Characteristics of Linear Graphs Revision Notes for HSC SSCE Mathematics Advanced

Characteristics of Linear Graphs

Introduction

A linear function is a function that can be expressed in the form $y = mx + c$ , where $m$ and $c$ are constants. The graph of a linear function is always a straight line. Linear functions are fundamental in mathematics because they model constant rates of change and are used extensively in real-world applications.

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Linear functions appear everywhere in real life - from calculating costs based on quantity, to understanding speed and distance relationships, to predicting trends over time. Mastering linear graphs provides a foundation for understanding many practical situations.

In this topic, you will learn to:

Identify the gradient and y-intercept from an equation in gradient-intercept form
Determine the x-intercept and y-intercept of a line from its equation
Graph a straight line using its gradient and y-intercept
Graph a straight line using its intercepts
Convert linear equations between gradient-intercept form and general form

The gradient

The gradient (also called slope) is a measure of the steepness of a line. It represents the rate of change in the vertical direction relative to the horizontal direction.

If $A(x_1, y_1)$ and $B(x_2, y_2)$ are two distinct points on a line, the gradient of the line (or line segment $AB$ ) is calculated using the formula:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

This formula tells us how much $y$ changes for every unit change in $x$ .

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Understanding Gradient Direction:

A positive gradient means the line slopes upward from left to right
A negative gradient means the line slopes downward from left to right
The larger the absolute value of the gradient, the steeper the line

Gradient-intercept form

The gradient-intercept form is a way of writing the equation of a straight line that makes it easy to identify both its steepness and where it crosses the y-axis.

Definition

chatImportant

The gradient-intercept form of a straight line is:

$y = mx + c$

where:

$m$ is the gradient of the line
$c$ is the y-intercept (the value of $y$ when $x = 0$ )
$(0, c)$ is the point where the line intersects the y-axis

This form is particularly useful for describing and graphing linear relationships because both key features of the line are immediately visible.

Finding the equation from gradient and y-intercept

When you know the gradient and y-intercept, you can write the equation directly by substituting these values into $y = mx + c$ .

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Worked Example 1: A straight line has gradient $3$ and y-intercept $4$ . Determine its equation in gradient-intercept form.

Solution:

Use the gradient-intercept form $y = mx + c$ , then identify the gradient $m$ and the y-intercept $c$ , and substitute the given values.

$y = mx + c$

$y = 3x + 4$

Substitute $m = 3$ and $c = 4$ .

Finding the equation from gradient and a point

When you know the gradient and one point on the line, you can find the y-intercept by substituting the point's coordinates into the equation and solving for $c$ .

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Worked Example 2: A straight line passes through the point $(2, 3)$ with a gradient of $-\frac{1}{2}$ . Determine its equation in gradient-intercept form.

Solution:

Substitute the coordinates of the point $(2, 3)$ and the value of the gradient $m$ into the gradient-intercept form $y = mx + c$ to determine the y-intercept $c$ .

Substituting the values of $m$ and $c$ , the equation is:

$y = mx + c$

$y = -\frac{1}{2}x + 4$

Verification:

To verify that the equation $y = -\frac{1}{2}x + 4$ is correct, substitute $x = 2$ and $y = 3$ into the equation:

$y = -\frac{1}{2}x + 4$

$3 = -\frac{1}{2} \times 2 + 4$

$3 = -1 + 4$

$3 = 3$

The left-hand side equals the right-hand side, confirming the point $(2, 3)$ lies on the line.

Finding the equation from two points

When you are given two points on the line, first calculate the gradient using the gradient formula, then use one of the points to find the y-intercept.

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Worked Example 3: Determine the equation of the straight line that passes through the points $(2, 3)$ and $(-1, 6)$ in gradient-intercept form $y = mx + c$ .

Solution:

First, calculate the gradient using the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$ , then use one point to determine the y-intercept $c$ .

Calculating the gradient:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{6 - 3}{-1 - 2}$

Substitute $(x_1, y_1) = (2, 3)$ and $(x_2, y_2) = (-1, 6)$ .

$m = \frac{3}{-3}$

Evaluate the numerator and denominator.

$m = -1$

Using the point $(2, 3)$ to determine the y-intercept $c$ :

Substituting back $m = -1$ and $c = 5$ :

$y = mx + c$

$y = (-1) \times x + 5$

$y = -x + 5$

The equation is $y = -x + 5$ .

Graphing using gradient-intercept form

The gradient-intercept form $y = mx + c$ is particularly useful for graphing because both key features of the line are immediately identifiable.

Graphing process

To graph a straight line using gradient-intercept form:

Identify and plot the y-intercept at the point $(0, c)$ , where the line crosses the y-axis
Use the gradient to find a second point on the line
Draw a straight line through both points, extending it in both directions

Interpreting the gradient

The gradient $m$ can be interpreted as a ratio of rise to run. This means:

The numerator represents the vertical change (rise)
The denominator represents the horizontal change (run)

For example, a gradient of $m = \frac{2}{1} = 2$ means move $1$ unit right and $2$ units up from any point on the line to locate another point.

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Working with Negative Gradients:

When the gradient is negative, the direction of movement changes. For instance, a gradient of $m = -\frac{2}{3}$ can be interpreted as:

Moving $3$ units to the right and $2$ units down, OR
Moving $3$ units to the left and $2$ units up

Either interpretation correctly positions a second point, as long as the ratio between rise and run is maintained.

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Worked Example 4: Graph the straight line represented by the equation $y = 2x - 3$ .

Solution:

Identify and plot the y-intercept, then use the gradient to determine a second point. Draw a straight line through the two points to complete the graph.

The y-intercept is $c = -3$ , so the point $(0, -3)$ should be plotted first.

The gradient is $m = 2 = \frac{2}{1}$ . From $(0, -3)$ , move $1$ unit right and $2$ units up to locate a second point at $(1, -1)$ .

Draw a line through the points $(0, -3)$ and $(1, -1)$ .

General form

The general form is an alternative way to express the equation of a straight line.

Definition

chatImportant

A straight line in general form is given as:

$ax + by + c = 0$

where $a$ , $b$ and $c$ are constants.

The general form does not immediately display the gradient or y-intercept, but it is suitable for graphing, particularly when the coefficients are integers and the intercepts are easy to identify.

Finding intercepts from general form

To graph a straight line from general form, you typically find the intercepts first.

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Finding Intercepts - Quick Method:

The x-intercept is found by setting $y = 0$ and solving for $x$ . This gives the point where the line crosses the x-axis.
The y-intercept is found by setting $x = 0$ and solving for $y$ . This gives the point where the line crosses the y-axis.

Once both intercepts are determined, plot the corresponding points on the Cartesian plane and draw a straight line through them, extending in both directions.

Converting between forms

Linear equations can be converted between gradient-intercept form and general form using algebraic manipulation.

Converting from gradient-intercept to general form:

To convert $y = mx + c$ to general form, rearrange all terms to one side of the equation so that it equals zero, ensuring the coefficient of $x$ is positive.

Converting from general form to gradient-intercept form:

To convert from general form to gradient-intercept form, rearrange the equation to make $y$ the subject.

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Worked Example 5: Consider the straight line equation $3x + 4y - 12 = 0$ .

a) Determine the x-intercept and y-intercept.

Solution:

Set $y = 0$ to determine the x-intercept, and $x = 0$ to determine the y-intercept.

To determine the x-intercept, substitute $y = 0$ :

$3x + 4y - 12 = 0$

$3x + 4 \times 0 - 12 = 0$

$3x - 12 = 0$

$3x = 12$

$x = 4$

Therefore, the x-intercept is at $(4, 0)$ .

To determine the y-intercept, substitute $x = 0$ :

3x + 4y - 12 = 0 \quad \text{Write the equation} \\ 0 + 4y - 12 = 0 \quad \text{Substitute } x = 0 \\ 4y - 12 = 0 \quad \text{Evaluate the multiplication} \\ 4y = 12 \quad \text{Add 12 to both sides} \\ y = 3 \quad \text{Divide both sides by 4}

Therefore, the y-intercept is at $(0, 3)$ .

Verification:

To verify, convert to gradient-intercept form:

3x + 4y - 12 = 0 \quad \text{Write the equation} \\ 4y = -3x + 12 \quad \text{Subtract } 3x - 12 \text{ from both sides} \\ y = -\frac{3}{4}x + 3 \quad \text{Divide both sides by 4}

The y-intercept is $(0, 3)$ and setting $y = 0$ gives $x = 4$ . This confirms the intercepts are correct.

b) Graph the straight line represented by the equation.

Solution:

Plot the x-intercept and y-intercept from part (a), then draw a straight line through the two points.

Plot the points $(4, 0)$ and $(0, 3)$ , and then draw a straight line through them.

Alternatively, you can use the gradient-intercept form found above to plot the y-intercept at $(0, 3)$ , then use the gradient $m = -\frac{3}{4}$ to find another point:

From the y-intercept at $(0, 3)$ , move $4$ units right and $3$ units down to locate and plot another point at $(4, 0)$ .

Summary

Linear functions have graphs that are always straight lines. The gradient-intercept form $y = mx + c$ immediately shows the gradient $m$ and y-intercept $c$ , making it useful for both interpreting and graphing straight-line equations.

The gradient represents the rate of change in $y$ with respect to $x$ . It can be calculated from two points using $m = \frac{y_2 - y_1}{x_2 - x_1}$ and interpreted as rise over run when graphing.

To graph using gradient-intercept form, plot the y-intercept at $(0, c)$ first, then use the gradient as a ratio to find a second point on the line. A straight line is drawn through both points.

The general form $ax + by + c = 0$ is an alternative way to express linear equations. To graph from general form, find the intercepts by setting $y = 0$ (for x-intercept) and $x = 0$ (for y-intercept), then plot these points and draw a straight line through them.

Linear equations can be converted between gradient-intercept form and general form using algebraic manipulation.

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Key Points to Remember:

The gradient-intercept form is $y = mx + c$ , where $m$ is the gradient and $c$ is the y-intercept
The gradient represents the rate of change and can be interpreted as rise over run
To find an equation from two points: calculate the gradient first, then substitute one point to find the y-intercept
To graph from gradient-intercept form: plot the y-intercept, then use the gradient to find a second point
To graph from general form $ax + by + c = 0$ : find both intercepts by setting $y = 0$ and $x = 0$ , then draw a line through them
You can convert between gradient-intercept form and general form by rearranging the equation algebraically

Characteristics of Linear Graphs (HSC SSCE Mathematics Advanced): Revision Notes