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The Rectangular Coordinate Systems and Graphs

Plot ordered pairs in a Cartesian coordinate system.

The Rectangular Coordinate Systems and Graphs

Plot points on a real number line (IA 1.4.7)
Plot points in a rectangular coordinate system (IA 3.1.1)

Locate Fractions and Decimals on the Number Line

We now want to include fractions and decimals on the number line.

In this example we will locate and plot the following points: \(\frac{1}{5},-\frac{4}{5},3,\frac{7}{4},-\frac{9}{2},-5\) and \(\frac{8}{3}\)

We’ll start with the whole numbers 3 and \(-5\) because they are the easiest to plot. Use zero as your starting point, move to the right for positive numbers and to the left for negative numbers.

The proper fractions listed are \(\frac{1}{5}\) and \(-\frac{4}{5}.\) We know the proper fraction \(\frac{1}{5}\) has value less than one and so would be located between 0 and 1. The denominator is 5, so imagine dividing the unit from 0 to 1 into 5 equal parts.

Similarly, \(-\frac{4}{5}\) is between 0 and \(-1.\) After dividing the unit into 5 equal parts we plot \(-\frac{4}{5}.\)

Finally, look at the improper fractions \(\frac{7}{4},-\frac{9}{2}\), and \(\frac{8}{3}.\) Locating these points may be easier if you change each of them to a mixed number or use your calculator to get a decimal approximation.

The figure below shows the number line with all the points plotted.

Try it.

\(-5,\sqrt{36},\frac{5}{3},3\frac{1}{2},-\frac{8}{10}\)

Try it.

\(-\pi ,\sqrt{25},1\frac{3}{4},0.8,2.55,-\frac{15}{16}\)

Try it.

\(e,\sqrt{5},-2\frac{1}{4},-0.75,-\frac{5}{2}\)

Condensed — the full section is in OpenStax College Algebra 2e.

Plotting Ordered Pairs in the Cartesian Coordinate System

An old story describes how seventeenth-century philosopher/mathematician René Descartes, while sick in bed, invented the system that has become the foundation of algebra. According to the story, Descartes was staring at a fly crawling on the ceiling when he realized that he could describe the fly’s location in relation to the perpendicular lines formed by the adjacent walls of his room. He viewed the perpendicular lines as horizontal and vertical axes. Further, by dividing each axis into equal unit lengths, Descartes saw that it was possible to locate any object in a two-dimensional plane using just two numbers—the displacement from the horizontal axis and the displacement from the vertical axis.

While there is evidence that ideas similar to Descartes’ grid system existed centuries earlier, it was Descartes who introduced the components that comprise the Cartesian coordinate system, a grid system having perpendicular axes. Descartes named the horizontal axis the x-axis and the vertical axis the y-axis.

The Cartesian coordinate system, also called the rectangular coordinate system, is based on a two-dimensional plane consisting of the x-axis and the y-axis. Perpendicular to each other, the axes divide the plane into four sections. Each section is called a quadrant; the quadrants are numbered counterclockwise as shown in

The center of the plane is the point at which the two axes cross. It is known as the origin, or point \((0,0).\) From the origin, each axis is further divided into equal units: increasing, positive numbers to the right on the x-axis and up the y-axis; decreasing, negative numbers to the left on the x-axis and down the y-axis. The axes extend to positive and negative infinity as shown by the arrowheads in .

Each point in the plane is identified by its x-coordinate, or horizontal displacement from the origin, and its y-coordinate, or vertical displacement from the origin. Together, we write them as an ordered pair indicating the combined distance from the origin in the form \((x,y).\) An ordered pair is also known as a coordinate pair because it consists of x- and y-coordinates. For example, we can represent the point \((3,-1)\) in the plane by moving three units to the right of the origin in the horizontal direction, and one unit down in the vertical direction. See .

Condensed — the full section is in OpenStax College Algebra 2e.

Graphing Equations by Plotting Points

We can plot a set of points to represent an equation. When such an equation contains both an x variable and a y variable, it is called an equation in two variables. Its graph is called a graph in two variables. Any graph on a two-dimensional plane is a graph in two variables.

Suppose we want to graph the equation \(y=2x-1.\) We can begin by substituting a value for x into the equation and determining the resulting value of y. Each pair of x- and y-values is an ordered pair that can be plotted. lists values of x from –3 to 3 and the resulting values for y.

\(x\)	\(y=2x-1\)	\((x,y)\)
\(-3\)	\(y=2(-3)-1=-7\)	\((-3,-7)\)
\(-2\)	\(y=2(-2)-1=-5\)	\((-2,-5)\)
\(-1\)	\(y=2(-1)-1=-3\)	\((-1,-3)\)
\(0\)	\(y=2(0)-1=-1\)	\((0,-1)\)
\(1\)	\(y=2(1)-1=1\)	\((1,1)\)
\(2\)	\(y=2(2)-1=3\)	\((2,3)\)
\(3\)	\(y=2(3)-1=5\)	\((3,5)\)

We can plot the points in the table. The points for this particular equation form a line, so we can connect them. See . This is not true for all equations.

Note that the x-values chosen are arbitrary, regardless of the type of equation we are graphing. Of course, some situations may require particular values of x to be plotted in order to see a particular result. Otherwise, it is logical to choose values that can be calculated easily, and it is always a good idea to choose values that are both negative and positive. There is no rule dictating how many points to plot, although we need at least two to graph a line. Keep in mind, however, that the more points we plot, the more accurately we can sketch the graph.

Example

Try it.

Graph the equation \(y=-x+2\) by plotting points.

Solution

First, we construct a table similar to . Choose x values and calculate y.

\(x\)	\(y=-x+2\)	\((x,y)\)
\(-5\)	\(y=-(-5)+2=7\)	\((-5,7)\)
\(-3\)	\(y=-(-3)+2=5\)	\((-3,5)\)
\(-1\)	\(y=-(-1)+2=3\)	\((-1,3)\)
\(0\)	\(y=-(0)+2=2\)	\((0,2)\)
\(1\)	\(y=-(1)+2=1\)	\((1,1)\)
\(3\)	\(y=-(3)+2=-1\)	\((3,-1)\)
\(5\)	\(y=-(5)+2=-3\)	\((5,-3)\)

Now, plot the points. Connect them if they form a line. See

Condensed — the full section is in OpenStax College Algebra 2e.

Graphing Equations with a Graphing Utility

Most graphing calculators require similar techniques to graph an equation. The equations sometimes have to be manipulated so they are written in the style \(y=_____.\) The TI-84 Plus, and many other calculator makes and models, have a mode function, which allows the window (the screen for viewing the graph) to be altered so the pertinent parts of a graph can be seen.

For example, the equation \(y=2x-20\) has been entered in the TI-84 Plus shown in a. In b, the resulting graph is shown. Notice that we cannot see on the screen where the graph crosses the axes. The standard window screen on the TI-84 Plus shows \(-10\le x\le 10,\) and \(-10\le y\le 10.\) See c.

By changing the window to show more of the positive x-axis and more of the negative y-axis, we have a much better view of the graph and the x- and y-intercepts. See a and b.

Example

Try it.

Use a graphing utility to graph the equation: \(y=-\frac{2}{3}x+\frac{4}{3}.\)

Solution

Enter the equation in the y= function of the calculator. Set the window settings so that both the x- and y- intercepts are showing in the window. See .

Finding

The intercepts of a graph are points at which the graph crosses the axes. The x-intercept is the point at which the graph crosses the x-axis. At this point, the y-coordinate is zero. The y-intercept is the point at which the graph crosses the y-axis. At this point, the x-coordinate is zero.

To determine the x-intercept, we set y equal to zero and solve for x. Similarly, to determine the y-intercept, we set x equal to zero and solve for y. For example, lets find the intercepts of the equation \(y=3x-1.\)

To find the x-intercept, set \(y=0.\)

\[\begin{array}{ll}\ y=3x-1 & \\ 0=3x-1 & \\ 1=3x & \\ \frac{1}{3}=x & \\ (\frac{1}{3},0) & x\text{-intercept}\end{array}\]

To find the y-intercept, set \(x=0.\)

\[\begin{array}{l}y=3x-1 \\ y=3(0)-1 \\ y=-1 \\ (0,-1)\ y\text{-intercept}\end{array}\]

We can confirm that our results make sense by observing a graph of the equation as in . Notice that the graph crosses the axes where we predicted it would.

Example

Try it.

Find the intercepts of the equation \(y=-3x-4.\) Then sketch the graph using only the intercepts.

Solution

Set \(y=0\) to find the x-intercept.

\[\begin{array}{l}\ y=-3x-4 \\ 0=-3x-4 \\ 4=-3x \\ -\frac{4}{3}=x \\ (-\frac{4}{3},0)\ x\text{-intercept}\end{array}\]

Set \(x=0\) to find the y-intercept.

\[\begin{array}{l}y=-3x-4 \\ y=-3(0)-4 \\ y=-4 \\ (0,-4)\ y\text{-intercept}\end{array}\]

Plot both points, and draw a line passing through them as in .

Using the Distance Formula

Derived from the Pythagorean Theorem, the distance formula is used to find the distance between two points in the plane. The Pythagorean Theorem, \({a}^{2}+{b}^{2}={c}^{2},\) is based on a right triangle where a and b are the lengths of the legs adjacent to the right angle, and c is the length of the hypotenuse. See .

The relationship of sides \(|{x}_{2}-{x}_{1}|\) and \(|{y}_{2}-{y}_{1}|\) to side d is the same as that of sides a and b to side c. We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example, \(|-3|=3.\) ) The symbols \(|{x}_{2}-{x}_{1}|\) and \(|{y}_{2}-{y}_{1}|\) indicate that the lengths of the sides of the triangle are positive. To find the length c, take the square root of both sides of the Pythagorean Theorem.

\[{c}^{2}={a}^{2}+{b}^{2}\to c=\sqrt{{a}^{2}+{b}^{2}}\]

It follows that the distance formula is given as

\[{d}^{2}={({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}\to d=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}\]

We do not have to use the absolute value symbols in this definition because any number squared is positive.

Example

Try it.

Find the distance between the points \((-3,-1)\) and \((2,3).\)

Solution

Let us first look at the graph of the two points. Connect the points to form a right triangle as in .

Then, calculate the length of d using the distance formula.

\[\begin{array}{l} \\ \begin{array}{l}d=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}} \\ d=\sqrt{{(2-(-3))}^{2}+{(3-(-1))}^{2}} \\ =\sqrt{{(5)}^{2}+{(4)}^{2}} \\ =\sqrt{25+16} \\ =\sqrt{41}\end{array}\end{array}\]

Condensed — the full section is in OpenStax College Algebra 2e.

Using the Midpoint Formula

When the endpoints of a line segment are known, we can find the point midway between them. This point is known as the midpoint and the formula is known as the midpoint formula. Given the endpoints of a line segment, \(({x}_{1},{y}_{1})\) and \(({x}_{2},{y}_{2}),\) the midpoint formula states how to find the coordinates of the midpoint \(M.\)

\[M=(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2})\]

A graphical view of a midpoint is shown in . Notice that the line segments on either side of the midpoint are congruent.

Example

Try it.

Find the midpoint of the line segment with the endpoints \((7,-2)\) and \((9,5).\)

Solution

Use the formula to find the midpoint of the line segment.

\[\begin{array}{l}(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2})=(\frac{7+9}{2},\frac{-2+5}{2}) \\ =(8,\frac{3}{2})\end{array}\]

Example

Try it.

The diameter of a circle has endpoints \((-1,-4)\) and \((5,-4).\) Find the center of the circle.

Solution

The center of a circle is the center, or midpoint, of its diameter. Thus, the midpoint formula will yield the center point.

\[\begin{array}{l}(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}) \\ (\frac{-1+5}{2},\frac{-4-4}{2})=(\frac{4}{2},-\frac{8}{2})=(2,-4)\end{array}\]

Key Concepts

We can locate, or plot, points in the Cartesian coordinate system using ordered pairs, which are defined as displacement from the x-axis and displacement from the y-axis. See .
An equation can be graphed in the plane by creating a table of values and plotting points. See .
Using a graphing calculator or a computer program makes graphing equations faster and more accurate. Equations usually have to be entered in the form y=_____. See .
Finding the x- and y-intercepts can define the graph of a line. These are the points where the graph crosses the axes. See .
The distance formula is derived from the Pythagorean Theorem and is used to find the length of a line segment. See and .
The midpoint formula provides a method of finding the coordinates of the midpoint dividing the sum of the x-coordinates and the sum of the y-coordinates of the endpoints by 2. 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.

\(-5,\sqrt{36},\frac{5}{3},3\frac{1}{2},-\frac{8}{10}\)
\(-\pi ,\sqrt{25},1\frac{3}{4},0.8,2.55,-\frac{15}{16}\)
\(e,\sqrt{5},-2\frac{1}{4},-0.75,-\frac{5}{2}\)
Plot each point in the rectangular coordinate system and identify the quadrant in which the point is located:
ⓐ \((-5,4)\) ⓑ \((-3,-4)\) ⓒ \((2,-3)\) ⓓ \((-2,3)\) ⓔ \((3,\frac{5}{2}).\)

Lafunua jibu

The first number of the coordinate pair is the x-coordinate, and the second number is the y-coordinate. To plot each point, sketch a vertical line through the x-coordinate and a horizontal line through the y-coordinate. Their intersection is the point.
ⓐ Since \(x=-5,\) the point is to the left of the y-axis. Also, since \(y=4,\) the point is above the x-axis. The point \((-5,4)\) is in Quadrant II.
ⓑ Since \(x=-3,\) the point is to the left of the y-axis. Also, since \(y=-4,\) the point is below the x-axis. The point \((-3,-4)\) is in Quadrant III.
ⓒ Since \(x=2,\) the point is to the right of the y-axis. Since \(y=-3,\) the point is below the x-axis. The point \((2,-3)\) is in Quadrant IV.
ⓓ Since \(x=-2,\) the point is to the left of the y-axis and since \(y=3\), the point is above the x-axis. The point \((-2,3)\) is in Quadrant II.
ⓔ Since \(x=3,\) the point is to the right of the y-axis. Since \(y=\frac{5}{2},\) the point is above the x-axis. (It may be helpful to write \(\frac{5}{2}\) as a mixed number or decimal.) The point \((3,\frac{5}{2})\) is in Quadrant I.
Plot the points \((-2,4),\) \((3,3),\) and \((0,-3)\) in the plane.

Lafunua jibu

To plot the point \((-2,4),\) begin at the origin. The x-coordinate is –2, so move two units to the left. The y-coordinate is 4, so then move four units up in the positive y direction.
To plot the point \((3,3),\) begin again at the origin. The x-coordinate is 3, so move three units to the right. The y-coordinate is also 3, so move three units up in the positive y direction.
To plot the point \((0,-3),\) begin again at the origin. The x-coordinate is 0. This tells us not to move in either direction along the x-axis. The y-coordinate is –3, so move three units down in the negative y direction. See the graph in .

Graph the equation \(y=-x+2\) by plotting points.

Lafunua jibu

First, we construct a table similar to . Choose x values and calculate y.

\(x\)	\(y=-x+2\)	\((x,y)\)
\(-5\)	\(y=-(-5)+2=7\)	\((-5,7)\)
\(-3\)	\(y=-(-3)+2=5\)	\((-3,5)\)
\(-1\)	\(y=-(-1)+2=3\)	\((-1,3)\)
\(0\)	\(y=-(0)+2=2\)	\((0,2)\)
\(1\)	\(y=-(1)+2=1\)	\((1,1)\)
\(3\)	\(y=-(3)+2=-1\)	\((3,-1)\)
\(5\)	\(y=-(5)+2=-3\)	\((5,-3)\)

Now, plot the points. Connect them if they form a line. See

Construct a table and graph the equation by plotting points: \(y=\frac{1}{2}x+2.\)

Lafunua jibu

\(x\)	\(y=\frac{1}{2}x+2\)	\((x,y)\)
\(-2\)	\(y=\frac{1}{2}(-2)+2=1\)	\((-2,1)\)
\(-1\)	\(y=\frac{1}{2}(-1)+2=\frac{3}{2}\)	\((-1,\frac{3}{2})\)
\(0\)	\(y=\frac{1}{2}(0)+2=2\)	\((0,2)\)
\(1\)	\(y=\frac{1}{2}(1)+2=\frac{5}{2}\)	\((1,\frac{5}{2})\)
\(2\)	\(y=\frac{1}{2}(2)+2=3\)	\((2,3)\)

Use a graphing utility to graph the equation: \(y=-\frac{2}{3}x+\frac{4}{3}.\)

Lafunua jibu

Enter the equation in the y= function of the calculator. Set the window settings so that both the x- and y- intercepts are showing in the window. See .
Find the intercepts of the equation \(y=-3x-4.\) Then sketch the graph using only the intercepts.

Lafunua jibu

Set \(y=0\) to find the x-intercept.
\[\begin{array}{l}\ y=-3x-4 \\ 0=-3x-4 \\ 4=-3x \\ -\frac{4}{3}=x \\ (-\frac{4}{3},0)\ x\text{-intercept}\end{array}\]
Set \(x=0\) to find the y-intercept.
\[\begin{array}{l}y=-3x-4 \\ y=-3(0)-4 \\ y=-4 \\ (0,-4)\ y\text{-intercept}\end{array}\]
Plot both points, and draw a line passing through them as in .
Find the intercepts of the equation and sketch the graph: \(y=-\frac{3}{4}x+3.\)

Lafunua jibu

x-intercept is \((4,0);\) y-intercept is \((0,3).\)
Find the distance between the points \((-3,-1)\) and \((2,3).\)

Lafunua jibu

Let us first look at the graph of the two points. Connect the points to form a right triangle as in .
Then, calculate the length of d using the distance formula.
\[\begin{array}{l} \\ \begin{array}{l}d=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}} \\ d=\sqrt{{(2-(-3))}^{2}+{(3-(-1))}^{2}} \\ =\sqrt{{(5)}^{2}+{(4)}^{2}} \\ =\sqrt{25+16} \\ =\sqrt{41}\end{array}\end{array}\]
Find the distance between two points: \((1,4)\) and \((11,9).\)

Lafunua jibu

\(\sqrt{125}=5\sqrt{5}\)
Let’s return to the situation introduced at the beginning of this section.
Tracie set out from Elmhurst, IL, to go to Franklin Park. On the way, she made a few stops to do errands. Each stop is indicated by a red dot in . Find the total distance that Tracie traveled. Compare this with the distance between her starting and final positions.
Lafunua jibu
The first thing we should do is identify ordered pairs to describe each position. If we set the starting position at the origin, we can identify each of the other points by counting units east (right) and north (up) on the grid. For example, the first stop is 1 block east and 1 block north, so it is at \((1,1).\) The next stop is 5 blocks to the east, so it is at \((5,1).\) After that, she traveled 3 blocks east and 2 blocks north to \((8,3).\) Lastly, she traveled 4 blocks north to \((8,7).\) We can label these points on the grid as in .
Next, we can calculate the distance. Note that each grid unit represents 1,000 feet.
- From her starting location to her first stop at \((1,1),\) Tracie might have driven north 1,000 feet and then east 1,000 feet, or vice versa. Either way, she drove 2,000 feet to her first stop.
- Her second stop is at \((5,1).\) So from \((1,1)\) to \((5,1),\) Tracie drove east 4,000 feet.
- Her third stop is at \((8,3).\) There are a number of routes from \((5,1)\) to \((8,3).\) Whatever route Tracie decided to use, the distance is the same, as there are no angular streets between the two points. Let’s say she drove east 3,000 feet and then north 2,000 feet for a total of 5,000 feet.
- Tracie’s final stop is at \((8,7).\) This is a straight drive north from \((8,3)\) for a total of 4,000 feet.
Next, we will add the distances listed in .
From/To Number of Feet Driven
\((0,0)\) to \((1,1)\) 2,000
\((1,1)\) to \((5,1)\) 4,000
\((5,1)\) to \((8,3)\) 5,000
\((8,3)\) to \((8,7)\) 4,000
Total 15,000
The total distance Tracie drove is 15,000 feet, or 2.84 miles. This is not, however, the actual distance between her starting and ending positions. To find this distance, we can use the distance formula between the points \((0,0)\) and \((8,7).\)
\[\begin{array}{l}d=\sqrt{{(8-0)}^{2}+{(7-0)}^{2}} \\ =\sqrt{64+49} \\ =\sqrt{113} \\ \approx 10.63\ \text{units}\end{array}\]
At 1,000 feet per grid unit, the distance between Elmhurst, IL, to Franklin Park is 10,630.14 feet, or 2.01 miles. The distance formula results in a shorter calculation because it is based on the hypotenuse of a right triangle, a straight diagonal from the origin to the point \((8,7).\) Perhaps you have heard the saying “as the crow flies,” which means the shortest distance between two points because a crow can fly in a straight line even though a person on the ground has to travel a longer distance on existing roadways.
Find the midpoint of the line segment with the endpoints \((7,-2)\) and \((9,5).\)

Lafunua jibu

Use the formula to find the midpoint of the line segment.
\[\begin{array}{l}(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2})=(\frac{7+9}{2},\frac{-2+5}{2}) \\ =(8,\frac{3}{2})\end{array}\]
Find the midpoint of the line segment with endpoints \((-2,-1)\) and \((-8,6).\)

Lafunua jibu

\((-5,\frac{5}{2})\)
The diameter of a circle has endpoints \((-1,-4)\) and \((5,-4).\) Find the center of the circle.

Lafunua jibu

The center of a circle is the center, or midpoint, of its diameter. Thus, the midpoint formula will yield the center point.
\[\begin{array}{l}(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}) \\ (\frac{-1+5}{2},\frac{-4-4}{2})=(\frac{4}{2},-\frac{8}{2})=(2,-4)\end{array}\]
Is it possible for a point plotted in the Cartesian coordinate system to not lie in one of the four quadrants? Explain.

Lafunua jibu

Answers may vary. Yes. It is possible for a point to be on the x-axis or on the y-axis and therefore is considered to NOT be in one of the quadrants.
Describe the process for finding the x-intercept and the y-intercept of a graph algebraically.
Describe in your own words what the y-intercept of a graph is.

Lafunua jibu

The y-intercept is the point where the graph crosses the y-axis.
When using the distance formula \(d=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}},\) explain the correct order of operations that are to be performed to obtain the correct answer.
\(y=-3x+6\)

Lafunua jibu

The x-intercept is \((2,0)\) and the y-intercept is \((0,6).\)
\(4y=2x-1\)
\(3x-2y=6\)

Lafunua jibu

The x-intercept is \((2,0)\) and the y-intercept is \((0,-3).\)
\(4x-3=2y\)
\(3x+8y=9\)

Lafunua jibu

The x-intercept is \((3,0)\) and the y-intercept is \((0,\frac{9}{8}).\)
\(2x-\frac{2}{3}=\frac{3}{4}y+3\)
\(4x+2y=8\)

Lafunua jibu

\(y=4-2x\)
\(3x-2y=6\)
\(2x=5-3y\)

Lafunua jibu

\(y=\frac{5-2x}{3}\)
\(x-2y=7\)
\(5y+4=10x\)

Lafunua jibu

\(y=2x-\frac{4}{5}\)
\(5x+2y=0\)
\((-4,1)\) and \((3,-4)\)

Lafunua jibu

\(d=\sqrt{74}\)
\((2,-5)\) and \((7,4)\)
\((5,0)\) and \((5,6)\)

Lafunua jibu

\(d=\sqrt{36}=6\)
\((-4,3)\) and \((10,3)\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

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

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

|x|

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

imaginary unit
i² = −1.

\log_b x,\ \ln x

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

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: The Rectangular Coordinate Systems and Graphs

Plot ordered pairs in a Cartesian coordinate system.
Graph equations by plotting points.
Graph equations with a graphing utility.
Find x-intercepts and y-intercepts.
Use the distance formula.
Use the midpoint formula.
Plot points on a real number line (IA 1.4.7)
Plot points in a rectangular coordinate system (IA 3.1.1)

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Jaribu kufanya mambo yako mwenyewe

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

Mengi zaidi katika Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value

From/To	Number of Feet Driven
\((0,0)\) to \((1,1)\)	2,000
\((1,1)\) to \((5,1)\)	4,000
\((5,1)\) to \((8,3)\)	5,000
\((8,3)\) to \((8,7)\)	4,000
Total	15,000