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Year 9

5.05 Gradient-intercept form

Lesson

Introduction

Let's have a quick recap of what we have learnt about straight lines on the xy-plane so far.

  • They have a gradient (slope)- a measure of how steep the line is.

  • They can be increasing (positive gradient) or decreasing (negative gradient).

  • They can be horizontal (zero gradient).

  • They can be vertical (gradient is undefined).

  • They have x-intercepts, y-intercepts or both an x and a y-intercept.

  • The gradient can be calculated using \dfrac{\text{rise}}{\text{run}} or \dfrac{y_2-y_1}{x_2-x_1}.

  • They have an equation of the form y=mx+c.

The gradient-intercept formula

The values of m and c have specific meanings.

Exploration

Explore for yourself what these values do by exploring on this interactive. Move the sliders and notice how m affects the gradient and c affects the y-intercept of the line.

Loading interactive...

What you will have found is that the value of m affects the gradient, and that the value of c affects the y-intercept.

For the gradient:

  • If m<0, the gradient is negative and the line is decreasing.

  • If m>0, the gradient is positive and the line is increasing.

  • If m=0, the gradient is 0 and the line is horizontal.

  • The larger the magnitude of m the steeper the line.

For the y-intercept:

  • If c is positive then the line cross the y-axis above the origin.

  • If c is negative then the line cross the y-axis below the origin.

An equation of the form y=mx+c is known as the gradient-intercept form of a line, as we can easily identify both the gradient, m, and the value of the y-intercept, c.

Examples

Example 1

Find the equation of a line whose gradient is -8 and crosses the y-axis at -9. Express your answer in the form y=mx+c.

Worked Solution
Create a strategy

Substitute the gradient for m and the value of the y-intercept for c.

Apply the idea

From the information we have m=-8 and c=-9. So the equation is: y=-8x-9

Idea summary

For all equations of the form y=mx+c:

  • The value of m is the gradient.

    • This means that as x increases by 1, the y-value increases by m (or decreases if m is negative).

  • The value of c is the y-intercept.

    • This means the line passes through the point (0,c).

Find the equation of a line

Sometimes we are given the graph of a line, and we are asked to find the equation of the line.

The first thing we want to do is find the  gradient of the line  , which we can do by using any two points (usually the intercepts). Using the coordinates of the two points, we can either use the gradient formula, or by calculating the rise and the run.

We can then identify the y-intercept, by looking where the line crosses the y-axis. Once we have identified these two features, we can write the equation of the line in the form y=mx+c.

Examples

Example 2

Consider the line shown on the coordinate-plane:

-4
-3
-2
-1
1
2
3
4
x
-1
1
2
3
4
5
6
7
y
a

Complete the table of values.

x-1012
y
Worked Solution
Create a strategy

Use the x-values given in the table and find the corresponding points on the line graph.

Apply the idea
-3
-2
-1
1
2
3
x
-3
-2
-1
1
2
3
4
5
6
7
8
y

The coordinates of the points are (-1,7), (0,4), (1,1), and (2,-2).

x-1012
y741-2
b

Linear relations can be written in the form y=mx+c. For this relationship, state the values of m and c.

Worked Solution
Create a strategy

Use the values from the table in part (a) where:

  • m is equal to the change in the y-values for every increase in the x-value by 1, and

  • c is the value of y when x=0.

Apply the idea
x-1012
y741-2

We can see that the y-values are decreasing by 3 for every increase in x-value by 1. So m=-3.

The value of y is 4 when x=0. So c=4.

c

Write the linear equation expressing the relationship between x and y.

Worked Solution
Create a strategy

Substitute the values found in part (b) into the equation y=mx+c.

Apply the idea

y=-3x+4

d

Find the coordinates for the point on the line where x=20.

Worked Solution
Create a strategy

Substitute the given value of x into the equation found in part (c).

Apply the idea
\displaystyle y\displaystyle =\displaystyle -3\times(20)+4Substitute x=20
\displaystyle =\displaystyle -60+4Evaluate the multiplication
\displaystyle =\displaystyle -56Evaluate

The coordinates are (20, -56).

Example 3

Consider the line shown on the coordinate-plane:

-2
-1
1
2
x
-2
-1
1
2
y
a

State the value of the y-intercept.

Worked Solution
Create a strategy

The y-intercept is the point where the line intersects the y-axis, when x=0.

Apply the idea
-2
-1
1
2
x
-2
-1
1
2
y

The line intersects at (0,-1) in the y-axis, so the y-intercept is y=-1.

b

By how much does the y-value change as the x-value increases by 1?

Worked Solution
Create a strategy

Move 1 unit to the right from the y-intercept, then count spaces need to move up to the next point on the line.

Apply the idea
-2
-1
1
2
x
-2
-1
1
2
y

From the y-intercept moving 1 unit to the right and \dfrac12 unit up gets us to the next point on the line.

So, the increase of y-value is \dfrac12.

c

Write the linear equation expressing the relationship between x and y.

Worked Solution
Create a strategy

Use the values found in part (a) and (b) to substitute to the equation y=mx+c, where:

  • m is the increase in y-values when x-value increases by 1, and

  • c is the y-intercept.

Apply the idea

We found from part (b) that the increase in the y-values or the m is \dfrac12.

In part (a), we found that the y-intercept or c is -1.

So, the linear equation is y=\dfrac12 x -1 or to simplify, y=\dfrac{x}{2}-1.

Idea summary

We can find the equation of a line of the form y=mx+c by finding the gradient for the value of m, and the point of y-intercept for the value of c.

Outcomes

ACMNA214

Find the distance between two points located on the Cartesian plane using a range of strategies, including graphing software

ACMNA294

Find the midpoint and gradient of a line segment (interval) on the Cartesian plane using a range of strategies, including graphing software

ACMNA215

Sketch linear graphs using the coordinates of two points and solve linear equations

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