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Graphs of Linear Functions

Graph linear functions.

Graphing Linear Functions

In Linear Functions, we 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. By graphing two functions, then, we can more easily compare their characteristics.

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. And the third is by using transformations of the identity function $f(x)=x.$

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. For example, given the function, $f(x)=2x,$ 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.

Example

Try it.

Graph $f(x)=-\frac{2}{3}x+5$ by plotting points.

Solution

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.

\[\begin{array}{lll}x=0 & & f(0)=-\frac{2}{3}(0)+5=5⇒(0,5) \\ x=3 & & f(3)=-\frac{2}{3}(3)+5=3⇒(3,3) \\ x=6 & & f(6)=-\frac{2}{3}(6)+5=1⇒(6,1)\end{array}\]

Plot the coordinate pairs and draw a line through the points. represents the graph of the function $f(x)=-\frac{2}{3}x+5.$

Condensed — the full section is in OpenStax Precalculus 2e.

Writing the Equation for a Function from the Graph of a Line

Recall that in Linear Functions, we 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 . We can see right away that the graph crosses the y-axis at the point $(0,\text{ 4})$ 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

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

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

\[y=2x+4\]

Example

Try it.

Match each equation of the linear functions with one of the lines in .

ⓐ $f(x)=2x+3$
ⓑ $g(x)=2x-3$
ⓒ $h(x)=-2x+3$
ⓓ $j(x)=\frac{1}{2}x+3$

Solution

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 $(0,\text{ 3})$ 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 $\frac{1}{2}$ 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 $(0,\text{ 3}),$ 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 as in .

Finding the

So far, we have been finding the y-intercepts of a function: the point at which the graph of the 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 the 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 a function $f(x)$ equal to zero and solve for the value of $x.$ For example, consider the function shown.

\[f(x)=3x-6\]

Set the function equal to 0 and solve for $x.$

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

The graph of the function crosses the x-axis at the point $(2,\text{ 0}).$

Example

Try it.

Find the x-intercept of $f(x)=\frac{1}{2}x-3.$

Solution

Set the function equal to zero to solve for $x.$

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

The graph crosses the x-axis at the point $(6,\text{ 0}).$

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 , we see that the output has a value of 2 for every input value. The change in outputs between any two points, therefore, is 0. In the slope formula, the numerator is 0, so the slope is 0. If we use $m=0$ in the equation $f(x)=mx+b,$ the equation simplifies to $f(x)=b.$ In other words, the value of the function is a constant. This graph represents the function $f(x)=2.$

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, such as the one in , has an x-intercept, but no y-intercept unless it’s the line $x=0.$ This graph represents the line $x=2.$

Example

Try it.

Write the equation of the line graphed in .

Solution

For any x-value, the y-value is $-4,$ so the equation is $y=-4.$

Example

Try it.

Write the equation of the line graphed in .

Solution

The constant x-value is $7,$ so the equation is $x=7.$

Determining Whether Lines are Parallel or Perpendicular

The two lines in 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.

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.

\[\begin{array}{ll}\begin{array}{l}f(x)=-2x+6 \\ f(x)=-2x-4\end{array}\}\ \text{parallel}\ & \begin{array}{l}f(x)=3x+2 \\ f(x)=2x+2\end{array}\}\ \text{not parallel}\end{array}\]

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

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. So, if ${m}_{1}$ and ${m}_{2}$ are negative reciprocals of one another, they can be multiplied together to yield –1.

\[{m}_{1}{m}_{2}=-1\]

To find the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is $\frac{1}{8},$ and the reciprocal of $\frac{1}{8}$ 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, assuming that the lines are neither horizontal nor vertical. The slope of each line below is the negative reciprocal of the other so the lines are perpendicular.

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

The product of the slopes is –1.

\[-4(\frac{1}{4})=-1\]

Condensed — the full section is in OpenStax Precalculus 2e.

Writing the Equation of a Line Parallel or Perpendicular to a Given Line

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 for example, we are given the following equation.

\[f(x)=3x+1\]

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

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

Suppose then we want to write the equation of a line that is parallel to $f$ and passes through the point $(1,\text{ 7}).$ 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 with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form.

\[\begin{array}{l}y-{y}_{1}=m(x-{x}_{1}) \\ y-7=3(x-1) \\ y-7=3x-3 \\ y=3x+4\end{array}\]

So $g(x)=3x+4$ is parallel to $f(x)=3x+1$ and passes through the point $(1,\text{ 7}).$

Example

Try it.

Find a line parallel to the graph of $f(x)=3x+6$ that passes through the point $(3,\text{ 0}).$

Solution

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

\[\begin{array}{l}g(x)=3x+b \\ 0=3(3)+b \\ b=-9\end{array}\]

The line parallel to $f(x)$ that passes through $(3,\text{ 0})$ is $g(x)=3x-9.$

Condensed — the full section is in OpenStax Precalculus 2e.

Solving a System of Linear Equations Using a Graph

A system of linear equations includes two or more linear equations. The graphs of two lines will intersect at a single point if they are not parallel. Two parallel lines can also intersect if they are coincident, which means they are the same line and they intersect at every point. For two lines that are not parallel, the single point of intersection will satisfy both equations and therefore represent the solution to the system.

To find this point when the equations are given as functions, we can solve for an input value so that $f(x)=g(x).$ In other words, we can set the formulas for the lines equal to one another, and solve for the input that satisfies the equation.

Example

Try it.

Find the point of intersection of the lines $h(t)=3t-4$ and $j(t)=5-t.$

Solution

Set $h(t)=j(t).$

\[\begin{array}{l}3t-4=5-t \\ 4t=9 \\ t=\frac{9}{4}\end{array}\]

This tells us the lines intersect when the input is $\frac{9}{4}.$

We can then find the output value of the intersection point by evaluating either function at this input.

\[\begin{array}{l}\begin{array}{l} \\ j(\frac{9}{4})=5-\frac{9}{4}\end{array} \\ =\frac{11}{4}\end{array}\]

These lines intersect at the point $(\frac{9}{4},\frac{11}{4}).$

Condensed — the full section is in OpenStax Precalculus 2e.

Key Concepts

Linear functions may be graphed by plotting points or by using the y-intercept and slope. See and .
Graphs of linear functions may be transformed by using shifts up, down, left, or right, as well as through stretches, compressions, and reflections. See .
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. See and .
Horizontal lines are written in the form, $f(x)=b.$ See .
Vertical lines are written in the form, $x=b.$ See .
Parallel lines have the same slope.
Perpendicular lines have negative reciprocal slopes, assuming neither is vertical. See .
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, $f(x)=mx+b,$ and using the $b$ that results. Similarly, the point-slope form of an equation can also be used. See .
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. See and .
A system of linear equations may be solved setting the two equations equal to one another and solving for $x.$ The y-value may be found by evaluating either one of the original equations using this x-value.
A system of linear equations may also be solved by finding the point of intersection on a graph. See and .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Graph $f(x)=-\frac{2}{3}x+5$ by plotting points.

ເປີດເຜີຍຄຳຕອບ

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.
\[\begin{array}{lll}x=0 & & f(0)=-\frac{2}{3}(0)+5=5⇒(0,5) \\ x=3 & & f(3)=-\frac{2}{3}(3)+5=3⇒(3,3) \\ x=6 & & f(6)=-\frac{2}{3}(6)+5=1⇒(6,1)\end{array}\]
Plot the coordinate pairs and draw a line through the points. represents the graph of the function $f(x)=-\frac{2}{3}x+5.$
Graph $f(x)=-\frac{3}{4}x+6$ by plotting points.
Graph $f(x)=-\frac{2}{3}x+5$ using the y-intercept and slope.

ເປີດເຜີຍຄຳຕອບ

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 $-\frac{2}{3}.$ This tells us that for each vertical decrease in the “rise” of $-2$ units, the “run” increases by 3 units in the horizontal direction. We can now graph the function by first plotting the y-intercept on the graph in . 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.
Find a point on the graph we drew in that has a negative x-value.

ເປີດເຜີຍຄຳຕອບ

Possible answers include $(-3,7),$ $(-6,9),$ or $(-9,11).$
Graph $f(x)=\frac{1}{2}x-3$ using transformations.

ເປີດເຜີຍຄຳຕອບ

The equation for the function shows that $m=\frac{1}{2}$ so the identity function is vertically compressed by $\frac{1}{2}.$ The equation for the function also shows that $b=-3$ so the identity function is vertically shifted down 3 units. First, graph the identity function, and show the vertical compression as in .
Then show the vertical shift as in .
Graph $f(x)=4+2x,$ using transformations.
Match each equation of the linear functions with one of the lines in .
1. ⓐ $f(x)=2x+3$
2. ⓑ $g(x)=2x-3$
3. ⓒ $h(x)=-2x+3$
4. ⓓ $j(x)=\frac{1}{2}x+3$
ເປີດເຜີຍຄຳຕອບ
Analyze the information for each function.
1. ⓐ 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 $(0,\text{ 3})$ so $f$ must be represented by Line I.
2. ⓑ 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.
3. ⓒ 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.
4. ⓓ This function has a slope of $\frac{1}{2}$ 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 $(0,\text{ 3}),$ 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 as in .
Find the x-intercept of $f(x)=\frac{1}{2}x-3.$

ເປີດເຜີຍຄຳຕອບ

Set the function equal to zero to solve for $x.$
\[\begin{array}{l}0=\frac{1}{2}x-3 \\ 3=\frac{1}{2}x \\ 6=x \\ x=6\end{array}\]
The graph crosses the x-axis at the point $(6,\text{ 0}).$
Find the x-intercept of $f(x)=\frac{1}{4}x-4.$

ເປີດເຜີຍຄຳຕອບ

$(16,\text{ 0})$
Write the equation of the line graphed in .

ເປີດເຜີຍຄຳຕອບ

For any x-value, the y-value is $-4,$ so the equation is $y=-4.$
Write the equation of the line graphed in .

ເປີດເຜີຍຄຳຕອບ

The constant x-value is $7,$ so the equation is $x=7.$
Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines.
\[\begin{array}{lll}f(x)=2x+3 & & h(x)=-2x+2 \\ g(x)=\frac{1}{2}x-4 & & \ j(x)=2x-6\end{array}\]

ເປີດເຜີຍຄຳຕອບ

Parallel lines have the same slope. Because the functions $f(x)=2x+3$ and $j(x)=2x-6$ each have a slope of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because −2 and $\frac{1}{2}$ are negative reciprocals, the equations, $g(x)=\frac{1}{2}x-4$ and $h(x)=-2x+2$ represent perpendicular lines.
Find a line parallel to the graph of $f(x)=3x+6$ that passes through the point $(3,\text{ 0}).$

ເປີດເຜີຍຄຳຕອບ

The slope of the given line is 3. If we choose the slope-intercept form, we can substitute $m=3,$ $x=3,$ and $f(x)=0$ into the slope-intercept form to find the y-intercept.
\[\begin{array}{l}g(x)=3x+b \\ 0=3(3)+b \\ b=-9\end{array}\]
The line parallel to $f(x)$ that passes through $(3,\text{ 0})$ is $g(x)=3x-9.$
Find the equation of a line perpendicular to $f(x)=3x+3$ that passes through the point $(3,\text{ 0}).$

ເປີດເຜີຍຄຳຕອບ

The original line has slope $m=3,$ so the slope of the perpendicular line will be its negative reciprocal, or $-\frac{1}{3}.$ Using this slope and the given point, we can find the equation for the line.
\[\begin{array}{l}g(x)=-\frac{1}{3}x+b \\ 0=-\frac{1}{3}(3)+b \\ 1=b \\ b=1\end{array}\]
The line perpendicular to $f(x)$ that passes through $(3,\text{ 0})$ is $g(x)=-\frac{1}{3}x+1.$
Given the function $h(x)=2x-4,$ write an equation for the line passing through $(0,0)$ that is
1. ⓐ parallel to $h(x)$
2. ⓑ perpendicular to $h(x)$
ເປີດເຜີຍຄຳຕອບ
1. ⓐ $f(x)=2x$
2. ⓑ $g(x)=-\frac{1}{2}x$
A line passes through the points $(-2,\text{ 6})$ and $(4,5).$ Find the equation of a perpendicular line that passes through the point $(4,5).$

ເປີດເຜີຍຄຳຕອບ

From the two points of the given line, we can calculate the slope of that line.
\[\begin{array}{l}{m}_{1}=\frac{5-6}{4-(-2)} \\ =\frac{-1}{6} \\ =-\frac{1}{6}\end{array}\]
Find the negative reciprocal of the slope.
\[\begin{array}{l}{m}_{2}=\frac{-1}{-\frac{1}{6}} \\ =-1(-\frac{6}{1}) \\ =6\end{array}\]
We can then solve for the y-intercept of the line passing through the point $(4,5).$
\[\begin{array}{l}g(x)=6x+b \\ 5=6(4)+b \\ 5=24+b \\ -19=b \\ b=-19\end{array}\]
The equation for the line that is perpendicular to the line passing through the two given points and also passes through point $(4,5)$ is
\[y=6x-19\]
A line passes through the points, $(-2,\text{-15})$ and $(2,-3).$ Find the equation of a perpendicular line that passes through the point, $(6,4).$

ເປີດເຜີຍຄຳຕອບ

$y=-\frac{1}{3}x+6$
Find the point of intersection of the lines $h(t)=3t-4$ and $j(t)=5-t.$

ເປີດເຜີຍຄຳຕອບ

Set $h(t)=j(t).$
\[\begin{array}{l}3t-4=5-t \\ 4t=9 \\ t=\frac{9}{4}\end{array}\]
This tells us the lines intersect when the input is $\frac{9}{4}.$
We can then find the output value of the intersection point by evaluating either function at this input.
\[\begin{array}{l}\begin{array}{l} \\ j(\frac{9}{4})=5-\frac{9}{4}\end{array} \\ =\frac{11}{4}\end{array}\]
These lines intersect at the point $(\frac{9}{4},\frac{11}{4}).$
Look at the graph in and identify the following for the function $j(t):$
1. ⓐ y-intercept
2. ⓑ x-intercept(s)
3. ⓒ slope
4. ⓓ Is $j(t)$ parallel or perpendicular to $h(t)$ (or neither)?
5. ⓔ Is $j(t)$ an increasing or decreasing function (or neither)?
6. ⓕ Write a transformation description for $j(t)$ from the identity toolkit function $f(x)=x.$
ເປີດເຜີຍຄຳຕອບ
1. ⓐ $(0,5)$
2. ⓑ $(5,\text{ 0})$
3. ⓒ Slope -1
4. ⓓ Neither parallel nor perpendicular
5. ⓔ Decreasing function
6. ⓕ Given the identity function, perform a vertical flip (over the t-axis) and shift up 5 units.
A company sells sports helmets. The company incurs a one-time fixed cost for $250,000. Each helmet costs $120 to produce, and sells for $140.
1. ⓐ Find the cost function, $C,$ to produce $x$ helmets, in dollars.
2. ⓑ Find the revenue function, $R,$ from the sales of $x$ helmets, in dollars.
3. ⓒ Find the break-even point, the point of intersection of the two graphs $C$ and $R.$
ເປີດເຜີຍຄຳຕອບ
1. ⓐ The cost function is the sum of the fixed cost, $250,000, and the variable cost, $120 per helmet. \[C(x)=120x+250,000\]
2. ⓑ The revenue function is the total revenue from the sale of $x$ helmets, $R(x)=140x.$
3. ⓒ The break-even point is the point of intersection of the graph of the cost and revenue functions. To find the x-coordinate of the coordinate pair of the point of intersection, set the two equations equal, and solve for $x.$ \[\begin{array}{l}\ C(x)=R(x) \\ 250,000+120x=140x \\ 250,000=20x \\ 12,500=x \\ x=12,500\end{array}\]
  To find $y,$ evaluate either the revenue or the cost function at 12,500.
  \[\begin{array}{l}R(12,500)=140(12,500) \\ =\$1,750,000\end{array}\]
The break-even point is $(12,500,1,750,000).$
If the graphs of two linear functions are parallel, describe the relationship between the slopes and the y-intercepts.

ເປີດເຜີຍຄຳຕອບ

The slopes are equal; y-intercepts are not equal.
If the graphs of two linear functions are perpendicular, describe the relationship between the slopes and the y-intercepts.
If a horizontal line has the equation $f(x)=a$ and a vertical line has the equation $x=a,$ what is the point of intersection? Explain why what you found is the point of intersection.

ເປີດເຜີຍຄຳຕອບ

The point of intersection is $(a,a).$ This is because for the horizontal line, all of the $y$ coordinates are $a$ and for the vertical line, all of the $x$ coordinates are $a.$ The point of intersection will have these two characteristics.
Explain how to find a line parallel to a linear function that passes through a given point.
Explain how to find a line perpendicular to a linear function that passes through a given point.

ເປີດເຜີຍຄຳຕອບ

First, find the slope of the linear function. Then take the negative reciprocal of the slope; this is the slope of the perpendicular line. Substitute the slope of the perpendicular line and the coordinate of the given point into the equation $y=mx+b$ and solve for $b.$ Then write the equation of the line in the form $y=mx+b$ by substituting in $m$ and $b.$
$\begin{array}{l}4x-7y=10 \\ 7x+4y=1\end{array}$
$\begin{array}{l}3y+x=12 \\ -y=8x+1\end{array}$

ເປີດເຜີຍຄຳຕອບ

neither parallel or perpendicular
$\begin{array}{l}3y+4x=12 \\ -6y=8x+1\end{array}$
$\begin{array}{l}6x-9y=10 \\ 3x+2y=1\end{array}$

ເປີດເຜີຍຄຳຕອບ

perpendicular
$\begin{array}{l}y=\frac{2}{3}x+1 \\ 3x+2y=1\end{array}$
$\begin{array}{l}y=\frac{3}{4}x+1 \\ -3x+4y=1\end{array}$

ເປີດເຜີຍຄຳຕອບ

parallel
$f(x)=-x+2$
$g(x)=2x+4$

ເປີດເຜີຍຄຳຕອບ

$(-2\text{, }0)$; $(0\text{, 4})$
$h(x)=3x-5$
$k(x)=-5x+1$

ເປີດເຜີຍຄຳຕອບ

$(\frac{1}{5}\text{, }0)$; $(0\text{, 1})$
$-2x+5y=20$
$7x+2y=56$

ເປີດເຜີຍຄຳຕອບ

$(8\text{, }0)$; $(0\text{, }28)$
- Line 1: Passes through $(0,6)$ and $(3,-24)$
- Line 2: Passes through $(-1,19)$ and $(8,-71)$
- Line 1: Passes through $(-8,-55)$ and $(10,\ 89)$
- Line 2: Passes through $(9,-44)$ and $(4,-14)$
ເປີດເຜີຍຄຳຕອບ

$\text{Line 1}:m=8$
$\text{Line 2}:\ m=-6$
$\text{Neither}$
- Line 1: Passes through $(2,3)$ and $(4,-1)$
- Line 2: Passes through $(6,3)$ and $(8,5)$

Symbols used here

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

imaginary unit
i² = −1.

\theta

theta
The usual name for an angle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

How to: Graphs of Linear Functions

Graph linear functions.
Write the equation for a linear function from the graph of a line.
Given the equations of two lines, determine whether their graphs are parallel or perpendicular.
Write the equation of a line parallel or perpendicular to a given line.
Solve a system of linear equations.
Choose a minimum of two input values.
Evaluate the function at each input value.
Use the resulting output values to identify coordinate pairs.

Questions people ask

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

ພະຍາຍາມເອງ

Parts of this page are adapted from OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

ເພີ່ມເຕີມໃນ Precalculus

Complex numbers Polynomial functions Rational functions Sequences and series The binomial theorem Conic sections Vectors Exponential and logarithmic functions Polynomial division and the remainder theorem Parametric equations and polar coordinates

Graphing Linear Functions

Writing the Equation for a Function from the Graph of a Line

Finding the

Describing Horizontal and Vertical Lines

Determining Whether Lines are Parallel or Perpendicular

Writing the Equation of a Line Parallel or Perpendicular to a Given Line

Solving a System of Linear Equations Using a Graph

Key Concepts

Practice (40)

Symbols used here

How to: Graphs of Linear Functions

Questions people ask

ພະຍາຍາມ​ເອງ

ເພີ່ມເຕີມໃນ Precalculus

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