Functions and their curves 195
Hence y = mx + c = ±1x + 0, i.e. y = x and y =−x
are asymptotes.
To determine any asymptotes parallel to the x- and
y-axes for the function x 3 − x y 2 + 2x − 9 =0:
Equating the coefficient of the highest power of x term
to zero gives 1 = 0 which is not an equation of a line.
Hence there is no asymptote parallel with the x-axis.
Equating the coefficient of the highest power of y term
to zero gives −x = 0 from which, x = 0.
Hence x = 0, y = x and y = − x are asymptotes for the
function x 3 − xy 2 + 2x − 9 =0.
Problem 12. Find the asymptotes for the function
y =
x 2 + 1
x
and sketch a graph of the function.
Rearranging y =
x 2 + 1
x
gives yx = x 2 + 1.
Equating the coefficient of the highest power x term to
zero gives 1 =0, hence there is no asymptote parallel to
the x-axis.
Equating the coefficient of the highest power y term to
zero gives x = 0.
Hence there is an asymptote at x = 0 (i.e. the
y-axis).
To determine any other asymptotes we substitute
y = mx + c into yx = x 2 + 1 which gives
(mx + c)x = x 2 + 1
i.e.
mx 2 + cx = x 2 + 1
and (m − 1)x 2 + cx − 1 = 0
Equating the coefficient of the highest power x term to
zero gives m − 1 = 0, from which m = 1.
Equating the coefficient of the next highest power x term
to zero gives c = 0. Hence y = mx + c = 1x + 0, i.e. y = x
is an asymptote.
A sketch of y =
x 2 + 1
x
is shown in Fig. 18.35.
It is possible to determine maximum/minimum points
on the graph (see Chapter 28).
Since
y =
x 2 + 1
x
=
x 2
x
+
1
x
= x + x −1
then
d y
dx
= 1 − x −2 = 1 −
1
x 2 = 0
for a turning point.
Hence 1 =
1
x 2 and x 2 = 1, from which, x = ±1.
When x = 1,
y =
x 2 + 1
x
=
1 + 1
1
= 2
and when x =−1,
y =
(−1) 2 + 1
−1
= −2
i.e. (1, 2) and (−1, −2) are the co-ordinates of the turning
points.
d 2 y
dx 2 = 2x −3 =
2
x 3 ; when x = 1,
d 2 y
dx 2 is positive,
which indicates a minimum point and when x =−1,
d 2 y
dx 2 is negative, which indicates a maximum point, as
shown in Fig. 18.35.
Now try the following exercise
Exercise 80 Further problems on
asymptotes
In Problems 1 to 3, determine the asymptotes
parallel to the x- and y-axes.
1. y =
x − 2
x + 1
[y = 1, x =−1]
2. y
2
=
x
x − 3
[x = 3, y = 1 and y =−1]
3. y =
x(x + 3)
(x + 2)(x + 1)
[x =−1, x =−2 and y = 1]
In Problems 4 and 5, determine all the asymptotes.
4. 8x − 10 + x 3 − x y 2 = 0
[x = 0, y = x and y =−x]
5. x 2 (y 2 − 16) = y
[y = 4, y =−4 and x = 0]
In Problems 6 and 7, determine the asymptotes and
sketch the curves.
6. y =
x 2 − x − 4
x + 1
x = −1, y = x − 2,
see Fig 18.40, page 202
7. x y 2 − x 2 y + 2x − y = 5
x = 0, y = 0, y = x,
see Fig. 18.41, page 202
Hence y = mx + c = ±1x + 0, i.e. y = x and y =−x
are asymptotes.
To determine any asymptotes parallel to the x- and
y-axes for the function x 3 − x y 2 + 2x − 9 =0:
Equating the coefficient of the highest power of x term
to zero gives 1 = 0 which is not an equation of a line.
Hence there is no asymptote parallel with the x-axis.
Equating the coefficient of the highest power of y term
to zero gives −x = 0 from which, x = 0.
Hence x = 0, y = x and y = − x are asymptotes for the
function x 3 − xy 2 + 2x − 9 =0.
Problem 12. Find the asymptotes for the function
y =
x 2 + 1
x
and sketch a graph of the function.
Rearranging y =
x 2 + 1
x
gives yx = x 2 + 1.
Equating the coefficient of the highest power x term to
zero gives 1 =0, hence there is no asymptote parallel to
the x-axis.
Equating the coefficient of the highest power y term to
zero gives x = 0.
Hence there is an asymptote at x = 0 (i.e. the
y-axis).
To determine any other asymptotes we substitute
y = mx + c into yx = x 2 + 1 which gives
(mx + c)x = x 2 + 1
i.e.
mx 2 + cx = x 2 + 1
and (m − 1)x 2 + cx − 1 = 0
Equating the coefficient of the highest power x term to
zero gives m − 1 = 0, from which m = 1.
Equating the coefficient of the next highest power x term
to zero gives c = 0. Hence y = mx + c = 1x + 0, i.e. y = x
is an asymptote.
A sketch of y =
x 2 + 1
x
is shown in Fig. 18.35.
It is possible to determine maximum/minimum points
on the graph (see Chapter 28).
Since
y =
x 2 + 1
x
=
x 2
x
+
1
x
= x + x −1
then
d y
dx
= 1 − x −2 = 1 −
1
x 2 = 0
for a turning point.
Hence 1 =
1
x 2 and x 2 = 1, from which, x = ±1.
When x = 1,
y =
x 2 + 1
x
=
1 + 1
1
= 2
and when x =−1,
y =
(−1) 2 + 1
−1
= −2
i.e. (1, 2) and (−1, −2) are the co-ordinates of the turning
points.
d 2 y
dx 2 = 2x −3 =
2
x 3 ; when x = 1,
d 2 y
dx 2 is positive,
which indicates a minimum point and when x =−1,
d 2 y
dx 2 is negative, which indicates a maximum point, as
shown in Fig. 18.35.
Now try the following exercise
Exercise 80 Further problems on
asymptotes
In Problems 1 to 3, determine the asymptotes
parallel to the x- and y-axes.
1. y =
x − 2
x + 1
[y = 1, x =−1]
2. y
2
=
x
x − 3
[x = 3, y = 1 and y =−1]
3. y =
x(x + 3)
(x + 2)(x + 1)
[x =−1, x =−2 and y = 1]
In Problems 4 and 5, determine all the asymptotes.
4. 8x − 10 + x 3 − x y 2 = 0
[x = 0, y = x and y =−x]
5. x 2 (y 2 − 16) = y
[y = 4, y =−4 and x = 0]
In Problems 6 and 7, determine the asymptotes and
sketch the curves.
6. y =
x 2 − x − 4
x + 1
x = −1, y = x − 2,
see Fig 18.40, page 202
7. x y 2 − x 2 y + 2x − y = 5
x = 0, y = 0, y = x,
see Fig. 18.41, page 202