21 Graphs
By studying this lesson you will be able to
• find the gradient of the graph of a straight line,
• draw the graph of a function of the form y = ax2 + b.
Graph of a function of the form y = mx + c
The graph of a function of the form y = mx + c is a straight line. The coefficient of
x, which is m, represents the gradient of the line and the constant term c represents
the intercept of the graph.
Review Exercise
1. Write down the gradient and intercept of the straight line represented by each of
the following equations.
21.1 Geometrical description of the gradient of a straight line
We defined the coefficient m of x in the equation y = mx + c as the gradient of
the straight line. Now by considering an example, let us see how the value of m
is represented geometrically. To do this, let us consider the straight line given by
y = 2x + 1. Let us use the following table of values to draw its graph.
x –2 0 2
y (= 2x + 1) –3 1 5
20 For free distribution
,Let us mark any three points on the straight line. For example, let us take the three
points as A (0, 1), B (2, 5) and C (5, 11).
y
C
11
10
^5" 11&
9
8
7
6
5 B ^5" 5&
^2" 5& Q
4
3
2
^0" 1& A 1 P ^2" 1&
x
−5 −4 −3 −2 −1 0 1 2 3 4 5
−1
−2
−3
−4
−5
−6
First let us consider the points A and B.
Let us draw a line from A, parallel to the x – axis, and a line from B, parallel to the
y – axis, and name the point of intersection of these lines as P. It is clear that the
coordinates of the point P are (2, 1).
Also, length of AP = 2 – 0
= 2
length of BP = 5 – 1
= 4
Vertical distance BP 4 = 2 .
Now for the points A and B, = =
Horizontal distance AP 2
We already know the gradiant of the straight line y = 2x + 1 is 2.
The quotient Vertical Distance for the point A and B is also 2.
Horizontal distance
Now let us consider another case.
As the second case let us consider the points B and C.
Let us draw a line from B, parallel to the x - axis, and a line from C, parallel to the
y - axis and name the point of intersection of these two lines as Q.
Then the coordinates of Q are (5, 5).
Length of BQ = 5 – 2
= 3
For free distribution 21
, Length of CQ = 11 – 5
=6
Vertical distance CQ 6
Now, for the points B and C, = BQ = 3 = 2 .
Horizontal distance
In both instances, the ratio of the vertical distance to the horizontal distance between
the two points under consideration is the gradient 2 of the straight line.
Accordingly, let us develop a formula to find the gradient of a straight line using its
graph. Let us consider any straight line with equation y = mx + c.
y
7
6 A
5
(x1, y1)
4
3
2
1 B (x2, y2)
−5 −4 −3 −2 −1 0 1 2 3 4 5 x
−1
−2
−3
−4
−5
−6
Let us consider any two points A (x1, y1) and B (x2, y2) on the straight line.
Since these two points lie on the straight line,
From and
The gradient of the straight line
22 For free distribution
By studying this lesson you will be able to
• find the gradient of the graph of a straight line,
• draw the graph of a function of the form y = ax2 + b.
Graph of a function of the form y = mx + c
The graph of a function of the form y = mx + c is a straight line. The coefficient of
x, which is m, represents the gradient of the line and the constant term c represents
the intercept of the graph.
Review Exercise
1. Write down the gradient and intercept of the straight line represented by each of
the following equations.
21.1 Geometrical description of the gradient of a straight line
We defined the coefficient m of x in the equation y = mx + c as the gradient of
the straight line. Now by considering an example, let us see how the value of m
is represented geometrically. To do this, let us consider the straight line given by
y = 2x + 1. Let us use the following table of values to draw its graph.
x –2 0 2
y (= 2x + 1) –3 1 5
20 For free distribution
,Let us mark any three points on the straight line. For example, let us take the three
points as A (0, 1), B (2, 5) and C (5, 11).
y
C
11
10
^5" 11&
9
8
7
6
5 B ^5" 5&
^2" 5& Q
4
3
2
^0" 1& A 1 P ^2" 1&
x
−5 −4 −3 −2 −1 0 1 2 3 4 5
−1
−2
−3
−4
−5
−6
First let us consider the points A and B.
Let us draw a line from A, parallel to the x – axis, and a line from B, parallel to the
y – axis, and name the point of intersection of these lines as P. It is clear that the
coordinates of the point P are (2, 1).
Also, length of AP = 2 – 0
= 2
length of BP = 5 – 1
= 4
Vertical distance BP 4 = 2 .
Now for the points A and B, = =
Horizontal distance AP 2
We already know the gradiant of the straight line y = 2x + 1 is 2.
The quotient Vertical Distance for the point A and B is also 2.
Horizontal distance
Now let us consider another case.
As the second case let us consider the points B and C.
Let us draw a line from B, parallel to the x - axis, and a line from C, parallel to the
y - axis and name the point of intersection of these two lines as Q.
Then the coordinates of Q are (5, 5).
Length of BQ = 5 – 2
= 3
For free distribution 21
, Length of CQ = 11 – 5
=6
Vertical distance CQ 6
Now, for the points B and C, = BQ = 3 = 2 .
Horizontal distance
In both instances, the ratio of the vertical distance to the horizontal distance between
the two points under consideration is the gradient 2 of the straight line.
Accordingly, let us develop a formula to find the gradient of a straight line using its
graph. Let us consider any straight line with equation y = mx + c.
y
7
6 A
5
(x1, y1)
4
3
2
1 B (x2, y2)
−5 −4 −3 −2 −1 0 1 2 3 4 5 x
−1
−2
−3
−4
−5
−6
Let us consider any two points A (x1, y1) and B (x2, y2) on the straight line.
Since these two points lie on the straight line,
From and
The gradient of the straight line
22 For free distribution
