REFRESHERLinear Graphs

INTERPRETING STRAIGHT-LINE GRAPHS

Equations and graphs are used to study the relationship between two variables such as distance and speed.

If a relationship exists between the variables, one can be said to be a function of the other. A function can be described by a table, a rule or a graph.

We are going to analyze linear functions.

Standard linear equation

y = mx + c

Where ‘m’ is the gradient and ‘c’ is the y-intercept.

Gradient

m =m =

200

4000.5 500

4001.25

400 m

500 m

200 m

400 m

The gradient (or slope) is the steepness of a line. The gradient gives us information about how

much one variable changes compared to another. The steeper the line the greater the gradient Gradient is simply the change in the vertical distance

(rise) over the change in the horizontal distance (run).

Gradient = m = Rise

Run

rise

rise

runrun

Gradient

Gradient = m = Rise

Run

run

rise

8

4

m = 8 4 = 2

The graph of y = 2x - 4

Gradient

Gradient = m = Rise

Run

run

rise

4

2

m = 4 2 = 2

The graph of y = 2x - 4

The gradient is the same at any point along a straight line

o If a line goes up from left to right,

then the slope has to be positive

o If a line goes down from left to right,

then the slope has to be negative

o Lines that are horizontal have zero slope.

o Vertical lines have no slope, or undefined

slope.

Gradient: Positive or negative?

Gradient: Positive or Negative?

Positive gradiento x- value increases o y- value increases

Negative gradiento x- value increases o y- value decreases

o Since the rise is simply the change in the vertical distance and the run is the change in the horizontal distance. The rise and run can be found by calculating the change between two points on a line.

12

12

xx

yy

x

y

run

risemslope

Gradient: Definition

oIn order to use this formula we need to know, or be able to find 2 points on the line. Eg (0,-4) & (2,0)

x1, y1 x2, y2

Gradient

m = 4- -4 4-0 = 8 4 = 2

The graph of y = 2x - 4

12

12

xx

yym

1. Choose two points on the graph

2. Substitute the values and calculate

(4,4)

(0,-4)

(0,-4) (4,4) x1, y1 x2, y2

Exercise 6.2 Q1, Q2, Q3, Q4, Q5, Q6 RHS

Gradient

m = 0 - 4 2 - 0 = -4 2 = -2

The graph of y = -2x + 4

12

12

xx

yym

1. Choose two points on the graph

2. Substitute the values and calculate

(2,0)

(0, 4)

(0,4) (2,0) x1, y1 x2, y2

Exercise 6.2 Q1, Q2, Q3, Q4, Q5, Q6 RHS

Intercepts The x-intercept occurs where the line cuts

the x-axis. At the x- intercept the y-value always equals to

0.

The y- intercept occurs where the line cuts

the y-axis. At the y-intercept the x-value is always equal to 0.

x-intercept

y-intercept

(2,0)

(0,-4)

Brain Storm!1. Explain how you could find the y-intercept by:

a) Looking at a graph

b) Looking at an equationy = 2x +3

c) Looking at a table of values

X 0 2 4

Y 4 8 12

Find the coordinates for the point where the line passes through the x-axis (the x-intercept), and for the point where the line passes through the y axis (the y-intercept)

Substitute y = 0 into the equation to find the x value for the x-intercept.Substitute x = 0 into the equation to find the y-value for the y-intercept.

Carefully plot the points and look at the graph. OR... Work out the rule used and use the method listed above under looking at an equation Rule: y = 2x +4

Find the Intercepts:

1. Find the x and y intercepts of the following graphs:a. b.

c. d.

x-intercept(2,0)

x-intercept(2,0)

x-intercept(0,0)

x-intercept(4,0)

y-intercept(0,-4)

y-intercept(0, 4)

y-intercept(0,0)

y-intercept(0, 4)

Find the interceptsExample 1. Find the x and y intercepts of thefollowing equation: y = 3 x + 2

x int: y = 00 = 3 x + 20-2 = 3 x +2 - 2-2 = 3 x 3 3X = - 2 3 x – intercept = -2/3 i.e. The line cuts the x –axis at (-2/3, 0)

y int: x = 0 y = 3 (0) + 2y = 0 + 2y = 2y-intercept = 2 i.e. the line cuts the y axis at (0,2)

Find the interceptsExample 2. Find the x and y intercepts of thefollowing equation: y = -2x + 1

x int: y = 00 = -2x + 10-1 = -2x + 1 - 1-1 = -2x -2 -2X = 1 2 x – intercept = ½ i.e. The line cuts the x –axis at ( ½ , 0)

y int: x = 0 y = -2(0) + 1y = 0 + 1y = 1y-intercept = 1 i.e. the line cuts the y axis at (0,1)

Find the intercepts

Example 1. Find the x and y intercepts of thefollowing equation: 2x + 3y = 6

x int: y = 02x + 3(0) = 62x + 0 = 62x = 6 2 2x = 6 2 x = 3x – intercept = 3 i.e. The line cuts the x –axis at (3, 0)

y int: x = 0 2 (0) + 3y = 60 + 3y = 63y = 6• 3y = 2y-intercept = 2 i.e. the line cuts the y axis at (0,2)

REMEMBER!! The x-intercept occurs where the line cuts

the x-axis. At the x- intercept the y-value always equals to

0.

The y- intercept occurs where the line cuts

the y-axis. At the y-intercept the x-value is always equal to 0.

x-intercept

y-intercept

(2,0)

(0,-4)

Sketching linear graphs using the x- and y-intercepts

State the equation Find the x- intercept (substitute 0 for y and

solve for x) Find the y- intercept (substitute 0 for x and

solve for y) Mark the x intercept and y intercept and rule

a straight line through them.

Sketch the linear equationExample 1. Find the x and y intercepts of the

following equation and sketch the line. y = 2 x + 4

x int: y = 00 = 2 x + 40-4 = 2 x +4 - 4-4 = 2 x 2 2X = - 4 2 X = -2x – intercept = -2 i.e. The line cuts the x –axis at (-2, 0)

y int: x = 0 y = 2 (0) + 4y = 0 + 4y = 4y-intercept = 4 i.e. the line cuts the y axis at (0,4)

STEP 1: FIND THE INTERCEPTS...

STEP 2: PLOT THE COORDINATES AND RULE A STRAIGHT LINE BETWEEN THEM...

Sketch the linear equationExample 1. Find the x and y intercepts of the

following equation and sketch the line. y = 2 x + 4

X int

(-2,0)

Y int

(0,4) x

y

y = 2 x + 4