136 Basic Engineering Mathematics
Ϫ4 Ϫ3 Ϫ2 Ϫ1
1 2 3
4 x
A
B
C
F
E
D
y ϭ
x ϩ
5
y ϭ
x ϩ
2
y ϭ
x Ϫ
3
y ϭ
x
9
y
8
7
6
5
4
3
2
1
Ϫ1
Ϫ2
Ϫ3
Ϫ4
Ϫ5
Ϫ6
Ϫ7
Figure 17.9
To find the gradient of any straight line, say, y = x − 3,
a horizontal and vertical component needs to be constructed. In Figure 17.9, AB is constructed vertically at
x = 4 and BC is constructed horizontally at y = −3.
The gradient of AC =
AB
BC
=
1 − (−3)
4 − 0
=
4
4
= 1
i.e. the gradient of the straight line y = x − 3 is 1, which
could have been deduced ‘on sight’ since y = 1x − 3
represents a straight line graph with gradient 1 and
y-axis intercept of −3.
The actual positioning of AB and BC is unimportant
because the gradient is also given by
DE
EF
=
−1 − (−2)
2 − 1
=
1
1
= 1
The slope or gradient of each of the straight lines in
Figure 17.9 is thus 1 since they are parallel to each
other.
Problem 3. Plot the following graphs on the same
axes between the values x = −3 to x = +3 and
determine the gradient and y-axis intercept of each.
(a) y = 3x
(b) y = 3x + 7
(c) y = −4x + 4
( d )y = −4x − 5
A table of co-ordinates is drawn up for each equation.
(a) y = 3x
x −3 −2 −1 0 1 2 3
y −9 −6 −3 0 3 6 9
(b) y = 3x + 7
x −3 −2 −1 0 1 2 3
y −2
1
4 7 10 13 16
(c) y = −4x + 4
x −3 −2 −1 0 1
2
3
y 16 12
8 4 0 −4 −8
(d) y = −4x − 5
x −3 −2 −1
0
1
2
3
y
7
3 −1 −5 −9 −13 −17
Each of the graphs is plotted as shown in Figure 17.10
and each is a straight line. y = 3x and y = 3x + 7 are
parallel to each other and thus have the same gradient.
The gradient of AC is given by
CB
BA
=
16 − 7
3 − 0
=
9
3
= 3
y 5
3 x 1
7
y 5 2 4 x 2 5
y 5 2 4 x 1 4
y 5
3 x
16
12
8
4
23
22
0
28
212
216
21
1
2
3 x
24
B
C
A
F
E
D
y
Figure 17.10
Hence, the gradients of both y = 3x and y = 3x + 7
are 3, which could have been deduced ‘on sight’.
y = −4x + 4 and y = −4x − 5 are parallel to each other
and thus have the same gradient. The gradient of DF is
Ϫ4 Ϫ3 Ϫ2 Ϫ1
1 2 3
4 x
A
B
C
F
E
D
y ϭ
x ϩ
5
y ϭ
x ϩ
2
y ϭ
x Ϫ
3
y ϭ
x
9
y
8
7
6
5
4
3
2
1
Ϫ1
Ϫ2
Ϫ3
Ϫ4
Ϫ5
Ϫ6
Ϫ7
Figure 17.9
To find the gradient of any straight line, say, y = x − 3,
a horizontal and vertical component needs to be constructed. In Figure 17.9, AB is constructed vertically at
x = 4 and BC is constructed horizontally at y = −3.
The gradient of AC =
AB
BC
=
1 − (−3)
4 − 0
=
4
4
= 1
i.e. the gradient of the straight line y = x − 3 is 1, which
could have been deduced ‘on sight’ since y = 1x − 3
represents a straight line graph with gradient 1 and
y-axis intercept of −3.
The actual positioning of AB and BC is unimportant
because the gradient is also given by
DE
EF
=
−1 − (−2)
2 − 1
=
1
1
= 1
The slope or gradient of each of the straight lines in
Figure 17.9 is thus 1 since they are parallel to each
other.
Problem 3. Plot the following graphs on the same
axes between the values x = −3 to x = +3 and
determine the gradient and y-axis intercept of each.
(a) y = 3x
(b) y = 3x + 7
(c) y = −4x + 4
( d )y = −4x − 5
A table of co-ordinates is drawn up for each equation.
(a) y = 3x
x −3 −2 −1 0 1 2 3
y −9 −6 −3 0 3 6 9
(b) y = 3x + 7
x −3 −2 −1 0 1 2 3
y −2
1
4 7 10 13 16
(c) y = −4x + 4
x −3 −2 −1 0 1
2
3
y 16 12
8 4 0 −4 −8
(d) y = −4x − 5
x −3 −2 −1
0
1
2
3
y
7
3 −1 −5 −9 −13 −17
Each of the graphs is plotted as shown in Figure 17.10
and each is a straight line. y = 3x and y = 3x + 7 are
parallel to each other and thus have the same gradient.
The gradient of AC is given by
CB
BA
=
16 − 7
3 − 0
=
9
3
= 3
y 5
3 x 1
7
y 5 2 4 x 2 5
y 5 2 4 x 1 4
y 5
3 x
16
12
8
4
23
22
0
28
212
216
21
1
2
3 x
24
B
C
A
F
E
D
y
Figure 17.10
Hence, the gradients of both y = 3x and y = 3x + 7
are 3, which could have been deduced ‘on sight’.
y = −4x + 4 and y = −4x − 5 are parallel to each other
and thus have the same gradient. The gradient of DF is