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Leto [7]
3 years ago
9

Help me pleaseeeeeeeeeeeee

Mathematics
2 answers:
Tanya [424]3 years ago
5 0
D Is You Answer! Vertical Asymptote
enyata [817]3 years ago
3 0
Your answer is D. Vertical Asymptote
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2n^2+3n-9= \\\\ a=2 \\ b=3 \\ c=-9 \\\\ \Delta= 3^2-4*2*(-9)=9+72\to\boxed{81} \\\\ x_1;x_2=\frac{-3 +\-\sqrt{81}}{2*2}= \frac{-3+/- 9}{4} \\\\ x_1=\frac{-3+9}{4}=\frac{6}{4}\to\boxed{\frac{3}{2}} \\\\ x_2=\frac{-3-9}{4}=\frac{-12}{4}\to\boxed{-4} \\\\ 2n^2+3n-9=(x-\frac{3}{2})(\x+4)
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3 years ago
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Morgarella [4.7K]

Answer:

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Step-by-step explanation:

In order to resolve this problem, we must have in mind that the negative numbers are smaller when they are more distant from 0, and that positive numbers are bigger when they are more distant from 0. So, the biggest number of growth is the one that is more distant from 0.

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