when I graphed it that you are indeed correct.

7. sweetburger

Wait actually according to my graphing calculator there is a large gap in the graph

8. sweetburger

ill screenshot it one second

9. MrNood

10. MrNood

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.

11. sweetburger

12. MrNood

soz - that was my initial typo but there are 2 zeroes in other values for ax^2

13. sweetburger

Yea, I see the 2 zeroe values with the correct equation. I understand where your answer came from now.

14. MrNood

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

15. sweetburger

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.

16. MrNood

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.

17. sweetburger

Yea the question seems to be misworded in some way or the asker typed it in wrong.