**SUBJECT: MATHEMATICS**

**CLASS: SS 3**

**DATE:**

**TERM: 1st TERM**

**REFERENCE TEXTS: **

- New General Mathematics for SS book 3 by J.B Channon
- Essential Mathematics for SS book 3
- Mathematics Exam Focus
- Waec and Jamb past Questions

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 ().

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.

- In moving from C to D

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

m = y2-y1x2-x1= 2-(-6)3-(-2)= 2+63+2= 85

- In moving from D to C

(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.

- (9,7) , (2,5)
- (2,5) , (4,5)
- (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 = 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.

- 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 = 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**

- 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?
- 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
- Find the gradients of the curves at the points indicated.
- a) y= 6x - x2 at x= 3 b) x2 – 6x + 5

**WEEKEND ASSIGNMENT**

- Find the gradient of the equation of line 2y – 10 = 2x A. 1 B. 2 C. 3 D. 4
- Find the gradient of the line joining (7,-2) and (-1,2) A. ½ B. – ½ C. 1/3 D. -1/3
- Find the equation of a straight line passing through (-3,-5) with gradient 2.
- 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.

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

**THEORY**

- 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.

X | -2 | 1- | 0 | 1 | 2 | 3 | 4 | 5 |

Y | 19 | -3 | -9 |

- 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:
- minimum value of y.
- 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.

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