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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:

\begin{equation*} \begin{aligned} f(x)\amp = -\frac{5}{4}x+b\amp \alert{\text{Use }(x,f(x))=(-3,6)}\\ \alert{6}\amp = -\frac{5}{4}(\alert{-3})+b\\ 6\amp =\frac{15}{4}+b\\ \frac{6\cdot \alert{4}}{1\cdot\alert{4}}-\frac{15}{4}\amp =b\\ \frac{24-15}{4}\amp =b\\ \frac{9}{4}\amp =b \end{aligned} \end{equation*}

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:

\begin{align*} y-y_1\amp =m(x-x_1)\\ y-(\alert{6})\amp =\alert{-\frac{5}{4}}(x-(\alert{-3}))\\ y-6\amp =-\frac{5}{4}(x+3)\\ y-6\amp=-\frac{5}{4}x-\frac{15}{4}\\ y\amp =-\frac{5}{4}x-\frac{15}{4}+6\\ y\amp =-\frac{5}{4}x+\frac{9}{4} \end{align*}

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:

Quadrants of the Plane

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.

Quadrants of the Plane

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.

Geometric Understanding

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 \(y=-\frac{1}{4}x+1\text{.}\)

Quadrants of the Plane

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 \(c\text{.}\)

Quadrants of the Plane

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 \((0,0)\text{.}\)

Quadrants of the Plane

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