Chapter 8 – Graphs and Functions:
8.1
Cartesian axes, coordinates and points
Pictorially we plot points and graphs in a plane (flat space) using a set of
Cartesian axes traditionally called the x and y axes to locate points on it.
See diagram below.
8
6
4
2
0
-6
-4
-2
-2
0
2
4
6
-4
-6
-8
The vertical axis is the y axis and the horizontal axis is the x axis. Any
point in this plane has a set of numbers called an ordered pair, which is a
set of coordinates that locate a point in the plane.
In the diagram above the top right point has coordinates (x=5,y=7) i.e. 5
units on the x axis and 7 units on the y axis. It is in quadrant 1. However
because we always write the x coordinate first we just write
(5,7) instead of (x=5,y=7).
The top right quadrant is quadrant 1, then going anticlockwise we get
the other 3 quadrants. For example the bottom left point has
coordinates (-5,-6) i.e. -5 units on the x axis and -6 units on the y axis. It
is in quadrant 3.
E8.1
A8.1
8.2
E8.2
Give the coordinates of the point in the
(i)
second quadrant
(ii)
fourth quadrant
The point in the
(i)
second quadrant is (-5,4).
(ii)
fourth quadrant is (3,-5).
The point (3,-2) is located 3 units to the right and 2 units below the
origin. It is in the 4th (bottom right) quadrant.
Complete the Table below on the location of a point relative to the
origin. The first row has been done for you.
Location
3 to the right and 2
below origin
Coordinates
(3,-2)
87
Quadrant
4th
(-5, 2)
At the origin
( , 4)
Junction of all 4
Boundary Between 1st
and 2nd.
(-2,0)
4 to the left and 2
above.
(5, )
Boundary between 1st
and 4th
Plot these points on the axes above.
A8.2
8.3
Location
3 to the right and 2
directly below origin
5 to the left, 2 below
At the origin
4 above the origin
Coordinates
(3,-2)
Quadrant
4th
(-5, 2)
(0,0)
(0, 4)
2 to the left
4 to the left and 2
above.
5 to the right
(-2,0)
(-4,2)
2nd
Junction of all 4
Boundary Between 1st
and 2nd.
Between 2nd And 3rd.
2nd
(5,0 )
Boundary between 1st
and 4th
Equations with infinite solutions
We know that x + 3 = 5 is an equation that we can solve by inspection.
There is only one solution,
x = 2.
Now consider the equation, x + y =10. It has two unknowns and we can
produce many solutions by inspection such as
x = 5 and y = 5, x = 0 and y = 10, x = 1 and y = 9. Also solutions with
negative values such as x = -4 and y = 14 etc.
Why does the first equation have one solution and the second equation
have an infinite number?
The reason is that the first equation has just one unknown, x whereas
the second equation has two unknowns x and y and yet we have just one
piece of information.
E8.3
How many solutions do the following equations have?
88
(i)
2x + 7 = 9. Identify it.
(ii)
5x + 6y = 10
Can you identify some of them?
A8.3
(i)
(ii)
2x + 7 = 9. Has 1 solution, x = 1.
5x + 2y = 10. Has an infinite number of solutions e.g.
(2,0), (0,5), (1,2.5) etc.
Can you identify some of them?
8.4
Since 1 equation with two unknowns has an infinite number of solutions
we can produce a graph by plotting the set of solutions on a set of
Cartesian axes.
Here are some solutions to the equation: y = 2x + 3
(0,3) because y = 3 when x = 0.
(2,7) because y = 7 when x = 2.
(-3, -3) because when x = -3, y = -3.
E8.4
Complete the following coordinates for some of the solutions of the
equation: y = 2x + 3
(-2, ) ; (-3, ) ; (-1, ) ; (1, ) ; ( , 4) ; ( , -3) ; (0, )
Plot the points on the axes given below.
14
12
10
8
6
4
2
0
-4
-2 0
2
4
6
-4
A8.4
Complete the following coordinates for some of the solutions of the
equation: y = 2x + 3
(-2,-1) ; (-3,-3) ; (-1,1) ; (1,5) ; (0.5, 4) ; (-3, -3) ; (0,3)
89
8.5
Linear graphs
What is the graph of the entire solution set? It is a straight line as shown
below:
14
12
10
8
6
4
2
-4
0
-2 0
-2
2
4
6
-4
E8.5
8.6
Graph of y =2x + 3
Examine the graph and identify the points where the graph crosses the y
axis and the x axis.
Graph crosses y axis at (0,3) and cross the x axis at (-1.5,0)
In graphing the equation, y = 2x + 3 let’s see what the significance of the
coefficient of 2 and and the constant term 3 is.
Note the 2 is the coefficient of the x term (linear term) and the 3 is the
constant term.
First consider the constant term, 3.
Notice that when x = 0 then y = 3. Hence (0,3) is a point on the graph.
Which quadrant is it in? It is on the boundary between the 1st and 2nd
quadrants i.e. it lies on the y axis. We call this point the y intercept. So
we say either the y intercept is 3 or we say the y intercept is the point
(0,3).
We say that in the graph y = mx + c, c is the y intercept.
E8.6
A8.6
8.7
What is the y intercept of the following graphs?
(i)
y = 2x + 4
(ii)
y = 2x – 3
(iii)
y = 2x?
The y intercept of
(i)
y = 2x + 4 is 4 i.e. the point (0,4).
(ii)
y = 2x – 3 is -3 i.e. the point (0,-3).
(iii)
y = 2x is 0 i.e. the point (0,0).
Now what is the significance of the coefficient of x of the RHS (2 in this
case) in the graph of
y = 2x + 3?
90
Let us look at two random points on the graph, say point A at (1,5) and
point B at (3,9)
We can by looking at the x coordinates of the two points determine that
B is 2 units to the right of A.
We write ∆x = x value of B – x value of A. = 3 – 1 =2
We can by looking at the y coordinates of the two points determine that
B is 4 units above A.
We write ∆y = y value of B – y value of A = 9 – 5 = 4
So in going from A to B the graph is climbing 4 units up for every 2 units
to the right. So we say the graph has a slope or gradient of ‘4 in 2’ which
we write as
y 4
gradient 
  2.
x 2
So the coefficient of the x term in the equation y = 2x + 3 is the gradient.
E8.7
Summarise:
constant term = 3 is the y intercept
coefficient of x term = the slope or gradient of the line given by
y
gradient 
x
Complete the Table below: The first row has been done for you.
Equation of line
y = 2x + 3
Slope
2
4
Y intercept
3
-1
-2
2
0
-4
y = 3x – 4
y = 4x
y = 5 – 2x
y=4
5
0
A8.7
Equation of line
y = 2x + 3
y = 4x – 1
y = 3x – 4
y = -2x + 2
y = 4x
y = -4
Slope
2
4
3
-2
4
0
91
Y intercept
3
-1
-4
2
0
-4
8.8
y = 5 – 2x
-2
5
y=4
0
4
y = 5x
5
0
What is the equation of the line that passes through the points
(0,3) and has a slope of -2?
We know that in general the equation is of the form:
y = mx + c,
where m is the slope (coefficient of x term) and y intercept is c (the
constant term).
Since the slope is -2 (i.e. the coefficient of x is -2) y = mx + c becomes y =
-2x + c.
Since the graph passes through (0,3) we know the y intercept is 3; So c is
3 and y = -2x + c becomes y = -2x + 3.
So the equation is y = -2x + 3 OR y = 3 – 2x.
E8.8
A8.8
8.9
What is the equation of the line that passes through the point
i.
(0 , -1) and has a slope of 2
ii.
(0, 2) and has a slope of -3
iii.
(0,0) and has a slope of 1
i.
ii.
iii.
y = 2x – 1
y = -3x + 2
y=x
What is the equation of the line that passes through the points
(0,3) and (2,5)?
We know that in general the equation is of the form:
y = mx + c,
where m is the slope (coefficient of x term) and y intercept is c (the
constant term).
Since the graph passes through (0,3), which is on the y axis, we know the
y intercept is 3; so c is 3, so y = mx + c becomes
y = mx + 3.
The point (2,5) is 2 units to the right (2 – 0) and 2 units above (5 – 3) the
point (0,3).
y 5 - 3 2
gradient 

  1.
x 2 - 0 2
So the slope is 2 in 2 i.e. m=1
92
so y = mx + 3 becomes y = x + 3.
So the equation of the line is y = x + 3
E8.9
What is the equation of the line that passes through the points
(0,-1) and (1,4)?
A8.9
(0,-1) tells us that the y intercept is -1 so we can write y = mx – 1.
(0,-1) and (1,4) tells us that the gradient is 5.
So y = 5x – 1.
Now consider the line that passes through the points (-2,4) and (3,-5).
8.10
We can’t determine the slope or the y intercept at sight easily:
So we proceed as follows:
gradient 
y - 5 - 4  9


.
x 3 - -2
5
So m= -9/5 in the equation y = mx + c.
So write: y  -
9x
 c.
5
Now remember (3,-5) is (x=3,y=-5). So substitute in the equation above
to get:
9(3)
c
5
9(3)  25  27
2
c  - 5 


5
5
5
-5 -
9x
9x 2
 c. becomes y   .
5
5 5
Find the equation of the line that passes through (-3, 6) and (2,1)
So our line y  E8.10
A8.10
gradient 
y 6 - 1
5


 1.
x 2 - 3  5
So m= -1 in the equation y = mx + c.
93
So write: y  - x  c.
Now remember (2,1) is (x=1,y=1). So substitute in the equation above to
get:
2  -1  c
8.11
c  3
So our line y  - x  c. becomes y  3 - x.
Consider the line y = 2x + 3 again. It has a slope of 2 and the y intercept is
3. What is the x intercept?
We know that at the x intercept y=0. If we substitute y=0 in y = 2x + 3
we get 0 = 2x + 3.
We can solve this equation to get x  - 32 .
So the x intercept is - 32 and the point of interception on the x axis is
( - 32 ,0 )
E8.11
A8.11
8.12
Check by seeing the graph in box 8.5.
For the graph y = 4 – 3x what is
i.
the slope ii. y intercept iii. x intercept?
y = 4 – 3x
i. slope = -3 ii. y intercept = 4 iii. 0 = 4 – 3x  x intercept = 4/3
Graphs of Quadratic functions
A quadratic function is function of the form :
y = ax2 + bx + c where a,b,c are constants.
There are three terms in the function the first is a term in x2 (called the
quadratic term), the second is a term in x (called a linear term) and the
third is a constant term.
This is a 2nd degree function, whereas linear functions are first degree
functions.
The simplest quadratic is y = x2 i.e. no linear term and no constant term.
We can complete the ordered pairs using the x values given below:
(-5,25) , (-4,16), (-3,9), (-2,4), (-1,1), (0,0), (1,1), (2,4), (3,9),
(4,16), (5,25).
Note the x values range from -5 to +5 but the y values are all positive
94
E8.12
because the square of a number is always +ve.
Complete the following points on this quadratic:
( ,25), ( ,0), ( ,25) so that the points are distinct (i.e. separate)
A8.12
8.13
( -5,25), (0,0), (5,25)
The graph is plotted below
30
25
20
15
10
5
0
-10
-5
0
5
10
E8.13
What is the y intercept?
What is the y intercept? the x intercept?
What is the axis of symmetry?
What is the minimum point?
What is the minimum y value?
A8.13
y intercept is 0, the x intercept is 0, the axis of symmetry is x = 0,
the minimum point is (0,0),
the minimum value is 0.
Now suppose we want to sketch (rather than draw) the quadratic,
y = x2 – 2x – 15
8.14
We can find the y intercept by putting x = 0 in
y = x2 – 2x – 15
to get y = -15. So the graph crosses the y axis at (0,-15).
Now to find the x intercepts we put y = 0 to get: 0 = x2 – 2x – 15.
Write this as
x2 – 2x – 15 = 0 and solve viz.
(x – 5)(x + 3) = 0 So x = -3 and x = 5.
The graph crosses the x axis at (-3,0) and (5,0).
The axis of symmetry is a vertical line bisecting the line joining
(-3,0) and (5,0).
95
How do we find it? The middle value of any two numbers is its arithmetic
average.
(-3  5)
x
The average of x = -3 and x = 5 is thus
2
x 1
The axis of symmetry is x = 1.
Now we get the minimum y value using the fact that minimum y value is
on the axis of symmetry, x = 1.
So putting x=1 in y = x2 – 2x – 15 we get
ymin = 12 – 2(1) – 15 = -17.
So the minimum point of the graph is (1,-17).
We can now sketch our graph by putting all this information together:
Point where graph crosses y axis. (0,-15)
Two axis points where graph
(-3,0) and (5,0)
crosses x axis.
Minimum point on the axis of
(1,-17)
symmetry
The sketch is confirmed by this drawing.
25
20
15
10
5
-10
-5
0
-5 0
5
10
-10
-15
-20
E8.14
Sketch the graph of the quadratic function y = x2 – 2x – 8
1. Find the y intercept by putting x = 0 in
y = x2 – 2x – 8
to get y = ….
2. Find the x intercepts by putting put y = 0 to get:
…. = x2 – 2x – 8 and solve viz.
(x )(x ) = 0 So x = ….. and x = …..
96
The two points of interception on the x axis are (…,….) and (…,…)
The axis of symmetry is a vertical line bisecting the line joining
(….,…) and (…,…)
The middle x value is its arithmetic average.
The average of x = … and x = … is thus x = …..
The axis of symmetry is x = …...
The minimum y value is on the axis of symmetry, x = …..
So putting x=…. in y = x2 – 2x – 8 we get
ymin = ….
So the minimum point of the graph is (….,….).
We can now sketch our graph by putting all this information together:
Point of interception on y axis
( , )
Two points of interception on x
( , ) and ( , )
axis
Minimum point on the axis of
( , )
symmetry
Sketch:
A8.14
Sketch the graph of the quadratic function y = x2 – 2x – 8
1. Find the y intercept by putting x = 0 in
y = x2 – 2x – 8
to get y = -8. So the y intercept is the point (0,-8).
2. Find the x intercepts by putting put y = 0 to get:
0 = x2 – 2x – 8 and solve viz.
(x – 4)(x + 2) = 0 So x = 4 and x = -2 are the x intercepts.
97
The two points of interception on the x axis are (4,0) and (-2,0).
The axis of symmetry is a vertical line bisecting the line joining
(4,0) and (-2,0).
The middle x value is its arithmetic average.
The average of x = 4 and x = -2 is thus x = (4 – 2)/2 = 1
i.e. the axis of symmetry is the vertical line, x = 1
The minimum y value is on the axis of symmetry, x = 1.
So putting x = 1 in y = x2 – 2x – 8 we get
ymin = (1)2 – 2(1) – 8 = 1 – 2 – 8 = -9.
So the minimum point of the graph is (1,-9).
We can now sketch our graph by putting all this information together:
Point of interception on y axis
(0,-8)
Two points of interception on x
(-2,0) and (0,4)
axis
Minimum point on the axis of
(1,-9)
symmetry
Sketch:
98
99
Exercise 8
1. Find the equation of the straight line with a gradient of 2 passing thru the point (1,1).
2. A line passes thru the points (0,3)and (3,0). Find Delta-x, Delta-y, the slope, and
the equation of the line.
3. A variable P changes from 20 to 30. This causes a dependent variable Q to change
from 100 to 25. Find Delta-P, delta-Q, the slope, the proportional change in P,
proportional change in Q and the ratio of the proportional change in P to the
proportional change in Q.
4.
(i)
(ii)
(iii)
(iv)
Sketch the graph of the straight line y = 2 – 3x
what is the gradient?
what is the y intercept?
what is the x intercept?
5 Sketch the quadratic y = x2 – 2x – 8
100