COORDINATE GEOMETRY
This Topic Involves:

Sketch graphs of rational functions defined by:
f ( x )  ax m  bx  n m, n  1, 2 and f ( x) 
1
;
ax  bx  c
2
their asymptotic behaviour and the nature and location of stationary points.
( x  h) 2 ( y  k ) 2

 1.
a2
b2

Sketch graphs of ellipses from the general Cartesian relation:

Sketch graphs of hyperbolas (including asymptotic behaviour) from the general
Cartesian relation:
( x  h) 2 ( y  k ) 2
( y  k ) 2 ( x  h) 2

 1 and

1
a2
b2
b2
a2
RATIONAL ALGEBRAIC FUNCTIONS
Rational functions are functions of the type, where P(x) and Q(x) are polynomials with no
common non-constant factors.
For Example:
Expressions such as y 
3x  2
x
2
, y
and y 
are all rational functions.
2
x7
x 1
5 x  3x  1
2
SKETCHING RATIONAL ALGEBRAIC FUNCTIONS
Depending on the manner in which equations are presented, rational functions may be
sketched by:

The addition of ordinates.

Considering key features such as axes-intercepts, stationary points, asymptotes etc.

The use of reciprocals.
Note:
To find the horizontal and/or oblique asymptote(s) of a rational algebraic function where the
degree of the polynomial in the numerator is equal to or greater than the degree of the
polynomial in the denominator, the equation of the function must first be re-written in
quotient/remainder form. This can be done using synthetic or polynomial long division.
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For Example:
x3
x  2x
Equation of horizontal asymptote is y  0 .
y
x
x 1
Must first be re-written in quotient/remainder form.
y
1
x  3x  7
Equation of horizontal asymptote is y  0 .
y
2  x2
x3
Must first be re-written in quotient/remainder form.
y
2
2
Equations of vertical asymptote are x  0 and x  2 .
Example: Re-write y 
2  x2
in quotient/remainder form.
x3
x  3
x3
 x2  0 x  2
 x 2  3x
3x  2
3x  9
7
 y
7
2  x2
= x  3 
x3
x3
By inspection:

The oblique asymptote is y   x  3 .

The equation of vertical asymptote is x  3 .
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SKETCH GRAPHS OF f  x   ax m  bx  n m, n  1, 2
Examples of such functions are:
5
x2
2
f ( x)  3x 2 
x
3
f ( x)  4 x 
x
1
f ( x)  x 2  2
x
f ( x)  x 
Power functions
Step 1: Sketch each individual function separately.
Step 2: Use addition of ordinates method to combine the two graphs. At select values of
x , add the y -values of each function together.
Add:
Intercept points
Intersection points
Maximum/minimum points
End points
Step 3: Determine the asymptotes of the original function.
Vertical asymptotes: The value(s) that makes the denominator 0.
Note the graphs of the type f ( x )  ax m  bx  n , vertical asymptote will always be
x  0 , as the denominator of the fractional part is always x .
To determine the non-vertical asymptote, cover up only the fractional part, the part
of the function you can still see is the equation of either an oblique asymptote (if is
in terms of x ) or a curved asymptote (if it is terms of x 2 ).
Step 4: Work out the axial intercepts and the turning points if required.
From the examples above:
Functions
f ( x)  x 
5
x2
2
x
3
f ( x)  4 x 
x
1
f ( x)  x 2  2
x
f ( x)  3 x 2 
Vertical asymptote
Non-vertical asymptote
x0
yx
x0
y  3 x 2
x0
y  4x
x0
y  x2
(See Questions 1 and 2)
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SUMMARY OF KEY FEATURES
ON SKETCH GRAPHS
Key Feature on
Sketch Graph
How To Find Key Feature
y  f (x) and solve y  f (0) directly for y .
Y-intercepts
Substitute x  0 into
X-intercepts
Substitute y  0 into y  f (x) and solve f ( x)  0 for x .
Stationary Points
x-coordinate: Solve
dy
 0 for x .
dx
y-coordinate: Substitute x-coordinate into
for y .
Nature of Stationary Points
Option 1: Examine the sign of
y  f ( x) and solve directly
dy
at a neighbouring point on the
dx
RHS
LHS
of the stationary point:

Maximum turning point:
the

LHS
RHS
LHS
dy
is
dx
Stationary point of inflexion:
Option 2: Examine the sign of

2
d y
dx 2
d2y
dx

2
2
d y
dx 2
at a neighbouring point on
 ve
 ve
at a neighbouring point on
of the stationary point.
neighbouring point on the

 ve
+ ve
of the stationary point.
Minimum turning point:
the

RHS
dy
is
dx
dy
is
dx
RHS
LHS
 ve
 ve
or
 ve
 ve
at a
of the stationary point.
d2y
at the stationary point:
dx 2
 0  minimum turning point.
 0  maximum turning point.
 0  the test fails – use option 1. (Refer to Page 10 and
Page 68 for point of inflection material, as found by the 2nd
derivative).
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Key Feature on
Sketch Graph
How To Find Key Feature
Determine the values of ( x  a, x  b,  ) that make y  f (x )
undefined. In particular:
Vertical Asymptotes.
(The vertical lines located at
values of x for which y is not
defined).

For a rational function, solve DENOMINATOR = 0 for x .

For a logarithmic function, solve ARGUMENT OF LOG = 0
for x .
Equations of vertical asymptotes are the vertical lines x  a , x  b , …
If y  g ( x )  b and lim g ( x)  0 and/or lim g ( x)  0 , the equation
Horizontal Asymptote.
x  
x  
(The horizontal line located at
the value that y approaches as
x   ).
of the horizontal asymptote is defined by the horizontal line y  b .
Oblique Asymptote.
If y  g ( x)  h( x) and
(The line or curve that y
approaches as x   ).
Note: If y 
lim g ( x)  0 and/or lim g ( x)  0 , the
x  
x  
equation of the oblique asymptote is defined by
y  h(x) .
polynomial 1
and the degree of polynomial 1 is EQUAL TO OR GREATER
polynomial 2
than the degree of polynomial 2, then polynomial long division must be performed in
order to find the horizontal and oblique asymptote:
h(x )
polynomial 2 ) polynomial 1

Remainder







polynomial 1
polynomial 2
 h(x ) 
Remainder
polynomial 2
Since the degree of the remainder is always less than the degree of polynomial 2,
lim
x  
remainder
polynomial 2
 0 . y  h( x) is therefore the equation of the oblique asymptote.
(See Question 3)
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SUMMARY OF GRAPHS OF POWER FUNCTIONS
The following table illustrates general shapes of different power functions. Be familiar with
these shapes but don’t try to learn them off by heart!
Note: f ( x)  ax m  bx  n m, n  1, 2 always have the vertical asymptote at x  0 .
Functions
a, b  0
Non-Vertical
Asymptote
Graph
Number of
Turning points
y
y  ax 
b
x
y  ax
x
2
y
y  ax 
b
x
y  ax
0
x
y
y  ax 
b
x
y  ax
0
x
y
y  ax 
b
x
y  ax
2
x
y
y  ax 2 
b
x
y  ax 2
x
1
y
y  ax 2 
b
x
y  ax 2
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Functions
a, b  0
Non-Vertical
Asymptote
Graph
Number of
Turning Points
y
y  ax 2 
b
x
y  ax 2
1
x
y
y  ax 2 
b
x
y  ax 2
1
x
y
y  ax 
b
x2
y  ax
1
x
y
y  ax 
b
x2
y  ax
x
1
y
y  ax 
b
x2
y  ax
1
x
y
y  ax 
b
x2
y  ax
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Functions
a, b  0
Non-Vertical
Asymptote
Graph
Number of
Turning Points
y
y  ax 2 
b
x2
y  ax 2
2
x
y
y  ax 2 
b
x2
y  ax 2
0
x
y
y  ax 2 
b
x2
y  ax 2
0
x
y
y  ax 2 
b
x2
y  ax 2
2
x
(See Questions 4 and 5)
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SKETCHING RECIPROCAL FUNCTIONS
k
where k is a constant then we can sketch the
f (x)
graph of f ( x ) and then use reciprocal theory to sketch the graph of y.
If the function is written in the form y 
The use of reciprocals simply means that every y value on the original function is
reciprocated. For example, if the function y  f (x ) has a y -intercept at (0, 3) then the
reciprocal function y 
1
has a y -intercept at
f ( x)
 1
 0,  .
 3
1
g ( x)
Feature on Graph of g ( x)
Feature on Graph of
x - intercept at x  a (which may or may
Vertical asymptote at x  a .
not also be a stationary point).
y - intercept at y  a .
y - intercept at y  1 .
a
Maximum turning point at (a, b), b  0 .
Minimum turning point at  a , 1 , b  0 .
Minimum turning point at (a, b), b  0 .
Maximum turning point at  a , 1 , b  0 .

b

b
Stationary point of inflexion at
(a, b), b  0 .
Stationary point of inflexion at  a , 1 , b  0 .
Vertical asymptote at x  a .
x - intercept at x  a (which may or may
Horizontal asymptote at y  a , a  0 .
not also be a stationary point).
Horizontal asymptote at y  0 :
Horizontal asymptote at y  1 , a  0 .
y  0
y  
y  0
y  
y  
y  
Horizontal asymptote at y  0 : y  0 

b
a
Horizontal asymptote at y  0 : y  0 
(See Questions 6 and 7)
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QUESTION 1
Sketch the graph of f ( x )  g ( x ) where f ( x ) and g ( x ) are given below:
y
y  f ( x)
25
20
15
10
5
– 10 – 9 – 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1
– 5
y  g ( x)
1
2
3
4
5
6
7
8
9
10
– 10
– 15
– 20
– 25
– 30
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x
QUESTION 2
Sketch the graph of y  3 x 2 
2
using the method of addition of ordinates.
x
Solution
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QUESTION 3
Sketch the graph of y 
x 2  4x  1
showing all significant features.
x
Solution
Division changes the expression to the form y  x  4 
1
.
x
Asymptotic Behaviour:
There is a vertical asymptote at x  0 . (Denominator = 0 at x  0 )
As x  0 "from above",
1
  so y  
x
As x  0 "from below",
1
  so y  
x
Oblique and horizontal asymptotes are found by investigating behaviour as x   .
1

 0  so y   x  4 
x
1

As x   ,  0  so y   x  4 
x
As x   ,
i.e. The line y  x  4 is an oblique asymptote.
Turning Points:
y 
x 2  4x  1
x


dy 2 x  4 x  x 2  4 x  1 1

dx
x2
(using the quotient rule)
dy
0
dx
2 x 2  4 x  x 2  4 x  10
x 2  10
x 1
Substituting these values into the original equation, the turning points occur at 1, 6
and  1, 2 .
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y
Intercepts:
8
There are no y-intercepts as x  0 is a vertical asymptote.
6
4
(-1,2)
2
The x-intercept is found in the usual way:
y  0  x 2  4 x  1 0
Therefore, x   2 
-6
3
(1,6)
-4
-2
2
4
6
-2
i.e. x  - 3.73 and - 0.27
-4
-6
-8
Note: Graphs must not touch or curl away from asymptotes.
(i.e. They must draw as close as possible to the asymptote without touching).
QUESTION 4
The graph of y 
1  x2
has:
6x
A
No straight line asymptotes.
B
y  6 x as its only straight line asymptote.
C
x  0 as its only straight line asymptote.
D
E
1
y  0 and y   x as its only straight line asymptotes.
6
1
x  0 and y   x as its only straight line asymptotes.
6
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x
QUESTION 5
Sketch the graph of y 
16 x3  1
showing all significant features.
2x2
Solution
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QUESTION 6
Sketch the graph of y 
1
showing all significant features.
x  x6
2
y
Solution
10
Note that the graph of y  x 2  x  6 :
5
(a)
Has X-intercepts (2, 0) and (3, 0) .
(b)
Y-intercept (0,  6) .
(c)
1
2
1
4
-3
-2
-1
1
2
3
4
x
Has a local minimum at  ,  6  .
Hence we see that the graph of y 
1
:
x  x6
Has vertical asymptotes at x  2 and x  3 .
(b)
Has a Y-intercept at  0,   .


-5
2
(a)
(c)
-4
-10
1
6
y
8
7
4 
1
Has a local maximum at  , 
.
 2 25 
6
5
4
3
2
1
-5
-4
-3
-2
-1
-1
-2
1
2
3
-3
-4
-5
-6
-7
-8
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4
x
QUESTION 7
Sketch the graph of y 
5
showing all significant features.
x  6 x  10
2
Solution
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