Step-by-step explanation:
Here we have
f
(
x
)
=
2
x
2
(
x
2
−
9
)
, which can be factorized as
f
(
x
)
=
2
x
2
(
x
+
3
)
(
x
−
3
)
As there is no common factor between numerator and denominator, there s no hole.
Further vertical asymptotes are
x
=
−
3
and
x
=
3
and as
f
(
x
)
=
2
x
2
(
x
2
−
9
)
=
2
1
−
9
x
2
, as
x
→
∞
,
f
(
x
)
→
2
, hence horizontal asymptote is
y
=
2
.
Observe that
f
(
−
x
)
=
f
(
x
)
and hence graph is symmetric w.r.t.
y
-axis. Further as
x
=
0
,
f
(
x
)
=
0
. Using calculas we can find that at
(
0
,
0
)
there is a local maxima as
d
y
d
x
=
−
36
x
(
x
2
−
9
)
2
and at
x
=
0
it is
0
. Further while for
x
<
−
3
and
x
>
3
, function is positive, for
−
3
<
x
<
3
function is negative.
Now take a few values of
x
say
{
−
10
,
−
7
,
−
4
,
−
2
,
−
1
,
1
,
2
,
4
,
7
,
10
}
and corresponding values of
f
(
x
)
are
{
2
18
91
,
2
9
20
,
4
4
7
,
−
1
3
5
,
−
1
4
,
−
1
4
,
−
1
3
5
,
4
4
7
,
2
9
20
,
2
18
91
}
Please provide the context of problem
I^2 is negative 1, but I see you have a negative sign there. So it is -(-1), which is 1.
