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Find the discontinuaties and zeros of (2x^2+5x-12)/(x+4)
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I got the discontinuity at (-4,-11) is that right? And I still need help with the zero(s).
Okay, I got the zero of 3/2, is that correct?
\[ \frac{2 x^2+5 x-12}{x+4}=\frac{(x+4) (2 x-3)}{x+4}=2 x-3 \]
x =2/3 is correct
x=-4 is the only removable discontinuity.
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How did you get the -11?
That's where the point is on the graph, I wasn't stating another discontinuity.
Ok. But -4 is a removable discontinuity.
Alright, thank you.
yw
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