Graphing Linear Equations

Slope-Intercept Form

A linear function of two variables can be written in slope-intercept form: $y = a x + b$
- $y$ is the dependent variable (on the vertical axis)
- $x$ is the independent variable (on the horizontal axis)
- $a$ is the slope of the line
  - $a = \frac{rise}{run} = \frac{change in y}{change in x} = \frac{Δ x}{Δ y}$
  - you may have been taught the slope as “ $m$ ”, this is just personal taste, but again, get used to using different letters!
- $b$ is the vertical-intercept, a constant number where the line crosses the vertical axis
  - if $y$ is the dependent variable, this is the “ $y$ -intercept”, $y$ where $x = 0$
Any point on the line has an $x$ -coordinate and $y$ -coordinate $(x_{i}, y_{i})$

Other Forms

A linear function can equivalently be expressed in the following form: $a_{1} x_{1} + a_{2} x_{2} = c$
- $x_{2}$ is the dependent variable (on the vertical axis)
- $x_{1}$ is the independent variable (on the horizontal axis)
- $c$ is a constant
This is a valid equation, but is difficult to visualize in the traditional graph as above. Simply solve for the dependent variable on the vertical axis $(x_{2}$ as if you were solving for $y)$ :

$\begin{aligned} a_{1} x_{1} + a_{2} x_{2} = c & Original \\ a_{2} x_{2} = c - a_{1} x_{1} & Subtracting x_{1} term \\ x_{2} = \frac{c}{a_{2}} - \frac{a_{1}}{a_{2}} x_{1} & Dividing by a_{2} \end{aligned}$

The vertical intercept is $\frac{c}{a_{2}}$
The horizontal intercept is $\frac{c}{a_{1}}$
The slope is $- \frac{a_{1}}{a_{2}}$
This is extremely useful for dealing with constraints in constrained optimization problems: budget constraints and isocost lines

Drawing a Graph From an Equation

If we already have a linear equation that we would like to graph, we can follow these steps:

Take the equation and plug in two values, e.g. if we have:

$p = \frac{1}{2} q + 4$

We can find two points on the graph. The easiest one to find is the vertical-intercept, where the line crosses the vertical axis, where $q = 0$ , so plug in $q = 0$ :

$\begin{aligned} p & = \frac{1}{2} (0) + 4 \\ p & = 4 \end{aligned}$

Thus, one point is $(0, 4)$ . Note that the constant in the function itself is the $p - i n t e r c e p t$ ! So one valid point will always be $(0, b)$ !

For our second point, let’s plug in $q = 2$ :

$\begin{aligned} p & = \frac{1}{2} (2) + 4 \\ p & = 5 \end{aligned}$

Thus, another point is $(2, 5)$

Now, plot the two points on the graph, and connect them with a line

Note: A quick shortcut to plot a line is to find the vertical intercept and plot that, and then find the next point using the slope. Here, start our line at 4 on the vertical axis, and then, as the slope is $\frac{1}{2}$ , for every one unit increase in $q$ , $p$ increases by $\frac{1}{2}$ . Our second point, (2,5), is a 2 unit increase in $q$ resulting in a $1$ unit increase in $p$ .

Finding an Equation from a Graph

In order to find the equation of an existing line, we follow these steps:

First, take two points on the line and find the slope, $a$ , between them. Let’s pick $(1, 6)$ and $(3, 2)$ .

$\begin{aligned} Slope & = \frac{r i s e}{r u n} \\ = \frac{(p_{2} - p_{1})}{(q_{2} - q_{1})} \\ = \frac{(2 - 6)}{(3 - 1)} \\ = \frac{- 4}{2} \\ = - 2 \end{aligned}$

There is a shortcut that we can use to find the slope faster by eye-balling the graph: When $q$ changes by 1, how many units does $p$ change? If we move from (1,6) to (2,5), $q$ increases by 1, but $p$ falls by 2. So the slope is $- 2$ . For every one unit increase in $q$ , $p$ changes by -2.

Now with the slope, we need to find the vertical intercept, or $b$ , we solve this by plugging in the slope and any point on the graph, we will use (1,6):

$\begin{aligned} p & = a q + b \\ (6) & = - 2 (1) + b \\ 6 & = - 2 + b \\ 8 & = b \end{aligned}$

Note, there is another easy way to eye-ball what this value is. It is simply that $p$ value where $q = 0$ , or at what $p$ value the graph crosses the vertical axis. We can see it is at 8.

Thus, we have the slope and the intercept, so our equation is: $p = - 2 q + 8$