F(x) =
4/x²+1
Asymptotes

vertical asymptotes at

x

=

±

1

horizontal asymptote at

y

=

0

Step-by-step explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve:

x

2

−

1

=

0

⇒

x

2

=

1

⇒

x

=

±

1

⇒

x

=

−

1

and

x

=

1

are the asymptotes

Horizontal asymptotes occur as

lim

x

→

±

∞

,

f

(

x

)

→

c

(a constant)

divide terms on numerator/denominator by the highest power of x, that is

x

2

f

(

x

)

=

4

x

x

2

x

2

x

2

−

1

x

2

=

4

x

1

−

1

x

2

as

x

→

±

∞

,

f

(

x

)

→

0

1

−

0

⇒

y

=

0

is the asymptote

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1 , denominator-degree 2) Hence there are no oblique asymptotes.

graph{(4x)/(x^2-1) [-10, 10, -5, 5]}