
## SectionModeling Linear Functions

### SubsectionEquations of Lines

Given the algebraic equation of a line, we can graph it in a number of ways. In this section, we will be given a geometric description of a line and find the algebraic equation. Finding an equation of a line can be accomplished in a number of ways. The following example makes use of slope-intercept form, $y=mx+b\text{,}$ or using function notation, $f(x)=mx+b\text{.}$ If we can determine the slope, $m\text{,}$ and the $y$-intercept, $(0,b)\text{,}$ we can then construct an equation.

###### Example79

Find an equation of the line passing through $(-3,6)$ and $(5,-4)\text{.}$

Solution

We begin by finding the slope. Given two points, we can find the slope using the slope formula.

\begin{equation*} \begin{aligned} m\amp = \frac{y_2-y_1}{x_2-x_1}\\ \amp =\frac{-4-(6)}{5-(-3)}\\ \amp =\frac{-4-6}{5+3}\\ \amp =\frac{-10}{8}\\ \amp = -\frac{5}{4} \end{aligned} \end{equation*}

Here $m=-\frac{5}{4}$ and we have

\begin{equation*} \begin{aligned} f(x)\amp =mx+b\\ f(x)\amp = -\frac{5}{4}x+b. \end{aligned} \end{equation*}

To find $b\text{,}$ substitute either one of the given points through which the line passes. Here we will use $(-3,6)\text{,}$ but $(5,-4)$ would work just as well:

Therefore, the equation of the line passing through the two given points is $f(x)=mx+b=-\frac{5}{4}x+\frac{9}{4}\text{.}$

Next, we outline an alternative method for finding equations of lines. Begin by applying the slope formula with a given point $(x_1,y_1)$ and a variable point $(x,y)\text{.}$

\begin{align*} m\amp =\frac{y-y_1}{x-x_1}\\ \frac{m}{1}\amp =\frac{y-y_1}{x-x_1}\\ m(x-x_1)\amp = y-y_1\\ y-y_1\amp = m(x-x_1) \end{align*}

Therefore, the equation of a nonvertical line can be written in point-slope form:

\begin{gather*} y-y_1=m(x-x_1) \end{gather*}

Point-slope form is particularly useful for finding an equation of a line given the slope and any ordered pair solution. After finding the slope, $-\frac{5}{4}$ in the previous example, we could use this form to find an equation. Using the point $(-3,6)\text{,}$ we substitute as follows:

Notice that we obtain the same linear function $f(x)=-\frac{5}{4}x+\frac{9}{4}\text{.}$

Note: Sometimes a variable is not expressed explicitly in terms of another; however, it is still assumed that one variable is dependent on the other. For example, the equation $2x+3y=6$ implicitly represents the function $f(x)=-\frac{2}{3}x+2\text{.}$ You should become comfortable with working with functions in either form.

###### Example80

Find an equation of the following linear function:

Solution

From the graph we can determine two points $(-1, -2)$ and $(4, 1)\text{.}$ Use these points to read the slope from the graph. The rise is $3$ units and the run is $5$ units.

Therefore, we have the slope and a point. (It does not matter which of the given points we use, the result will be the same.) We will use the point $(-1,-2)$ with our slope $m=\frac{3}{5}\text{.}$

We will use point-slope form to determine an equation of the line.

\begin{equation*} \begin{aligned} y-y_1\amp =m(x-x_1)\\ y-(\alert{-2})\amp -\alert{\frac{3}{5}}(x-(\alert{-1}))\\ y+2\amp = \frac{3}{5}(x+1)\\ y+2\amp = \frac{3}{5}x+\frac{3}{5}\\ y\amp =\frac{3}{5}x+\frac{3}{5}-2\\ y\amp =\frac{3}{5}x-\frac{7}{5} \end{aligned} \end{equation*}

An equation of the line is given by $y=\frac{3}{5}x-\frac{7}{5}$ or $f(x)=\frac{3}{5}x-\frac{7}{5}\text{.}$

Parallel lines are lines in the same plane that never intersect. Two different, non-vertical, lines in the same plane with slopes $m_1$ and $m_2$ are parallel if their slopes are the same, $m_1=m_2\text{.}$

###### Example81

Find an equation of the line passing through $(4,1)$ and parallel to $x-2y=-2\text{.}$

Solution

To find the slope of the given line, solve for $y\text{.}$

\begin{equation*} \begin{aligned} x-2y\amp =-2\\ -2y\amp = -x-2\\ y\amp =\frac{-x-2}{-2}\\ y\amp = \frac{-x}{-2}-\frac{2}{-2}\\ y\amp =\frac{1}{2}x+1 \end{aligned} \end{equation*}

Here the given line has slope $m_1=\frac{1}{2}$ and thus the slope of a parallel line is given by $m_2=\frac{1}{2}\text{.}$ Since we are given a point and we now have the slope, we will choose to use point-slope form of a linear equations to determine the slope-intercept form of the equation.

Our point is $(4,1)$ and our slope is $m=\frac{1}{2}\text{.}$

\begin{align*} y-y_1\amp =m(x-x_1)\\ y-(1)\amp =\frac{1}{2}(x-4)\\ y-1\amp =\frac{1}{2}x-\frac{4}{2}\\ y-1\alert{+1}\amp =\frac{1}{2}x-2\alert{+1}\\ y\amp = \frac{1}{2}x-1 \end{align*}

The equation of the line is given by $y=\frac{1}{2}x-1\text{,}$ or $f(x)=\frac{1}{2}x-1\text{.}$

It is important to have a geometric understanding of this question. We were asked to find an equation of a line parallel to another line passing through a certain point.

Through the point $(3,-\frac{1}{2})$ we have a parallel line, $y=\frac{1}{2}x-1\text{,}$ shown as a magenta line. Notice that the slope is the same as the given line, $y=\frac{1}{2}x+1\text{,}$ but the $y$-intercept is different.

Perpendicular lines are lines in the same plane that intersect at right angles (90 degrees). Two nonvertical lines, in the same plane with slopes $m_1$ and $m_2\text{,}$ are perpendicular if the product of their slopes is $-1\text{,}$ $m_1\cdot m_2=-1\text{.}$ We can solve for $m_1$ and obtain $m_1=-\frac{1}{m_2}\text{.}$ In this form, we see that perpendicular lines have slopes that are negative reciprocals, or opposite reciprocals. In general, given non-zero real numbers $a$ and $b\text{,}$ if the slope of the first line is given by $m_1=\frac{a}{b}\text{,}$ then the slope of the perpendicular line is $m_2=-\frac{b}{a}\text{.}$

For example, the opposite reciprocal of $m_1=-\frac{3}{5}$ is $m_2=\frac{5}{3}\text{.}$ We can verify that two slopes produce perpendicular lines if their product is $-1\text{.}$

\begin{gather*} m_1\cdot m_2=-\frac{3}{5}\cdot\frac{5}{3}=-\frac{15}{15}=-1 \end{gather*}
###### Example82

Find an equation of the line passing through $(-5,-2)$ and perpendicular to the graph of $x+4y=4\text{.}$

Solution

To find the slope of the given line, solve for $y\text{.}$

\begin{equation*} \begin{aligned} x+4y\amp =4\\ 4y\amp = -x+4\\ y\amp = \frac{-x+4}{4}\\ y\amp = -\frac{x}{4}+\frac{4}{4}\\ y\amp = -\frac{1}{4}x+1 \end{aligned} \end{equation*}

The given line has slope $m_1=-\frac{1}{4}$ and thus, $m_2=+\frac{4}{1}=4\text{.}$ Substitute this slope and the given point into point-slope form.

Our point is $(-5,-2)$ and our slope is $m_2=4\text{.}$

\begin{equation*} \begin{aligned} y-y_1\amp =m(x-x_1)\\ y-(-2)\amp = 4(x-(-5))\\ y+2\amp = 4(x+5)\\ y+2\amp = 4x+20\\ y\amp = 4x+18 \end{aligned} \end{equation*}

The equation of the perpendicular line is given by $y=4x+18\text{,}$ or $f(x)=4x+18\text{.}$

Geometrically, we see that the graph of $y=4x+18\text{,}$ shown as the dashed line in the graph, passes through $(-5,-2)$ and is perpendicular to the graph of $y=-\frac{1}{4}x+1\text{.}$

Find an equation of the line passing through $(-5,-2)$ and perpendicular to $\frac{1}{3}x-\frac{1}{2}y=-2\text{.}$

Solution

The equation of the perpendicular line is given by $y=-\frac{3}{2}x-\frac{19}{2}\text{.}$

### SubsectionElementary Functions

In this section we graph two elementary functions that will be used throughout this course. Each function is graphed by plotting points. For this course $f(x)=y$ and thus $f(x)$ and $y$ can be used interchangeably.

Any function of the form $f(x)=c\text{,}$ where $c$ is any real number, is called a constant function. Constant functions are linear and can be written in the form $f(x)=0x+c\text{.}$ In this form, it is clear that the slope is 0 and the $y$-intercept is $(0,c)\text{.}$ Evaluating any value for $x\text{,}$ such as $x=2\text{,}$ will result in $c\text{.}$

The graph of a constant function is a horizontal line. The domain consists of all real numbers $\mathbb{R}=(-\infty,\infty)$ and the range consists of the single value ${c}\text{.}$

We next define the identity function $f(x)=x\text{.}$ Evaluating any value for $x$ will result in the same value. For example, $f(0)=0$ and $f(2)=2\text{.}$ The identity function is linear, $f(x)=1x+0\text{,}$ with slope $m=1$ and $y$-intercept $(0,0)\text{.}$

The domain and range both consist of all real numbers, $\mathbb{R}=(-\infty,\infty)\text{.}$