STRAIGHT LINE GRAPHS

Subject: 

MATHEMATICS

Term:

FIRST TERM

Week:

WEEK 9

Class:

SS 2

Topic:

STRAIGHT LINE GRAPHS

Previous lesson: 

The pupils have previous knowledge of

SIMULTANEOUS EQUATIONS INVOLVING ONE LINEAR AND ONE QUADRATIC

that was taught as a topic in the previous lesson

Behavioural objectives:

At the end of the lesson, the learners will be able to

• solve simple word problem questions on simultaneous equation
• Solve Simultaneous Equations Involving One linear and One quadratic.
• Solve Simultaneous Equations Using Graphical Method

Instructional Materials:

• Wall charts
• Pictures
• Related Online Video
• Flash Cards

Methods of Teaching:

• Class Discussion
• Group Discussion
• Asking Questions
• Explanation
• Role Modelling
• Role Delegation

Reference Materials:

• Scheme of Work
• Online Information
• Textbooks
• Workbooks

Content:

WEEK NINE

TOPIC: STRAIGHT LINE GRAPHS

CONTENT

• Gradient of a Straight Line.
• Gradient of a Curve.
• Drawing of Tangents to a Curve.

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 ().

y

M

B

2 units up

A

C

4 units across

x

L

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 =

also in ABC, tan

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 ().

y

p

E

D

3 units down

F

O

Q

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 =

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,

y

B

S

R

W

Q

U

P

T

x

A

We have:

This means the gradient of the line is given by:

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

Q

A(x₁, y₁)

y

P

O

x

B(x₂, y₂)

(x₂, x₁)

(y₂, y₁)

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 =

=

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).

4

D(3. 2)

2

4

3

2

1

-3

-1

-2

8 units

-2

-4

C(-2. -6)

-6

5 units

-8

Draw a straight line to pass through the points.

Gradient =

Method 2

We can calculate the gradient in the following 2 ways.

1. In moving from C to D

(x₁, y₁) = (-2, -6) and (x₂, y₂) = (3, 2)

m =

1. In moving from D to C

(x_1, y₁) = (3, 2) and (x₂, y₂) = (-2, -6)

m =

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 =

Evaluation

Find the gradients of the line joining the following pairs of points.

1. (9,7) , (2,5)

2. (2,5) , (4,5)

3. (2,3) , (6,-5)

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 =

Choose three easy values and then make a table of values as shown below.

When x = 0, y =

When x = 2, y =

When x = 4, y =

The graph of 3x + 2y = 8 is shown below.

(b) Choose two easy points such as P and Q on the line.

2

y

P

4

3

2

1

x

5

-1

3

4

2

1

-1

Q

-2

Gradient of PQ =

Evaluation

Using three convenient points, draw the graph of the following linear equations and then find their gradients.

1. 2x-y-6=0 2.) 5y+4x=20 3.) 3x-2y=9

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 =

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

y

10

T

5

P

x

-1

-3

-2

3

2

1

U

S

Q

R

-5

Gradient =

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

1. A straight line passes through the points (3,k) and (-3,2k). If the gradient of the line is -2/3, find the value of k. What is the equation line?

2. Sketch the following graphs using gradient-intercept method.

a) y= 0.5x – 3 b) y= 5x c) y = x/4 – 3 d) 2y-10 = 2x

3. Find the gradients of the curves at the points indicated.

a) y= 6x – x² at x= 3 b) x² – 6x + 5

WEEKEND ASSIGNMENT

1. Find the gradient of the equation of line 2y – 10 = 2x A. 1 B. 2 C. 3 D. 4

2. Find the gradient of the line joining (7,-2) and (-1,2) A. ½ B. – ½ C. 1/3 D. -1/3

3. Find the equation of a straight line passing through (-3,-5) with gradient 2.

A. y =3x-1 B. y=2x-1C. y=2x-1 D. y=3x+1

Given that 3y-6x +15=0, use the information to answer questions 4 and 5.

4. Find the gradient of the line. A. 5 B. -5 C. 2 D. -2

5. Find the intercept of the line. A. 5 B. -5 C. 2 D. -2

THEORY

1. Draw the graph of y= 2x-3 using convenient points and scale. Hence , find the gradient of the line at any convenient point.

2a) Copy and complete the following table of values for the relation y= 2x² – 7x-3.

b) Using 2cm to 1unit on the x-axis and 2cm to 5units on the y-axis, draw the graph of y= 2x² -7x-3 for -2x≤5.

c) From your graph, find the:

i. minimum value of y.

ii. the equation of the line of symmetry.

iii. the gradient of the curve at x=1.

Reading Assignment

New General Mathematics for SSS2, pages 190-192, exercise 16d.

Conclusion

The class teacher wraps up or concludes the lesson by giving out short notes to summarize the topic that he or she has just taught.

The class teacher also goes round to make sure that the notes are well copied or well written by the pupils.

He or she makes the necessary corrections when and where the needs arise.