Answer:
vertical asymptotes at
x
=
±
1
horizontal asymptote at
y
=
0
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]}