Math, asked by rksharmabhu1245, 17 days ago

find the asymptotes of the curve : y = 2x^2 - 1 + 3x^3 / 3 - 2x^2

Answers

Answered by skdamindaminsa
0

Answer:

Vertical Asymptotes:

x

=

0

x

=

3

2

Horizontal Asymptote:

y

=

1

Explanation:

y

=

2

x

2

+

1

3

x

2

x

2

=

2

x

2

+

1

2

x

2

+

3

x

=

2

x

2

+

1

x

(

2

x

+

3

)

Verical Asymptotes

Since denominator could not be 0

we find the possible values of x that would make the equation in the denominator 0

x

(

2

x

+

3

)

=

0

Therefore

x

=

0

(

2

x

+

3

)

=

0

x

=

3

2

are vertical asymptotes.

Horizontal asymptotes

Since the degree of numerator and denominator is the same, we have an horizontal asymptotes

y

2

x

2

2

x

2

=

1

y

=

1

is a horizontal asymptotes for

x

±

graph{-(2x^2+1)/(x(2x+3)) [-25.66, 25.65, -12.83, 12.82]}

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