Lesson Notes By Weeks and Term - Senior Secondary School 2
STRAIGHT LINE GRAPHS
STRAIGHT LINE GRAPHS
SUBJECT: MATHEMATICS
CLASS: SS 2
DATE:
TERM: 1st TERM
REFERENCE BOOKS
WEEK NINE
TOPIC: STRAIGHT LINE GRAPHS
CONTENT
GRADIENT OF A STRAIGHT LINE
The gradient (or slope) of a straight line is a measure of the steepness of the line.
The gradient of a line may be positive or negative.
Positive gradient (uphill slope)
Consider line LM shown in the diagram below. The line slopes upwardsto the right and it makes an acute angle of with the positive x-axis, so tan is positive. The gradient of the line can be found by choosing any two convenient points such as A and B on the line. In moving from A to B, x increases () and y also incrases ().
i.e. increase in x = horizontal distance = AC
increase in y = vertical distance = BC
the gradient of a line is represented by letter m.
the gradient of a line LM is given by:
m = increase in y from A to B increase in x from A to B= BCAC= 24= 12
also in ABC, tan = BCAC= 24= 12
it follows that the gradient of line AB = tan . When a line slopes upwards (uphill) to the right, the gradient of the line is positive.
Negative gradient (downhill slope)
In the diagram below, line PQ slopes downwards and it makes an obtuse angle with positive x-axis, so tan is negative. Again, to find the gradient of the line, we choose two convenient points such as D and F on the line. In moving from D to F, x increase () and y decreases ().
i.e. increase in x = horizontal distance = DE and decrease in y = vertical distance = EF.
The gradient, m of line PQ is given by:
m = increase in y from D to Fincrease in x from D to F= EFDE= -34= -34
Also the gradient of line PQ = tan
When a line slopes downwards to the right (i.e. downhill) the gradient is negative.
For example, in the diagram below, the slope goes up 3 units for every 4 units across. Since triangles PQT, QRU and RSW are similar,
We have: QTPT= RUQU= SWRW= 34
This means the gradient of the line is given by:
m=QTPT or m= RUQU or m= SWRW
Where the letter ‘m’ represents gradient.
Calculating the Gradient of a Line
The gradient of a straight line can be calculated from any two points on the line.
Let the two points on line PQ be A and B. if the coordinates of point A are (x1, y 1) and the coordinates
Gradients of lines and curves
of point B are (x2, y2), then in moving from A to B, the increase in x (or change in x) is AC and the increase in y (or change in y) is CB, i.e. AC = x2 – x1 and CB = y2 – y1,
Thus, the gradient, m of the line PQ is given by:
m = increase in y increase in x = difference in y coordinate difference in x coordinate
= CBAC= y2-y1x2-x1
Exercise
Calculate the gradient of the line joining the points C(-2, -6) and D(3, 2) and.
Solution
Method 1
Plot the points C(-2, -6) and D(3, 2).
Draw a straight line to pass through the points.
Gradient = increase in yincrease in x= EDCE= 85
Method 2
We can calculate the gradient in the following 2 ways.
(x1, y1) = (-2, -6) and (x2, y2) = (3, 2)
m = y2-y1x2-x1= 2-(-6)3-(-2)= 2+63+2= 85
(x1, y1) = (3, 2) and (x2, y2) = (-2, -6)
m = y2-y1x2-x1= -6-2-2-3= -8-5= 85
Notice that the answer is the same in obht cases, therefore, it does not matter which point we call the first or the second.
Example
Find the gradient of the line joining (-4, 6) and (3, 0)
Solution
Let m = gradient,
(x1, y1) = (-4, 6) and (x2, y2) = (3 , 0)
m = y2-y1x2-x1= 0-63-(-4)= -63+4= -67= -67
Evaluation
Find the gradients of the line joining the following pairs of points.
Drawing the Graphs of Straight Lines
Example
(a) Draw the graph of 3x + 2y = 8
(b) Find the gradient of the line.
Solution
(a) First make y the subject.
3x + 2y = 8
2y = 8 – 3x
y = 8-3x2
Choose three easy values and then make a table of values as shown below.
When x = 0, y = 8-02=4
When x = 2, y = 8-62= 22=1
When x = 4, y =8-122= -42= -2
x | 0 | 2 | 4 |
y | 4 | 1 | -2 |
The graph of 3x + 2y = 8 is shown below.
(b) Choose two easy points such as P and Q on the line.
Gradient of PQ = increase in yincrease in x= -RQPR
-64= -32
Evaluation
Using three convenient points, draw the graph of the following linear equations and then find their gradients.
GRADIENT OF A CURVE
Finding the gradient of a straight line is constant at any point on the line. However, the gradient of a curve changes continuously as we move along the curve. In the diagram below, the gradient at P is not equal to the gradient at S. to find the gradient of a curve, draw a tangent to the curve, draw a tangent to the curve at the point your require to find the gradient. For example, the gradient of curve at point P is the same as the gradient of the tangent PQ. Also the gradient of the curve at S is the same as the gradient of the tangent ST.
The diagram above represents the graph of the function y =2x2 + x – 5.
The gradients at P and S can be found as follows:
Gradient at P = gradient of tangent PQ. By constructing a suitable right-angled triangle with hypotenuse PQ, the gradient is Gradient = PRPQ= -71= -7
Remember that the gradient is negative because the tangent slopes downwards from left to right.
Gradient at S = gradient of tangent ST.
By constructing a suitable right-angled triangle with hypotenuse ST, the gradient is
Gradient = TUSU= 102= 5
Remember that the gradient is positive because the tangent slopes upwards from left to right.
Note: This method only gives approximate answer. However, the more accurate your graphs are, the more accurate your answers will be.
Evaluation
Draw the graphs of the following functions and use the graphs to find the gradients at indicated points.
1) y= x2 –x-2 at x= -1
2) y= x2-3x-4=0 at x = 4
GENERAL EVALUATION/ REVISION QUESTIONS
WEEKEND ASSIGNMENT
Given that 3y-6x +15=0, use the information to answer questions 4 and 5.
THEORY
2a) Copy and complete the following table of values for the relation y= 2x2 – 7x-3.
X | -2 | 1- | 0 | 1 | 2 | 3 | 4 | 5 |
Y | 19 | -3 | -9 |
iii. the gradient of the curve at x=1.
Reading Assignment
New General Mathematics for SSS2, pages 190-192, exercise 16d.
© Lesson Notes All Rights Reserved 2023