# 1.1 The straight line

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## Functions of the form $y=ax+q$

Functions with a general form of $y=ax+q$ are called straight line functions. In the equation, $y=ax+q$ , $a$ and $q$ are constants and have different effects on the graph of the function. The general shape of the graph of functions of this form is shown in [link] for the function $f\left(x\right)=2x+3$ .

## Investigation : functions of the form $y=ax+q$

1. On the same set of axes, plot the following graphs:
1. $a\left(x\right)=x-2$
2. $b\left(x\right)=x-1$
3. $c\left(x\right)=x$
4. $d\left(x\right)=x+1$
5. $e\left(x\right)=x+2$
Use your results to deduce the effect of different values of $q$ on the resulting graph.
2. On the same set of axes, plot the following graphs:
1. $f\left(x\right)=-2·x$
2. $g\left(x\right)=-1·x$
3. $h\left(x\right)=0·x$
4. $j\left(x\right)=1·x$
5. $k\left(x\right)=2·x$
Use your results to deduce the effect of different values of $a$ on the resulting graph.

You may have noticed that the value of $a$ affects the slope of the graph. As $a$ increases, the slope of the graph increases. If $a>0$ then the graph increases from left to right (slopes upwards). If $a<0$ then the graph increases from right to left (slopes downwards). For this reason, $a$ is referred to as the slope or gradient of a straight-line function.

You should have also found that the value of $q$ affects where the graph passes through the $y$ -axis. For this reason, $q$ is known as the y-intercept .

These different properties are summarised in [link] .

 $a>0$ $a<0$ $q>0$ $q<0$

## Domain and range

For $f\left(x\right)=ax+q$ , the domain is $\left\{x:x\in \mathbb{R}\right\}$ because there is no value of $x\in \mathbb{R}$ for which $f\left(x\right)$ is undefined.

The range of $f\left(x\right)=ax+q$ is also $\left\{f\left(x\right):f\left(x\right)\in \mathbb{R}\right\}$ because there is no value of $f\left(x\right)\in \mathbb{R}$ for which $f\left(x\right)$ is undefined.

For example, the domain of $g\left(x\right)=x-1$ is $\left\{x:x\in \mathbb{R}\right\}$ because there is no value of $x\in \mathbb{R}$ for which $g\left(x\right)$ is undefined. The range of $g\left(x\right)$ is $\left\{g\left(x\right):g\left(x\right)\in \mathbb{R}\right\}$ .

## Intercepts

For functions of the form, $y=ax+q$ , the details of calculating the intercepts with the $x$ and $y$ axis are given.

The $y$ -intercept is calculated as follows:

$\begin{array}{ccc}\hfill y& =& ax+q\hfill \\ \hfill {y}_{int}& =& a\left(0\right)+q\hfill \\ & =& q\hfill \end{array}$

For example, the $y$ -intercept of $g\left(x\right)=x-1$ is given by setting $x=0$ to get:

$\begin{array}{ccc}\hfill g\left(x\right)& =& x-1\hfill \\ \hfill {y}_{int}& =& 0-1\hfill \\ & =& -1\hfill \end{array}$

The $x$ -intercepts are calculated as follows:

$\begin{array}{ccc}\hfill y& =& ax+q\hfill \\ \hfill 0& =& a·{x}_{int}+q\hfill \\ \hfill a·{x}_{int}& =& -q\hfill \\ \hfill {x}_{int}& =& -\frac{q}{a}\hfill \end{array}$

For example, the $x$ -intercepts of $g\left(x\right)=x-1$ is given by setting $y=0$ to get:

$\begin{array}{ccc}\hfill g\left(x\right)& =& x-1\hfill \\ \hfill 0& =& {x}_{int}-1\hfill \\ \hfill {x}_{int}& =& 1\hfill \end{array}$

## Turning points

The graphs of straight line functions do not have any turning points.

## Axes of symmetry

The graphs of straight-line functions do not, generally, have any axes of symmetry.

## Sketching graphs of the form $f\left(x\right)=ax+q$

In order to sketch graphs of the form, $f\left(x\right)=ax+q$ , we need to determine three characteristics:

1. sign of $a$
2. $y$ -intercept
3. $x$ -intercept

Only two points are needed to plot a straight line graph. The easiest points to use are the $x$ -intercept (where the line cuts the $x$ -axis) and the $y$ -intercept.

For example, sketch the graph of $g\left(x\right)=x-1$ . Mark the intercepts.

Firstly, we determine that $a>0$ . This means that the graph will have an upward slope.

The $y$ -intercept is obtained by setting $x=0$ and was calculated earlier to be ${y}_{int}=-1$ . The $x$ -intercept is obtained by setting $y=0$ and was calculated earlier to be ${x}_{int}=1$ .

Draw the graph of $y=2x+2$

1. To find the intercept on the y-axis, let $x=0$

$\begin{array}{ccc}\hfill y& =& 2\left(0\right)+2\hfill \\ & =& 2\hfill \end{array}$
2. For the intercept on the x-axis, let $y=0$

$\begin{array}{ccc}\hfill 0& =& 2x+2\hfill \\ \hfill 2x& =& -2\hfill \\ \hfill x& =& -1\hfill \end{array}$

## Exercise: straight line graphs

1. List the $y$ -intercepts for the following straight-line graphs:
1. $y=x$
2. $y=x-1$
3. $y=2x-1$
4. $y+1=2x$
2. Give the equation of the illustrated graph below:
3. Sketch the following relations on the same set of axes, clearly indicating the intercepts with the axes as well as the co-ordinates of the point of interception of the graph: $x+2y-5=0$ and $3x-y-1=0$

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