writing equations of linear functions homework 3

Module 7: Linear and Absolute Value Functions

Graphing and writing equations of linear functions, learning outcomes.

• Graph linear functions by plotting points, using the slope and y-intercept, and using transformations.
• Write the equation of a linear function given its graph.
• Match linear functions with their graphs.
• Find the x-intercept of a function given its equation.
• Find the equations of vertical and horizontal lines.
• Determine whether lines are parallel or perpendicular given their equations.
• Find equations of lines that are parallel or perpendicular to a given line.
• Graph an absolute value function.
• Find the intercepts of an absolute value function.

We can now describe a variety of characteristics that explain the behavior of linear functions. We will use this information to analyze a graphed line and write an equation based on its observable properties. From evaluating the graph, what can you determine about this linear function?

• initial value (y-intercept)?
• one or two points?
• increasing or decreasing?
• vertical or horizontal?

In this section, you will practice writing linear function equations using the information you’ve gathered. We will also practice graphing linear functions using different methods and predict how the graphs of linear functions will change when parts of the equation are altered.

[latex]\\[/latex]

Graphing linear functions.

We previously saw that that the graph of a linear function is a straight line. We were also able to see the points of the function as well as the initial value from a graph.

There are three basic methods of graphing linear functions. The first is by plotting points and then drawing a line through the points. The second is by using the y- intercept and slope. The third is applying transformations to the identity function [latex]f\left(x\right)=x[/latex].

Graphing a Function by Plotting Points

To find points of a function, we can choose input values, evaluate the function at these input values, and calculate output values. The input values and corresponding output values form coordinate pairs. We then plot the coordinate pairs on a grid. In general we should evaluate the function at a minimum of two inputs in order to find at least two points on the graph of the function. For example, given the function [latex]f\left(x\right)=2x[/latex], we might use the input values 1 and 2. Evaluating the function for an input value of 1 yields an output value of 2 which is represented by the point (1, 2). Evaluating the function for an input value of 2 yields an output value of 4 which is represented by the point (2, 4). Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error.

How To: Given a linear function, graph by plotting points.

• Choose a minimum of two input values.
• Evaluate the function at each input value.
• Use the resulting output values to identify coordinate pairs.
• Plot the coordinate pairs on a grid.
• Draw a line through the points.

Example: Graphing by Plotting Points

Graph [latex]f\left(x\right)=-\frac{2}{3}x+5[/latex] by plotting points.

Begin by choosing input values. This function includes a fraction with a denominator of 3 so let’s choose multiples of 3 as input values. We will choose 0, 3, and 6.

Evaluate the function at each input value and use the output value to identify coordinate pairs.

[latex]\begin{array}{llllll}x=0& & f\left(0\right)=-\frac{2}{3}\left(0\right)+5=5\Rightarrow \left(0,5\right)\\ x=3& & f\left(3\right)=-\frac{2}{3}\left(3\right)+5=3\Rightarrow \left(3,3\right)\\ x=6& & f\left(6\right)=-\frac{2}{3}\left(6\right)+5=1\Rightarrow \left(6,1\right)\end{array}[/latex]

Plot the coordinate pairs and draw a line through the points. The graph below is of the function [latex]f\left(x\right)=-\frac{2}{3}x+5[/latex].

Analysis of the Solution

The graph of the function is a line as expected for a linear function. In addition, the graph has a downward slant which indicates a negative slope. This is also expected from the negative constant rate of change in the equation for the function.

Graph [latex]f\left(x\right)=-\frac{3}{4}x+6[/latex] by plotting points.

Graphing a Linear Function Using y-intercept and Slope

Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. The first characteristic is its y- intercept which is the point at which the input value is zero. To find the y- intercept , we can set [latex]x=0[/latex] in the equation.

The other characteristic of the linear function is its slope,  m , which is a measure of its steepness. Recall that the slope is the rate of change of the function. The slope of a linear function is equal to the ratio of the change in outputs to the change in inputs. Another way to think about the slope is by dividing the vertical difference, or rise, between any two points by the horizontal difference, or run. The slope of a linear function will be the same between any two points. We encountered both the y- intercept and the slope in Linear Functions.

Let’s consider the following function.

[latex]f\left(x\right)=\frac{1}{2}x+1[/latex]

The slope is [latex]\frac{1}{2}[/latex]. Because the slope is positive, we know the graph will slant upward from left to right. The y- intercept is the point on the graph when x  = 0. The graph crosses the y -axis at (0, 1). Now we know the slope and the y -intercept. We can begin graphing by plotting the point (0, 1) We know that the slope is rise over run, [latex]m=\frac{\text{rise}}{\text{run}}[/latex]. From our example, we have [latex]m=\frac{1}{2}[/latex], which means that the rise is 1 and the run is 2. Starting from our y -intercept (0, 1), we can rise 1 and then run 2 or run 2 and then rise 1. We repeat until we have multiple points, and then we draw a line through the points as shown below.

A General Note: Graphical Interpretation of a Linear Function

In the equation [latex]f\left(x\right)=mx+b[/latex]

• b  is the y -intercept of the graph and indicates the point (0, b ) at which the graph crosses the y -axis.
• m  is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope:

[latex]m=\frac{\text{change in output (rise)}}{\text{change in input (run)}}=\frac{\Delta y}{\Delta x}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[/latex]

Do all linear functions have y -intercepts?

Yes. All linear functions cross the y-axis and therefore have y-intercepts. (Note: A vertical line parallel to the y-axis does not have a y-intercept. Keep in mind that a vertical line is the only line that is not a function.)

How To: Given the equation for a linear function, graph the function using the y -intercept and slope.

• Evaluate the function at an input value of zero to find the y- intercept.
• Identify the slope.
• Plot the point represented by the y- intercept.
• Use [latex]\frac{\text{rise}}{\text{run}}[/latex] to determine at least two more points on the line.
• Draw a line which passes through the points.

Example: Graphing by Using the y- intercept and Slope

Graph [latex]f\left(x\right)=-\frac{2}{3}x+5[/latex] using the y- intercept and slope.

Evaluate the function at x  = 0 to find the y- intercept. The output value when x  = 0 is 5, so the graph will cross the y -axis at (0, 5).

According to the equation for the function, the slope of the line is [latex]-\frac{2}{3}[/latex]. This tells us that for each vertical decrease in the “rise” of [latex]–2[/latex] units, the “run” increases by 3 units in the horizontal direction. We can now graph the function by first plotting the y -intercept. From the initial value (0, 5) we move down 2 units and to the right 3 units. We can extend the line to the left and right by repeating, and then draw a line through the points.

The graph slants downward from left to right which means it has a negative slope as expected.

Find a point on the graph we drew in the previous example: Graphing by Using the y -intercept and Slope that has a negative x -value.

Possible answers include [latex]\left(-3,7\right)[/latex], [latex]\left(-6,9\right)[/latex], or [latex]\left(-9,11\right)[/latex].

Graphing a Linear Function Using Transformations

Another option for graphing is to use transformations on the identity function [latex]f\left(x\right)=x[/latex]. A function may be transformed by a shift up, down, left, or right. A function may also be transformed using a reflection, stretch, or compression.

Vertical Stretch or Compression

In the equation [latex]f\left(x\right)=mx[/latex], the m  is acting as the vertical stretch or compression of the identity function. When m  is negative, there is also a vertical reflection of the graph. Notice that multiplying the equation [latex]f\left(x\right)=x[/latex] by m  stretches the graph of f  by a factor of m  units if m  > 1 and compresses the graph of f  by a factor of m  units if 0 < m  < 1. This means the larger the absolute value of m , the steeper the slope.

Vertical stretches and compressions and reflections on the function [latex]f\left(x\right)=x[/latex].

Vertical Shift

In [latex]f\left(x\right)=mx+b[/latex], the b  acts as the vertical shift , moving the graph up and down without affecting the slope of the line. Notice that adding a value of b  to the equation of [latex]f\left(x\right)=x[/latex] shifts the graph of  f  a total of b  units up if b  is positive and | b | units down if b  is negative.

This graph illustrates vertical shifts of the function [latex]f\left(x\right)=x[/latex].

Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of linear functions. Although this may not be the easiest way to graph this type of function, it is still important to practice each method.

How To: Given the equation of a linear function, use transformations to graph the linear function in the form [latex]f\left(x\right)=mx+b[/latex].

• Graph [latex]f\left(x\right)=x[/latex].
• Vertically stretch or compress the graph by a factor m .
• Shift the graph up or down b  units.

Example: Graphing by Using Transformations

Graph [latex]f\left(x\right)=\frac{1}{2}x - 3[/latex] using transformations.

The equation for the function shows that [latex]m=\frac{1}{2}[/latex] so the identity function is vertically compressed by [latex]\frac{1}{2}[/latex]. The equation for the function also shows that [latex]b=-3[/latex], so the identity function is vertically shifted down 3 units.

First, graph the identity function, and show the vertical compression.

The function [latex]y=x[/latex] compressed by a factor of [latex]\frac{1}{2}[/latex].

Then, show the vertical shift.

The function [latex]y=\frac{1}{2}x[/latex] shifted down 3 units.

Graph [latex]f\left(x\right)=4+2x[/latex], using transformations.

In Example: Graphing by Using Transformations, could we have sketched the graph by reversing the order of the transformations?

No. The order of the transformations follows the order of operations. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. This is why we performed the compression first. For example, following order of operations, let the input be 2.

[latex]\begin{array}{l}f\text{(2)}=\frac{\text{1}}{\text{2}}\text{(2)}-\text{3}\hfill \\ =\text{1}-\text{3}\hfill \\ =-\text{2}\hfill \end{array}[/latex]

Writing Equations of Linear Functions

We previously wrote the equation for a linear function from a graph. Now we can extend what we know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at the graph below. We can see right away that the graph crosses the y -axis at the point (0, 4), so this is the y -intercept.

Then we can calculate the slope by finding the rise and run. We can choose any two points, but let’s look at the point (–2, 0). To get from this point to the y- intercept, we must move up 4 units (rise) and to the right 2 units (run). So the slope must be:

[latex]m=\frac{\text{rise}}{\text{run}}=\frac{4}{2}=2[/latex]

Substituting the slope and y- intercept into slope-intercept form of a line gives:

[latex]y=2x+4[/latex]

How To: Given THE graph of A linear function, find the equation to describe the function.

• Identify the y- intercept from the graph.
• Choose two points to determine the slope.
• Substitute the y- intercept and slope into slope-intercept form of a line.

Example: Matching Linear Functions to Their Graphs

Match each equation of a linear function with one of the lines in the graph below.

• [latex]f\left(x\right)=2x+3[/latex]
• [latex]g\left(x\right)=2x - 3[/latex]
• [latex]h\left(x\right)=-2x+3[/latex]
• [latex]j\left(x\right)=\frac{1}{2}x+3[/latex]

Analyze the information for each function.

• This function has a slope of 2 and a y -intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. We can use two points to find the slope, or we can compare it with the other functions listed. Function g  has the same slope, but a different y- intercept. Lines I and III have the same slant because they have the same slope. Line III does not pass through (0, 3) so f  must be represented by line I.
• This function also has a slope of 2, but a y -intercept of –3. It must pass through the point (0, –3) and slant upward from left to right. It must be represented by line III.
• This function has a slope of –2 and a y- intercept of 3. This is the only function listed with a negative slope, so it must be represented by line IV because it slants downward from left to right.
• This function has a slope of [latex]\frac{1}{2}[/latex] and a y- intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. Lines I and II pass through (0, 3), but the slope of j  is less than the slope of f  so the line for j  must be flatter. This function is represented by Line II.

Now we can re-label the lines.

Finding the x -intercept of a Line

So far we have been finding the y- intercepts of functions: the point at which the graph of a function crosses the y -axis. A function may also have an x -intercept, which is the x -coordinate of the point where the graph of a function crosses the x -axis. In other words, it is the input value when the output value is zero.

To find the x -intercept, set the function f ( x ) equal to zero and solve for the value of x . For example, consider the function shown:

[latex]f\left(x\right)=3x - 6[/latex]

Set the function equal to 0 and solve for x .

[latex]\begin{array}{l}0=3x - 6\hfill \\ 6=3x\hfill \\ 2=x\hfill \\ x=2\hfill \end{array}[/latex]

The graph of the function crosses the x -axis at the point (2, 0).

Do all linear functions have x -intercepts?

No. However, linear functions of the form y  = c , where c is a nonzero real number are the only examples of linear functions with no x -intercept. For example, y  = 5 is a horizontal line 5 units above the x -axis. This function has no x -intercepts .

A General Note: x -intercept

The x -intercept of a function is the value of x  where  f ( x ) = 0. It can be found by solving the equation 0 = mx  + b .

Example: Finding an x -intercept

Find the x -intercept of [latex]f\left(x\right)=\frac{1}{2}x - 3[/latex].

Set the function equal to zero to solve for x .

[latex]\begin{array}{l}0=\frac{1}{2}x - 3\\ 3=\frac{1}{2}x\\ 6=x\\ x=6\end{array}[/latex]

The graph crosses the x -axis at the point (6, 0).

A graph of the function is shown below. We can see that the x -intercept is (6, 0) as expected.

The graph of the linear function [latex]f\left(x\right)=\frac{1}{2}x – 3[/latex].

Find the x -intercept of [latex]f\left(x\right)=\frac{1}{4}x - 4[/latex].

[latex]\left(16,\text{ 0}\right)[/latex]

Describing Horizontal and Vertical Lines

There are two special cases of lines on a graph—horizontal and vertical lines. A horizontal line indicates a constant output or y -value. In the graph below, we see that the output has a value of 2 for every input value. The change in outputs between any two points is 0. In the slope formula, the numerator is 0, so the slope is 0. If we use m  = 0 in the equation [latex]f\left(x\right)=mx+b[/latex], the equation simplifies to [latex]f\left(x\right)=b[/latex]. In other words, the value of the function is a constant. This graph represents the function [latex]f\left(x\right)=2[/latex].

A horizontal line representing the function [latex]f\left(x\right)=2[/latex].

A vertical line indicates a constant input or x -value. We can see that the input value for every point on the line is 2, but the output value varies. Because this input value is mapped to more than one output value, a vertical line does not represent a function. Notice that between any two points, the change in the input values is zero. In the slope formula, the denominator will be zero, so the slope of a vertical line is undefined.

Notice that a vertical line has an x -intercept but no y- intercept unless it’s the line x = 0. This graph represents the line x = 2.

The vertical line [latex]x=2[/latex] which does not represent a function.

A General Note: Horizontal and Vertical Lines

Lines can be horizontal or vertical.

A horizontal line is a line defined by an equation of the form [latex]f\left(x\right)=b[/latex] where [latex]b[/latex] is a constant.

A vertical line is a line defined by an equation of the form [latex]x=a[/latex] where [latex]a[/latex] is a constant.

Example: Writing the Equation of a Horizontal Line

Write the equation of the line graphed below.

For any x -value, the y -value is [latex]–4[/latex], so the equation is [latex]y=–4[/latex].

Example: Writing the Equation of a Vertical Line

The constant x -value is 7, so the equation is [latex]x=7[/latex].

• Write the equation of the function passing through the points [latex](2,6)[/latex] and [latex](4,4)[/latex] in slope-intercept form.
• Write the equation of a function whose slope is 2 and passes through the point [latex](-1,0)[/latex]
• Write the equation of a function whose slope is undefined.

Parallel and Perpendicular Lines

The two lines in the graph below are parallel lines : they will never intersect. Notice that they have exactly the same steepness which means their slopes are identical. The only difference between the two lines is the y -intercept. If we shifted one line vertically toward the y -intercept of the other, they would become the same line.

Parallel lines.

We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the y -intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel.

Unlike parallel lines, perpendicular lines do intersect. Their intersection forms a right or 90-degree angle. The two lines below are perpendicular.

Perpendicular lines.

Perpendicular lines do not have the same slope. The slopes of perpendicular lines are different from one another in a specific way. The slope of one line is the negative reciprocal of the slope of the other line. The product of a number and its reciprocal is 1. If [latex]{m}_{1}\text{ and }{m}_{2}[/latex] are negative reciprocals of one another, they can be multiplied together to yield [latex]-1[/latex].

[latex]{m}_{1}*{m}_{2}=-1[/latex]

To find the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is [latex]\frac{1}{8}[/latex], and the reciprocal of [latex]\frac{1}{8}[/latex] is 8. To find the negative reciprocal, first find the reciprocal and then change the sign.

As with parallel lines, we can determine whether two lines are perpendicular by comparing their slopes. The slope of each line below is the negative reciprocal of the other so the lines are perpendicular.

[latex]\begin{array}{ll}f\left(x\right)=\frac{1}{4}x+2\hfill & \text{negative reciprocal of }\frac{1}{4}\text{ is }-4\hfill \\ f\left(x\right)=-4x+3\hfill & \text{negative reciprocal of }-4\text{ is }\frac{1}{4}\hfill \end{array}[/latex]

The product of the slopes is –1.

[latex]-4\left(\frac{1}{4}\right)=-1[/latex]

A General Note: Parallel and Perpendicular Lines

Two lines are parallel lines if they do not intersect. The slopes of the lines are the same.

[latex]f\left(x\right)={m}_{1}x+{b}_{1}\text{ and }g\left(x\right)={m}_{2}x+{b}_{2}\text{ are parallel if }{m}_{1}={m}_{2}[/latex].

If and only if [latex]{b}_{1}={b}_{2}[/latex] and [latex]{m}_{1}={m}_{2}[/latex], we say the lines coincide. Coincident lines are the same line.

Two lines are perpendicular lines if they intersect at right angles.

[latex]f\left(x\right)={m}_{1}x+{b}_{1}\text{ and }g\left(x\right)={m}_{2}x+{b}_{2}\text{ are perpendicular if }{m}_{1}*{m}_{2}=-1,\text{ and }{m}_{2}=-\frac{1}{{m}_{1}}[/latex].

Example: Identifying Parallel and Perpendicular Lines

Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines.

[latex]\begin{array}{l}f\left(x\right)=2x+3\hfill & \hfill & h\left(x\right)=-2x+2\hfill \\ g\left(x\right)=\frac{1}{2}x - 4\hfill & \hfill & j\left(x\right)=2x - 6\hfill \end{array}[/latex]

Parallel lines have the same slope. Because the functions [latex]f\left(x\right)=2x+3[/latex] and [latex]j\left(x\right)=2x - 6[/latex] each have a slope of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because −2 and [latex]\frac{1}{2}[/latex] are negative reciprocals, the equations, [latex]g\left(x\right)=\frac{1}{2}x - 4[/latex] and [latex]h\left(x\right)=-2x+2[/latex] represent perpendicular lines.

A graph of the lines is shown below.

The graph shows that the lines [latex]f\left(x\right)=2x+3[/latex] and [latex]j\left(x\right)=2x – 6[/latex] are parallel, and the lines [latex]g\left(x\right)=\frac{1}{2}x – 4[/latex] and [latex]h\left(x\right)=-2x+2[/latex] are perpendicular.

Writing Equations of Parallel Lines

If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line.

Suppose we are given the following function:

[latex]f\left(x\right)=3x+1[/latex]

We know that the slope of the line is 3. We also know that the y- intercept is (0, 1). Any other line with a slope of 3 will be parallel to f ( x ). The lines formed by all of the following functions will be parallel to f ( x ).

[latex]\begin{array}{l}g\left(x\right)=3x+6\hfill \\ h\left(x\right)=3x+1\hfill \\ p\left(x\right)=3x+\frac{2}{3}\hfill \end{array}[/latex]

Suppose then we want to write the equation of a line that is parallel to f  and passes through the point (1, 7). We already know that the slope is 3. We just need to determine which value for b  will give the correct line. We can begin by using point-slope form of an equation for a line. We can then rewrite it in slope-intercept form.

[latex]\begin{array}{l}y-{y}_{1}=m\left(x-{x}_{1}\right)\hfill \\ y - 7=3\left(x - 1\right)\hfill \\ y - 7=3x - 3\hfill \\ \text{}y=3x+4\hfill \end{array}[/latex]

So [latex]g\left(x\right)=3x+4[/latex] is parallel to [latex]f\left(x\right)=3x+1[/latex] and passes through the point (1, 7).

How To: Given the equation of a linear function, write the equation of a line WHICH passes through a given point and is parallel to the given line.

• Find the slope of the function.
• Substitute the slope and given point into point-slope or slope-intercept form.

Example: Finding a Line Parallel to a Given Line

Find a line parallel to the graph of [latex]f\left(x\right)=3x+6[/latex] that passes through the point (3, 0).

The slope of the given line is 3. If we choose slope-intercept form, we can substitute [latex]m=3[/latex], [latex]x=3[/latex], and [latex]f(x)=0[/latex] into slope-intercept form to find the y- intercept.

[latex]\begin{array}{l}g\left(x\right)=3x+b\hfill \\ \text{}0=3\left(3\right)+b\hfill \\ \text{}b=-9\hfill \end{array}[/latex]

The line parallel to  f ( x ) that passes through (3, 0) is [latex]g\left(x\right)=3x - 9[/latex].

We can confirm that the two lines are parallel by graphing them. The figure below shows that the two lines will never intersect.

Writing Equations of Perpendicular Lines

We can use a very similar process to write the equation of a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following function:

[latex]f\left(x\right)=2x+4[/latex]

The slope of the line is 2, and its negative reciprocal is [latex]-\frac{1}{2}[/latex]. Any function with a slope of [latex]-\frac{1}{2}[/latex] will be perpendicular to  f ( x ). The lines formed by all of the following functions will be perpendicular to  f ( x ).

[latex]\begin{array}{l}g\left(x\right)=-\frac{1}{2}x+4\hfill \\ h\left(x\right)=-\frac{1}{2}x+2\hfill \\ p\left(x\right)=-\frac{1}{2}x-\frac{1}{2}\hfill \end{array}[/latex]

As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose that we want to write the equation of a line that is perpendicular to  f ( x ) and passes through the point (4, 0). We already know that the slope is [latex]-\frac{1}{2}[/latex]. Now we can use the point to find the y -intercept by substituting the given values into the slope-intercept form of a line and solving for b .

[latex]\begin{array}{l}g\left(x\right)=mx+b\hfill \\ 0=-\frac{1}{2}\left(4\right)+b\hfill \\ 0=-2+b\hfill \\ 2=b\hfill \\ b=2\hfill \end{array}[/latex]

The equation for the function with a slope of [latex]-\frac{1}{2}[/latex] and a y- intercept of 2 is

[latex]g\left(x\right)=-\frac{1}{2}x+2[/latex].

So [latex]g\left(x\right)=-\frac{1}{2}x+2[/latex] is perpendicular to [latex]f\left(x\right)=2x+4[/latex] and passes through the point (4, 0). Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.

A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not –1. Doesn’t this fact contradict the definition of perpendicular lines?

No. For two perpendicular linear functions, the product of their slopes is –1. However, a vertical line is not a function so the definition is not contradicted.

How To: Given the equation of a linear function, write the equation of a line WHICH passes through a given point and is Perpendicular to the given line.

• Find the slope of the given function.
• Determine the negative reciprocal of the slope.
• Substitute the new slope and the values for x  and y  from given point into [latex]g\left(x\right)=mx+b[/latex].
• Solve for b .
• Write the equation of the line.

Example: Finding the Equation of a Perpendicular Line

Find the equation of a line perpendicular to [latex]f\left(x\right)=3x+3[/latex] that passes through the point (3, 0).

The original line has slope [latex]m=3[/latex], so the slope of the perpendicular line will be its negative reciprocal, or [latex]-\frac{1}{3}[/latex]. Using this slope and the given point, we can find the equation for the line.

[latex]\begin{array}{l}g\left(x\right)=-\frac{1}{3}x+b\hfill \\ 0=-\frac{1}{3}\left(3\right)+b\hfill \\ \text{ }1=b\hfill \\ b=1\hfill \end{array}[/latex]

The line perpendicular to  f ( x ) that passes through (3, 0) is [latex]g\left(x\right)=-\frac{1}{3}x+1[/latex].

A graph of the two lines is shown below.

• For what y-intercept will the graph of[latex]f(x)[/latex] pass through the point [latex](-2,5)[/latex]?
• Add a new function that uses the slope m that will create a line that is perpendicular to the function [latex]f(x)=mx-2[/latex].
• For what y-intercept will the new function pass through the point [latex](4,1)[/latex], and still be perpendicular to [latex]f(x)[/latex]
• When the y-intercept is [latex](0,3)[/latex] the function will be [latex]f(x) = mx+3[/latex] and the function will pass through the point [latex](-2,5)[/latex].
• [latex]f(x)=\frac{-1}{m}x-2[/latex] for example. Any value for the y-intercept will work.
• The y-intercept [latex](0,-3)[/latex] will give a line perpendicular to [latex]f(x)[/latex] that passes through the point [latex](4,1)[/latex].

How To: Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point.

• Determine the slope of the line passing through the points.

Find the negative reciprocal of the slope.

• Use slope-intercept form or point-slope form to write the equation by substituting the known values.

Example: Finding the Equation of a Line going through a point and Perpendicular to a Given Line

A line passes through the points (–2, 6) and (4, 5). Find the equation of a line that is perpendicular and passes through the point (4, 5).

From the two points of the given line, we can calculate the slope of that line.

[latex]\begin{array}{l}{m}_{1}=\frac{5 - 6}{4-\left(-2\right)}\hfill \\ {m}_{1}=\frac{-1}{6}\hfill \\ {m}_{1}=-\frac{1}{6}\hfill \end{array}[/latex]

[latex]\begin{array}{l}{m}_{2}=\frac{-1}{-\frac{1}{6}}\hfill \\ {m}_{2}=-1\left(-\frac{6}{1}\right)\hfill \\ {m}_{2}=6\hfill \end{array}[/latex]

We can then solve for the y- intercept of the line passing through the point (4, 5).

[latex]\begin{array}{l}g\left(x\right)=6x+b\hfill \\ 5=6\left(4\right)+b\hfill \\ 5=24+b\hfill \\ -19=b\hfill \\ b=-19\hfill \end{array}[/latex]

The equation of the line that passes through the point (4, 5) and is perpendicular to the line passing through the two given points is [latex]y=6x - 19[/latex].

A line passes through the points, (–2, –15) and (2, –3). Find the equation of a perpendicular line that passes through the point, (6, 4).

[latex]y=-\frac{1}{3}x+6[/latex]

Absolute Value Functions

Distances in deep space can be measured in all directions. As such, it is useful to consider distance in terms of absolute values. (credit: “s58y”/Flickr)

Until the 1920s, the so-called spiral nebulae were believed to be clouds of dust and gas in our own galaxy some tens of thousands of light years away. Then, astronomer Edwin Hubble proved that these objects are galaxies in their own right at distances of millions of light years. Today, astronomers can detect galaxies that are billions of light years away. Distances in the universe can be measured in all directions. As such, it is useful to consider distance as an absolute value function. In this section, we will investigate absolute value functions .

Understanding Absolute Value

Recall that in its basic form [latex]\displaystyle{f}\left({x}\right)={|x|}[/latex], the absolute value function, is one of our toolkit functions. The absolute value function is commonly thought of as providing the distance a number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign.

A General Note: Absolute Value Function

The absolute value function can be defined as a piecewise function

[latex]f(x) =\begin{cases}x ,\ x \geq 0 \\ -x , x < 0\\ \end{cases} [/latex]

Example: Determine a Number within a Prescribed Distance

Describe all values [latex]x[/latex] within or including a distance of 4 from the number 5.

We want the distance between [latex]x[/latex] and 5 to be less than or equal to 4. We can draw a number line to represent the condition to be satisfied.

The distance from [latex]x[/latex] to 5 can be represented using [latex]|x - 5|[/latex]. We want the values of [latex]x[/latex] that satisfy the condition [latex]|x - 5|\le 4[/latex].

[latex]\displaystyle{-4}\le{x - 5}[/latex] [latex]\displaystyle{1}\le{x}[/latex] And: [latex]\displaystyle{x-5}\le{4}[/latex] [latex]\displaystyle{x}\le{9}[/latex]

So [latex]|x - 5|\le 4[/latex] is equal to [latex]1\le x\le 9[/latex].

However, mathematicians generally prefer absolute value notation.

Describe all values [latex]x[/latex] within a distance of 3 from the number 2.

[latex]|x - 2|\le 3[/latex]

Example: Resistance of a Resistor

Electrical parts, such as resistors and capacitors, come with specified values of their operating parameters: resistance, capacitance, etc. However, due to imprecision in manufacturing, the actual values of these parameters vary somewhat from piece to piece even when they are supposed to be the same. The best that manufacturers can do is to try to guarantee that the variations will stay within a specified range, often [latex]\pm 1\%,\pm5\%,[/latex] or [latex]\displaystyle\pm10\%[/latex].

Suppose we have a resistor rated at 680 ohms, [latex]\pm 5\%[/latex]. Use the absolute value function to express the range of possible values of the actual resistance.

5% of 680 ohms is 34 ohms. The absolute value of the difference between the actual and nominal resistance should not exceed the stated variability, so, with the resistance [latex]R[/latex] in ohms,

[latex]|R - 680|\le 34[/latex]

Students who score within 20 points of 80 will pass a test. Write this as a distance from 80 using absolute value notation.

Using the variable [latex]p[/latex] for passing, [latex]|p - 80|\le 20[/latex].

The most significant feature of the absolute value graph is the corner point at which the graph changes direction. This point is shown at the origin .

The graph below is of [latex]y=2\left|x - 3\right|+4[/latex]. The graph of [latex]y=|x|[/latex] has been shifted right 3 units, vertically stretched by a factor of 2, and shifted up 4 units. This means that the corner point is located at [latex]\left(3,4\right)[/latex] for this transformed function.

Example: Writing an Equation for an Absolute Value Function

Write an equation for the function graphed below.

The basic absolute value function changes direction at the origin, so this graph has been shifted to the right 3 units and down 2 units from the basic toolkit function.

We also notice that the graph appears vertically stretched, because the width of the final graph on a horizontal line is not equal to 2 times the vertical distance from the corner to this line, as it would be for an unstretched absolute value function. Instead, the width is equal to 1 times the vertical distance.

From this information we can write the equation

[latex]\begin{array}{l}f\left(x\right)=2\left|x - 3\right|-2,\hfill & \text{treating the stretch as a vertical stretch, or}\hfill \\ f\left(x\right)=\left|2\left(x - 3\right)\right|-2,\hfill & \text{treating the stretch as a horizontal compression}.\hfill \end{array}[/latex]

Note that these equations are algebraically the same—the stretch for an absolute value function can be written interchangeably as a vertical or horizontal stretch or compression.

If we couldn’t observe the stretch of the function from the graphs, could we algebraically determine it?

Yes. If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor by putting in a known pair of values for [latex]x[/latex] and [latex]f\left(x\right)[/latex].

[latex]f\left(x\right)=a|x - 3|-2[/latex]

Now substituting in the point (1, 2)

[latex]\begin{array}{l}2=a|1 - 3|-2\hfill \\ 4=2a\hfill \\ a=2\hfill \end{array}[/latex]

Write the equation for the absolute value function that is horizontally shifted left 2 units, vertically flipped, and vertically shifted up 3 units.

[latex]f\left(x\right)=-|x+2|+3\\[/latex]

Do the graphs of absolute value functions always intersect the vertical axis? The horizontal axis?

Yes, they always intersect the vertical axis. The graph of an absolute value function will intersect the vertical axis when the input is zero.

No, they do not always intersect the horizontal axis. The graph may or may not intersect the horizontal axis depending on how the graph has been shifted and reflected. It is possible for the absolute value function to intersect the horizontal axis at zero, one, or two points.

(a) The absolute value function does not intersect the horizontal axis. (b) The absolute value function intersects the horizontal axis at one point. (c) The absolute value function intersects the horizontal axis at two points.

Find the Intercepts of an Absolute Value Function

Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point.

How To: Given the formula for an absolute value function, find the horizontal intercepts of its graph.

• Isolate the absolute value term.
• Use [latex]|A|=B[/latex] to write [latex]A=B[/latex] or [latex]\mathrm{-A}=B[/latex], assuming [latex]B>0[/latex].
• Solve for [latex]x[/latex].

Example: Finding the Zeros of an Absolute Value Function

For the function [latex]f\left(x\right)=|4x+1|-7[/latex] , find the values of [latex]x[/latex] such that [latex]\text{ }f\left(x\right)=0[/latex] .

[latex]\begin{array}{l}0=|4x+1|-7\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \text{Substitute 0 for }f\left(x\right).\hfill \\ 7=|4x+1|\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \text{Isolate the absolute value on one side of the equation}.\hfill \\ \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ 7=4x+1\hfill & \text{or}\hfill & \hfill & \hfill & \hfill & -7=4x+1\hfill & \text{Break into two separate equations and solve}.\hfill \\ 6=4x\hfill & \hfill & \hfill & \hfill & \hfill & -8=4x\hfill & \hfill \\ \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ x=\frac{6}{4}=1.5\hfill & \hfill & \hfill & \hfill & \hfill & \text{ }x=\frac{-8}{4}=-2\hfill & \hfill \end{array}[/latex]

The function’s output is 0 when [latex]x=1.5[/latex] or [latex]x=-2[/latex].

For the function [latex]f\left(x\right)=|2x - 1|-3[/latex], find the values of [latex]x[/latex] such that [latex]f\left(x\right)=0[/latex].

[latex]x=-1[/latex] or [latex]x=2[/latex]

Key Concepts

• Linear functions may be graphed by plotting points or by using the y -intercept and slope.
• Graphs of linear functions may be transformed by shifting the graph up, down, left, or right as well as using stretches, compressions, and reflections.
• The y -intercept and slope of a line may be used to write the equation of a line.
• The x -intercept is the point at which the graph of a linear function crosses the x -axis.
• Horizontal lines are written in the form, [latex]f(x)=b[/latex].
• Vertical lines are written in the form, [latex]x=b[/latex].
• Parallel lines have the same slope.
• Perpendicular lines have negative reciprocal slopes, assuming neither is vertical.
• A line parallel to another line, passing through a given point, may be found by substituting the slope value of the line and the x – and y -values of the given point into the equation [latex]f\left(x\right)=mx+b[/latex] and using the b  that results. Similarly, point-slope form of an equation can also be used.
• A line perpendicular to another line, passing through a given point, may be found in the same manner, with the exception of using the negative reciprocal slope.
• The absolute value function is commonly used to measure distances between points.
• Applied problems, such as ranges of possible values, can also be solved using the absolute value function.
• The graph of the absolute value function resembles the letter V. It has a corner point at which the graph changes direction.
• In an absolute value equation, an unknown variable is the input of an absolute value function.
• If the absolute value of an expression is set equal to a positive number, expect two solutions for the unknown variable.
• An absolute value equation may have one solution, two solutions, or no solutions.
• An absolute value inequality is similar to an absolute value equation but takes the form [latex]|A|<B,|A|\le B,|A|>B,\text{ or }|A|\ge B[/latex]. It can be solved by determining the boundaries of the solution set and then testing which segments are in the set.
• Absolute value inequalities can also be solved graphically.
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4.3: Modeling with Linear Functions

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Learning Objectives

• Build linear models from verbal descriptions.
• Model a set of data with a linear function.

Emily is a college student who plans to spend a summer in Seattle. She has saved $3,500 for her trip and anticipates spending $400 each week on rent, food, and activities. How can we write a linear model to represent her situation? What would be the x-intercept, and what can she learn from it? To answer these and related questions, we can create a model using a linear function. Models such as this one can be extremely useful for analyzing relationships and making predictions based on those relationships. In this section, we will explore examples of linear function models.

Identifying Steps to Model and Solve Problems

When modeling scenarios with linear functions and solving problems involving quantities with a constant rate of change , we typically follow the same problem strategies that we would use for any type of function. Let’s briefly review them:

Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system. Carefully read the problem to identify important information. Look for information that provides values for the variables or values for parts of the functional model, such as slope and initial value. Carefully read the problem to determine what we are trying to find, identify, solve, or interpret. Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table, or even finding a formula for the function being used to model the problem. When needed, write a formula for the function. Solve or evaluate the function using the formula. Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically. Clearly convey your result using appropriate units, and answer in full sentences when necessary.

Building Linear Models

Now let’s take a look at the student in Seattle. In her situation, there are two changing quantities: time and money. The amount of money she has remaining while on vacation depends on how long she stays. We can use this information to define our variables, including units.

• Output: \(M\), money remaining, in dollars
• Input: \(t\), time, in weeks

So, the amount of money remaining depends on the number of weeks: \(M(t)\)

We can also identify the initial value and the rate of change.

• Initial Value: She saved $3,500, so $3,500 is the initial value for M.
• Rate of Change: She anticipates spending $400 each week, so –$400 per week is the rate of change, or slope.

Notice that the unit of dollars per week matches the unit of our output variable divided by our input variable. Also, because the slope is negative, the linear function is decreasing. This should make sense because she is spending money each week.

The rate of change is constant , so we can start with the linear model \(M(t)=mt+b\). Then we can substitute the intercept and slope provided.

To find the x-intercept, we set the output to zero, and solve for the input.

\[\begin{align*} 0&=−400t+3500 \\ t&=\dfrac{3500}{400} \\ &=8.75 \end{align*}\]

The x-intercept is 8.75 weeks. Because this represents the input value when the output will be zero, we could say that Emily will have no money left after 8.75 weeks.

When modeling any real-life scenario with functions, there is typically a limited domain over which that model will be valid—almost no trend continues indefinitely. Here the domain refers to the number of weeks. In this case, it doesn’t make sense to talk about input values less than zero. A negative input value could refer to a number of weeks before she saved $3,500, but the scenario discussed poses the question once she saved $3,500 because this is when her trip and subsequent spending starts. It is also likely that this model is not valid after the x-intercept, unless Emily will use a credit card and goes into debt. The domain represents the set of input values, so the reasonable domain for this function is \(0{\leq}t{\leq}8.75\).

In the above example, we were given a written description of the situation. We followed the steps of modeling a problem to analyze the information. However, the information provided may not always be the same. Sometimes we might be provided with an intercept. Other times we might be provided with an output value. We must be careful to analyze the information we are given, and use it appropriately to build a linear model.

Using a Given Intercept to Build a Model

Some real-world problems provide the y-intercept, which is the constant or initial value. Once the y-intercept is known, the x-intercept can be calculated. Suppose, for example, that Hannah plans to pay off a no-interest loan from her parents. Her loan balance is $1,000. She plans to pay $250 per month until her balance is $0. The y-intercept is the initial amount of her debt, or $1,000. The rate of change, or slope, is -$250 per month. We can then use the slope-intercept form and the given information to develop a linear model.

\[\begin{align*} f(x)&=mx+b \\ &=-250x+1000 \end{align*}\]

Now we can set the function equal to 0, and solve for \(x\) to find the x-intercept.

\[\begin{align*} 0&=-250+1000 \\ 1000&=250x \\ 4&=x \\ x&=4 \end{align*}\]

The x-intercept is the number of months it takes her to reach a balance of $0. The x-intercept is 4 months, so it will take Hannah four months to pay off her loan.

Using a Given Input and Output to Build a Model

Many real-world applications are not as direct as the ones we just considered. Instead they require us to identify some aspect of a linear function. We might sometimes instead be asked to evaluate the linear model at a given input or set the equation of the linear model equal to a specified output.

• Identify the input and output values.
• Convert the data to two coordinate pairs.
• Find the slope.
• Write the linear model.
• Use the model to make a prediction by evaluating the function at a given x-value.
• Use the model to identify an x-value that results in a given y-value.
• Answer the question posed.

Example \(\PageIndex{1}\): Using a Linear Model to Investigate a Town’s Population

A town’s population has been growing linearly. In 2004 the population was 6,200. By 2009 the population had grown to 8,100. Assume this trend continues.

• Predict the population in 2013.
• Identify the year in which the population will reach 15,000.

The two changing quantities are the population size and time. While we could use the actual year value as the input quantity, doing so tends to lead to very cumbersome equations because the y-intercept would correspond to the year 0, more than 2000 years ago!

To make computation a little nicer, we will define our input as the number of years since 2004:

• Input: \(t\), years since 2004
• Output: \(P(t)\), the town’s population

To predict the population in 2013 (\(t=9\)), we would first need an equation for the population. Likewise, to find when the population would reach 15,000, we would need to solve for the input that would provide an output of 15,000. To write an equation, we need the initial value and the rate of change, or slope.

To determine the rate of change, we will use the change in output per change in input.

\[m=\dfrac{\text{change in output}}{\text{change in input}}\]

The problem gives us two input-output pairs. Converting them to match our defined variables, the year 2004 would correspond to \(t=0\), giving the point \((0,6200)\). Notice that through our clever choice of variable definition, we have “given” ourselves the y-intercept of the function. The year 2009 would correspond to \(t=5\), giving the point \((5,8100)\).

The two coordinate pairs are \((0,6200)\) and \((5,8100)\). Recall that we encountered examples in which we were provided two points earlier in the chapter. We can use these values to calculate the slope.

\[\begin{align*} m&=\dfrac{8100-6200}{5-0}\\ &=\dfrac{1900}{5} \\ &=380 \text{ people per year} \end{align*}\]

We already know the y-intercept of the line, so we can immediately write the equation:

\[P(t)=380t+6200\]

To predict the population in 2013, we evaluate our function at \(t=9\).

\[\begin{align*} P(9)&=380(9)+6,200 \\ &=9,620 \end{align*}\]

If the trend continues, our model predicts a population of 9,620 in 2013.

To find when the population will reach 15,000, we can set \(P(t)=15000\) and solve for \(t\).

\[\begin{align*} 15000&=380t+6200 \\ 8800&=380t \\ t&{\approx}23.158 \end{align*}\]

Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere around the year 2027.

Exercise \(\PageIndex{1A}\)

A company sells doughnuts. They incur a fixed cost of $25,000 for rent, insurance, and other expenses. It costs $0.25 to produce each doughnut.

• Write a linear model to represent the cost C of the company as a function of \(x\), the number of doughnuts produced.
• Find and interpret the y-intercept.

a. \(C(x)=0.25x+25,000\) b. The y-intercept is \((0,25,000)\). If the company does not produce a single doughnut, they still incur a cost of $25,000.

Exercise \(\PageIndex{1B}\)

A city’s population has been growing linearly. In 2008, the population was 28,200. By 2012, the population was 36,800. Assume this trend continues.

• Predict the population in 2014.
• Identify the year in which the population will reach 54,000.

a. 41,100 b. 2020

Using a Diagram to Model a Problem

It is useful for many real-world applications to draw a picture to gain a sense of how the variables representing the input and output may be used to answer a question. To draw the picture, first consider what the problem is asking for. Then, determine the input and the output. The diagram should relate the variables. Often, geometrical shapes or figures are drawn. Distances are often traced out. If a right triangle is sketched, the Pythagorean Theorem relates the sides. If a rectangle is sketched, labeling width and height is helpful.

Example \(\PageIndex{2}\): Using a Diagram to Model Distance Walked

Anna and Emanuel start at the same intersection. Anna walks east at 4 miles per hour while Emanuel walks south at 3 miles per hour. They are communicating with a two-way radio that has a range of 2 miles. How long after they start walking will they fall out of radio contact?

In essence, we can partially answer this question by saying they will fall out of radio contact when they are 2 miles apart, which leads us to ask a new question:

“How long will it take them to be 2 miles apart?”

In this problem, our changing quantities are time and position, but ultimately we need to know how long will it take for them to be 2 miles apart. We can see that time will be our input variable, so we’ll define our input and output variables.

• Input: \(t\), time in hours.
• Output: \(A(t)\), distance in miles, and \(E(t)\), distance in miles

Because it is not obvious how to define our output variable, we’ll start by drawing a picture such as Figure \(\PageIndex{3}\).

• Initial Value: They both start at the same intersection so when \(t=0\), the distance traveled by each person should also be 0. Thus the initial value for each is 0.
• Rate of Change: Anna is walking 4 miles per hour and Emanuel is walking 3 miles per hour, which are both rates of change. The slope for \(A\) is 4 and the slope for \(E\) is 3.

Using those values, we can write formulas for the distance each person has walked.

\[A(t)=4t\]

\[E(t)=3t\]

For this problem, the distances from the starting point are important. To notate these, we can define a coordinate system, identifying the “starting point” at the intersection where they both started. Then we can use the variable, \(A\), which we introduced above, to represent Anna’s position, and define it to be a measurement from the starting point in the eastward direction. Likewise, can use the variable, \(E\), to represent Emanuel’s position, measured from the starting point in the southward direction. Note that in defining the coordinate system, we specified both the starting point of the measurement and the direction of measure.

We can then define a third variable, \(D\), to be the measurement of the distance between Anna and Emanuel. Showing the variables on the diagram is often helpful, as we can see from Figure \(\PageIndex{4}\).

Recall that we need to know how long it takes for \(D\), the distance between them, to equal 2 miles. Notice that for any given input \(t\), the outputs \(A(t)\), \(E(t)\), and \(D(t)\) represent distances.

Using the Pythagorean Theorem, we get:

\[\begin{align*} d(t)^2&=A(t)^2+E(t)^2 \\ &=(4t)^2+(3t)^2 \\ &=16t^2+9t^2 \\ &=25t^2 \\ D(t)&=\pm\sqrt{25t^2} &\text{Solve for $D(t)$ using the square root} \\ &= \pm 5|t| \end{align*}\]

In this scenario we are considering only positive values of \(t\), so our distance \(D(t)\) will always be positive. We can simplify this answer to \(D(t)=5t\). This means that the distance between Anna and Emanuel is also a linear function. Because D is a linear function, we can now answer the question of when the distance between them will reach 2 miles. We will set the output \(D(t)=2\) and solve for \(t\).

\[\begin{align*} D(t)&=2 \\ 5t&=2 \\ t&=\dfrac{2}{5}=0.4 \end{align*}\]

They will fall out of radio contact in 0.4 hours, or 24 minutes.

Yes. Sketch the figure and label the quantities and unknowns on the sketch.

Example \(\PageIndex{3}\): Using a Diagram to Model Distance between Cities

There is a straight road leading from the town of Westborough to Agritown 30 miles east and 10 miles north. Partway down this road, it junctions with a second road, perpendicular to the first, leading to the town of Eastborough. If the town of Eastborough is located 20 miles directly east of the town of Westborough, how far is the road junction from Westborough?

It might help here to draw a picture of the situation. See Figure \(\PageIndex{5}\). It would then be helpful to introduce a coordinate system. While we could place the origin anywhere, placing it at Westborough seems convenient. This puts Agritown at coordinates \((30, 10)\), and Eastborough at \((20,0)\).

Using this point along with the origin, we can find the slope of the line from Westborough to Agritown:

\[m=\dfrac{10-0}{30-0}=\dfrac{1}{3}\]

The equation of the road from Westborough to Agritown would be

\[W(x)=\dfrac{1}{3}x\]

From this, we can determine the perpendicular road to Eastborough will have slope \(m=–3\). Because the town of Eastborough is at the point \((20, 0)\), we can find the equation:

\[\begin{align*} E(x)&=−3x+b \\ 0&=−3(20)+b &\text{Substitute in $(20, 0)$} \\ b&=60 \\ E(x)&=−3x+60 \end{align*}\]

We can now find the coordinates of the junction of the roads by finding the intersection of these lines. Setting them equal,

\[\begin{align*} \dfrac{1}{3}x&=−3x+60 \\ \dfrac{10}{3}x&=60 \\ 10x&=180 \\ x&=18 &\text{Substituting this back into $W(x)$} \\ y&=W(18) \\ &= \dfrac{1}{3}(18) \\&=6 \end{align*}\]

The roads intersect at the point \((18,6)\). Using the distance formula, we can now find the distance from Westborough to the junction.

\[\begin{align*} \text{distance}&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ &=\sqrt{(18-0)^2+(6-0)^2} \\ &\approx 18.743 \text{miles} \end{align*}\]

One nice use of linear models is to take advantage of the fact that the graphs of these functions are lines. This means real-world applications discussing maps need linear functions to model the distances between reference points.

Exercise \(\PageIndex{2}\)

There is a straight road leading from the town of Timpson to Ashburn 60 miles east and 12 miles north. Partway down the road, it junctions with a second road, perpendicular to the first, leading to the town of Garrison. If the town of Garrison is located 22 miles directly east of the town of Timpson, how far is the road junction from Timpson?

21.15 miles

Building Systems of Linear Models

Real-world situations including two or more linear functions may be modeled with a system of linear equations . Remember, when solving a system of linear equations, we are looking for points the two lines have in common. Typically, there are three types of answers possible, as shown in Figure \(\PageIndex{6}\).

• Identify the input and output of each linear model.
• Identify the slope and y-intercept of each linear model.
• Find the solution by setting the two linear functions equal to another and solving for \(x\),or find the point of intersection on a graph.

Example \(\PageIndex{4}\): Building a System of Linear Models to Choose a Truck Rental Company

Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile[1]. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mile. When will Keep on Trucking, Inc. be the better choice for Jamal?

The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions.

• Input: \(d\), distance driven in miles
• Outputs: \(K(d):\) cost, in dollars, for renting from Keep on Trucking

\(M(d):\) cost, in dollars, for renting from Move It Your Way

• Initial Value: Up-front fee: \(K(0)=20\) and \(M(0)=16\)
• Rate of Change: \(K(d)=\dfrac{$0.59}{\text{mile}}\) and \(P(d)=\dfrac{$0.63}{\text{mile}}\)

A linear function is of the form \(f(x)=mx+b\). Using the rates of change and initial charges, we can write the equations

\[K(d)=0.59d+20 \nonumber\]

\[M(d)=0.63d+16 \nonumber\]

Using these equations, we can determine when Keep on Trucking, Inc. will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Keep on Trucking, Inc. will cost less, or when \(K(d)<M(d)\). The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the \(K(d)\) function is smaller.

These graphs are sketched in Figure \(\PageIndex{7}\), with \(K(d)\) in blue.

To find the intersection, we set the equations equal and solve:

\[\begin{align*} K(d)&=M(d) \\ 0.59d+20&=0.63d+16 \\ 4&=0.04d \\ 100&=d \\ d&=100 \end{align*}\]

This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that \(K(d)\) is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is \(d>100\).

Key Concepts

• We can use the same problem strategies that we would use for any type of function.
• When modeling and solving a problem, identify the variables and look for key values, including the slope and y-intercept.
• Draw a diagram, where appropriate.
• Check for reasonableness of the answer.
• Linear models may be built by identifying or calculating the slope and using the y-intercept.
• The x-intercept may be found by setting \(y=0\), which is setting the expression \(mx+b\) equal to 0.
• The point of intersection of a system of linear equations is the point where the x- and y-values are the same.
• A graph of the system may be used to identify the points where one line falls below (or above) the other line.

1 Rates retrieved Aug 2, 2010 from www.budgettruck.com and http://www.uhaul.com/