Module 3: Constructing and interpreting linear graphs

Chapter 16

The gradient of a straight line We already know from the core unit that the

equation of a straight line can be stated by finding the gradient and the vertical intercept.

The gradient measures the slope of the line. The value of the gradient is the ratio between

the rise and the run. The rise is the change in the vertical values

between any two points. The run is the change in the horizontal values

between the same two points.

x

y

8 6 4 2 0 2 4 6 8

4

4

rise

run =

84

= 2

8

rise = 8

4run = 4

Gradient =

Example

x

y

0

rise

run =

y2 y1x2 x1

y2 y1

rise = y2 y1

x2 x1run = x2 x1

Gradient =

(x1, y1)

(x2, y2)

Formula to calculate the gradient from two points

x

y

8 6 4 2 0 2 4 6 8

4

4

To calculate the gradient of

the line, choose any two

points on the line.

4 (4)4 0

=84

= 2

Gradient =

e.g., (x1, y1) = (0, 4)

x2 x1

y2 y1

=

4 (4)

4 0

and (x2, y2) = (4, 4)

(4, 4)

(0, 4)

Example

x

y

8 6 4 2 0 2 4 6 8

4

4

To calculate the gradient of the line, choose any two points on the line.

4 44 0

=8

4= 2

Gradient =

e.g., (x1, y1) = (0, 4)

x2 x1

y2 y1

=4

and (x2, y2) = (4, 4)

(4, 4)

(0, 4)

8

Example

Questions

Exercise 16A Pages 402 – 403 All

x

y

0

(x2, y2)

(x1, y1)

y values increase

x

y(x2, y2)

(x1, y1)

0

Positive gradient:

as x values increase

y values decrease

Negative gradient:

as x values increase

Positive and negative gradients

The general equation of a straight line

The general equation of a straight line is y = mx + c, where m is the gradient of the line and c is equal to the y-axis intercept.

This form, expressing the relation in terms of y, is called the gradient form.

Example

Find the gradient and y-axis intercept of the graph of

y = 3x − 4. Gradient = 3 Y intercept = -4 y = -2x + 4. y = 6x − 5. 2y = 8x + 6

Sketching the equation of a straight line If we are given the rule of a straight

line, we can sketch the graph using the gradient and the y-axis intercept.

1. Mark in the y intercept first2. Draw the gradient from this point

Sketch the graph of y = 3x +1 Sketch the graph of 3y + 6x = 9 Sketch the graph of 3x – y = 6

Parallel Lines

Lines that have the same gradient will be parallel.

Questions

Exercise 16B Pages 405 All

Finding the equation of a straight line given the gradient and the y intercept y = mx + c A line cuts the y axis at 3 and has a

gradient of -2. What is its equation. y = -2x + 3

Find the equation of the line that passes through the point ( 3, 2 ) and has a gradient of -4.

y = mx + c 2 = -4 * 3 + c 2 = -12 + c c = 14 y = -4x + 14

Finding the equation of a straight line given the gradient and a point.

Finding the equation of a straight line given two points

Find the equation of the straight line passing through the points (1, -2) and (3, 2)

First, find the gradientrise

run =

y2 y1Gradient =

x2 x1

Gradient = 2 - -2

3 - 1

Gradient =4

2

Gradient = 2

Finding the equation of a straight line given two points Then use the gradient and one of the

points to find the equation m = 2 and ( 3 , 2 ) y = mx + c 2 = 2 * 3 + c 2 = 6 + c c = -4 y = 2x – 4

Finding the equation of a straight line given the graph

Graphs of vertical and horizontal lines

All horizontal lines will have an equation of the form y = c where c is the vertical intercept.

All vertical lines will have an equation of the form x = k where k is the horizontal intercept

Questions

Exercise 16C Page 408 - 409 All

The intercept form of a linear equation

Not all equations will be given to you the form y = mx + c

An alternative notation is, ax + by = c where a, b & c are constants.

The best way to sketch an equation when it is given in this form is to find the x and y intercepts.

Example

Sketch the equation of 3x – 2y = 6

To find the x intercept, let y = 0

3x – 2 * 0 = 6

3x = 6

x = 2

To find the y intercept, let x = 0

3 * 0 – 2y = 6

- 2y = 6

y = -3

Exercises

Exercise 16D Page 410 Question 1

Linear Models

Example

Construct a rule for the Cost of hiring the taxi. C = 3.20 + 1.60k How much will a journey of 10km cost? $19.20 How far can you travel with $30? 16.75km Sketch the graph of the relationship between cost

and the number of kilometres travelled for journeys from 0 to 20 kilometres.

What does the vertical intercept represent? What does the gradient represent?

Exercise

Exercise 16E Pages 411 – 412 All

Solving Simultaneous Equations bythe graphical method Simultaneous equations involves

finding the point of intersection of two linear equations at the same time

Solve the following two equations simultaneously

y = 3x – 4 y = -2x + 1

Solving Simultaneous Equations by the graphical method

Solving Simultaneous Equations by the graphical method

Solving Simultaneous Equations by the graphical method

Solving Simultaneous Equations by the graphical method

Solving Simultaneous Equations bythe graphical method Solve graphically the simultaneous

equations x – y = 5 and x = 2

Solving Simultaneous Equations using the calculator

TI 89

Solve(y = 3x - 4 and y = -2x + 1 ,{x,y})

Using a Calculator to solve Simultaneous Equations

x + y = 3 2x – y = -9

Calculator

x + y = 3 2x – y = -9

Calculator

x + y = 3 2x – y = -9

Calculator

2x - 5y = 11 5x + 3y = 12

Questions

Exercise 16F Page 416, as many as you need.

Practical applications of simultaneous equations

The perimeter of a rectangle is 48 cm. If the length of the rectangle is three times the width, determine its dimensions.

Let w represent the width. Let l represent the length 2l + 2w = 48 l = 3w Use the substitution method to solve.

Practical applications of simultaneous equations

Two families went to the theatre. The first family bought tickets for 3 adults and 5

children and paid $73.50 The second family bought tickets for 2 adults and 3

children and paid $46.50 Let a represent the cost of an adult ticket Let c represent the cost of a child ticket. Express these statements as equations. 3 a + 5 c = 73.50 2 a + 3 c = 46.50 Calculator

Questions

Exercise 16G Page 417-418 all

Break Even Analysis

You decide to go into business making calculators.

Each week you find that there are always fixed costs that are independent of the cost of making the actual calculator i.e. wages, rent, insurance, utilities, etc.

There are also ongoing costs to buy the parts required for calculator.

Break even analysis

You find that the fixed costs of producing the calculator is $1500 and the cost for the parts for each calculator is $30.

Represent this relationship as an equation where ‘C’ represents the total cost per week and ‘n’ represents the number of calculators made in that week.

C = 1500 + 30n

Break even analysis

The calculators are sold which provides revenue for your business.

You decide to sell the calculators for $150. Represent this relationship as an equation

where ‘R’ represents the revenue and ‘n’ represents the number of calculators sold in that week.

R = 150n

Break even analysis

The break even point is when the revenue earned is equal to the costs incurred.

R = C Before the break even point your

business will be making a loss. After the break even point your

business will be making a profit

Break even analysis

The break even point can be found algebraically or graphically.

Algebraically

Revenue = Cost 150n = 30n + 1500 120n = 1500 n = 12.5 The Revenue and Costs at the breakeven

point will both be $1875 The business will begin to make a profit

when you sell 13 calculators.

Graphically

Graphically C = 1500 + 30n

Graphically R = 150n

The Profit Function

The profit made is equal to the Revenue earned less the costs incurred.

Profit = Revenue – Costs Profit = 150n – ( 30n + 1500) Profit = 150n – 30n – 1500 Profit = 120n – 1500 Calculate the profit made when you sell 35

calculators. Profit = 120 * 35 – 1500 Profit = $2700

Questions

Exercise 16H Page 419 All