Functions and their curves 193
Problem 9. Determine the asymptotes parallel to
the x- and y-axes for the function
x 2 y 2 = 9(x 2 + y 2 ).
Asymptotes parallel to the x-axis:
Rearranging x 2 y 2 = 9(x 2 + y 2 ) gives
x 2 y 2 − 9x 2 − 9y 2 = 0
hence x 2 (y 2 − 9) − 9y 2 = 0
Equating the coefficient of the highest power of x to zero
gives y 2 − 9 = 0 from which, y 2 = 9 and y =±3.
Asymptotes parallel to the y-axis:
Since x 2 y 2 − 9x 2 − 9y 2 = 0
then
y 2 (x 2 − 9) − 9x 2 = 0
Equating the coefficient of the highest power of y to zero
gives x 2 − 9 = 0 from which, x 2 = 9 and x =±3.
Hence asymptotes occur at y =±3 and x =±3.
Other asymptotes
To determine asymptotes other than those parallel to
x- and y-axes a simple procedure is:
(i) substitute y = mx + c in the given equation
(ii) simplify the expression
(iii) equate the coefficients of the two highest powers
of x to zero and determine the values of m and c.
y = mx + c gives the asymptote.
Problem 10. Determine the asymptotes for the
function: y(x + 1) = (x − 3)(x + 2) and sketch the
curve.
Following the above procedure:
(i) Substituting y = mx + c into
y(x + 1) = (x − 3) (x + 2) gives:
(mx + c)(x + 1) = (x − 3)(x + 2)
(ii) Simplifying gives
mx
2
+ mx + cx + c = x
2
− x − 6
and (m − 1)x 2 + (m + c + 1)x + c + 6 =0
(iii) Equating the coefficient of the highest power
of x to zero gives m − 1 = 0 from which,
m = 1.
Equating the coefficient of the next highest power
of x to zero gives m + c + 1 =0.
and since m = 1, 1 + c + 1 = 0 from which,
c =−2.
Hence y = mx + c = 1x − 2.
i.e. y = x − 2 is an asymptote.
To determine any asymptotes parallel to the x-axis:
Rearranging y(x + 1) = (x − 3)(x + 2)
gives
yx + y = x 2 − x − 6
The coefficient of the highest power of x (i.e. x 2 ) is 1.
Equating this to zero gives 1 =0 which is not an equation
of a line. Hence there is no asymptote parallel to the
x-axis.
To determine any asymptotes parallel to the y-axis:
Since y(x + 1) = (x − 3)(x + 2) the coefficient of
the highest power of y is x + 1. Equating this to
zero gives x + 1 = 0, from which, x =−1. Hence x =−1
is an asymptote.
When x = 0, y(1) = (−3)(2), i.e. y =−6.
When y = 0, 0 =(x − 3)(x + 2), i.e. x = 3 and x =−2.
A sketch of the function y(x + 1) = (x − 3)(x + 2) is
shown in Fig. 18.34.
Problem 11. Determine the asymptotes for the
function x
3
− x y
2
+ 2x − 9 =0.
Following the procedure:
(i) Substituting y = mx + c gives
x 3 − x(mx + c) 2 + 2x − 9 =0.
(ii) Simplifying gives
x 3 − x[m 2 x 2 + 2mcx + c 2 ] + 2x − 9 = 0
i.e. x 3 − m 2 x 3 − 2mcx 2 − c 2 x + 2x − 9 = 0
and x
3
(1 − m
2
) − 2mcx
2
− c
2 x + 2x − 9 = 0
(iii) Equating the coefficient of the highest power of x
(i.e. x 3 in this case) to zero gives 1 −m 2 = 0, from
which, m =±1.
Equating the coefficient of the next highest power
of x (i.e. x 2 in this case) to zero gives −2mc = 0,
from which, c = 0.
Problem 9. Determine the asymptotes parallel to
the x- and y-axes for the function
x 2 y 2 = 9(x 2 + y 2 ).
Asymptotes parallel to the x-axis:
Rearranging x 2 y 2 = 9(x 2 + y 2 ) gives
x 2 y 2 − 9x 2 − 9y 2 = 0
hence x 2 (y 2 − 9) − 9y 2 = 0
Equating the coefficient of the highest power of x to zero
gives y 2 − 9 = 0 from which, y 2 = 9 and y =±3.
Asymptotes parallel to the y-axis:
Since x 2 y 2 − 9x 2 − 9y 2 = 0
then
y 2 (x 2 − 9) − 9x 2 = 0
Equating the coefficient of the highest power of y to zero
gives x 2 − 9 = 0 from which, x 2 = 9 and x =±3.
Hence asymptotes occur at y =±3 and x =±3.
Other asymptotes
To determine asymptotes other than those parallel to
x- and y-axes a simple procedure is:
(i) substitute y = mx + c in the given equation
(ii) simplify the expression
(iii) equate the coefficients of the two highest powers
of x to zero and determine the values of m and c.
y = mx + c gives the asymptote.
Problem 10. Determine the asymptotes for the
function: y(x + 1) = (x − 3)(x + 2) and sketch the
curve.
Following the above procedure:
(i) Substituting y = mx + c into
y(x + 1) = (x − 3) (x + 2) gives:
(mx + c)(x + 1) = (x − 3)(x + 2)
(ii) Simplifying gives
mx
2
+ mx + cx + c = x
2
− x − 6
and (m − 1)x 2 + (m + c + 1)x + c + 6 =0
(iii) Equating the coefficient of the highest power
of x to zero gives m − 1 = 0 from which,
m = 1.
Equating the coefficient of the next highest power
of x to zero gives m + c + 1 =0.
and since m = 1, 1 + c + 1 = 0 from which,
c =−2.
Hence y = mx + c = 1x − 2.
i.e. y = x − 2 is an asymptote.
To determine any asymptotes parallel to the x-axis:
Rearranging y(x + 1) = (x − 3)(x + 2)
gives
yx + y = x 2 − x − 6
The coefficient of the highest power of x (i.e. x 2 ) is 1.
Equating this to zero gives 1 =0 which is not an equation
of a line. Hence there is no asymptote parallel to the
x-axis.
To determine any asymptotes parallel to the y-axis:
Since y(x + 1) = (x − 3)(x + 2) the coefficient of
the highest power of y is x + 1. Equating this to
zero gives x + 1 = 0, from which, x =−1. Hence x =−1
is an asymptote.
When x = 0, y(1) = (−3)(2), i.e. y =−6.
When y = 0, 0 =(x − 3)(x + 2), i.e. x = 3 and x =−2.
A sketch of the function y(x + 1) = (x − 3)(x + 2) is
shown in Fig. 18.34.
Problem 11. Determine the asymptotes for the
function x
3
− x y
2
+ 2x − 9 =0.
Following the procedure:
(i) Substituting y = mx + c gives
x 3 − x(mx + c) 2 + 2x − 9 =0.
(ii) Simplifying gives
x 3 − x[m 2 x 2 + 2mcx + c 2 ] + 2x − 9 = 0
i.e. x 3 − m 2 x 3 − 2mcx 2 − c 2 x + 2x − 9 = 0
and x
3
(1 − m
2
) − 2mcx
2
− c
2 x + 2x − 9 = 0
(iii) Equating the coefficient of the highest power of x
(i.e. x 3 in this case) to zero gives 1 −m 2 = 0, from
which, m =±1.
Equating the coefficient of the next highest power
of x (i.e. x 2 in this case) to zero gives −2mc = 0,
from which, c = 0.