Open Resources for Community College Algebra

Subsection 9.1.1 Cartesian Coordinates

A Cartesian coordinate system is usually represented as a graph where there are two “axes” (straight lines extending infinitely), one horizontal and one vertical. The horizontal axis is usually labeled “\(x\)”, and the vertical axis is usually labeled “\(y\)”. The scale marks on the axes are used to define locations on the plane that they span. An address is a pair of “coordinates” written as in this example: \((3,2)\text{.}\) The first coordinate tells you where the location is with respect to the horizontal axis, and the second coordinate tells you a location with respect to the vertical axis. The point \((3,2)\) is marked in Figure 9.1.2.

epspdfpngsvgtex

Example 9.1.3.

On paper, sketch a Cartesian coordinate system with the axes scaled using regularly spaced ticks and labels, and then plot the following points: \((2,3),(-5,1),(0,-3),(4,0)\text{.}\)

Explanation

epspdfpngsvgtex

Note that negative numbers in the first coordinate mean that a point is left of the \(y\)-axis. And similarly, negative numbers in the second coordinate mean that a point is below the \(x\)-axis.

Subsection 9.1.2 Graphing Equations by Plotting Points

When you have an equation in the form

it suggests that we could substitute in different values for \(x\) and get back different values for \(y\text{.}\) Pairing these \(x\)- and \(y\)-values together, we can plot points and create a “graph” of the equation. Creating a graph of a given equation in \(x\) and \(y\) is the basic objective of this section. Sometimes the equation has special features that give you a shortcut for creating a graph, for example as discussed in Section 3.5. However, here we want to focus on the universal approach of just substituting in values for \(x\) and seeing what comes out.

Example 9.1.5.

Rheema helped plant a lovely Douglas Fir in a local park volunteering with Portland's Friends of Trees ¹friendsoftrees.org/. The tree they planted was 4 ft tall when they planted it. Rheema watched the tree grow over the next few years and noticed that every year, the tree grew about 1.5 ft. So, the height of the tree can be found by using the formula \(y=1.5x+4\text{,}\) where \(x\)-values represent the number of years since the tree was planted. Let's make a graph of this equation by making a table of values. The most straightforward method to graph any equation is to build a table of \(x\)- and \(y\)-values, and then plot the points.

\(x\)	\(y=1.5x+4\)	Point	Interpretation
\(0\)	\(1.5(0)+4\) \({}=4\)	\((0,4)\)	When the tree was planted, the tree was 4 ft tall.
\(2\)	\(1.5(2)+4\) \({}=7\)	\((2,7)\)	Two years after tree was planted, the tree was 7 ft tall.
\(4\)	\(1.5(4)+4\) \({}=10\)	\((4,10)\)	Four years after tree was planted, the tree was 10 ft tall.
\(6\)	\(1.5(6)+4\) \({}=13\)	\((6,13)\)	Six years after tree was planted, the tree was 13 ft tall.
\(8\)	\(1.5(8)+4\) \({}=16\)	\((8,16)\)	Eight years after tree was planted, the tree was 16 ft tall.

epspdfpngsvgtex

Example 9.1.8.

Make a graph of the linear equation \(y=\frac{5}{2}x-3\) by building a table of \(x\)- and \(y\)-values and plotting the points.

Explanation

To create an easy-to-graph table of values, we should examine the formula and notice that if all of the \(x\)-values were multiples of \(2\text{,}\) then the fraction in the equation would cancel nicely and leave us with integer \(y\)-values.

\(x\)	\(y=\frac{5}{2}x-3\)	Point
\(-4\)	\(\frac{5}{2}(\substitute{-4})-3\) \({}=-13\)	\((-4,-13)\)
\(-2\)	\(\frac{5}{2}(\substitute{-2})-3\) \({}=-8\)	\((-2,-8)\)
\(0\)	\(\frac{5}{2}(\substitute{0})-3\) \({}=-3\)	\((0,-3)\)
\(2\)	\(\frac{5}{2}(\substitute{2})-3\) \({}=2\)	\((2,2)\)
\(4\)	\(\frac{5}{2}(\substitute{4})-3\) \({}=7\)	\((4,7)\)
\(6\)	\(\frac{5}{2}(\substitute{6})-3\) \({}=12\)	\((6,12)\)

epspdfpngsvgtex

Example 9.1.11.

Create a table of ordered pairs and then make a plot of the equation \(y=-\frac{2}{5}x-3\text{.}\) Note that this equation is a linear equation, and we can see that the slope is negative. Therefore we should expect to see a downward sloping line as we view it from left to right. If we don't see that in the end, it suggests some mistake was made.

Explanation

This time, with the slope having denominator \(5\text{,}\) it is wise to use multiples of \(5\) as the \(x\)-values.

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

epspdfpngsvgtex

Even when the equation is not a linear equation, this method (making a table of points) will work to help create a graph.

Example 9.1.13.

Create a table of ordered pairs and then make a plot of the equation \(y=\frac{6}{x^2-2x+2}\text{.}\)

Explanation

The form of this equation is not one that we recognize yet. But the general approach for making a graph is still going to work out.

\(x\)	\(y=\frac{6}{x^2-2x+2}\)	Point
\(-2\)	\(\frac{6}{(\substitute{-2})^2-2(\substitute{-2})+2}\) \({}=0.6\)	\((-2,0.6)\)
\(-1\)	\(\frac{6}{(\substitute{-1})^2-2(\substitute{-1})+2}\) \({}=1.2\)	\((-1,1.2)\)
\(0\)	\(\frac{6}{(\substitute{0})^2-2(\substitute{0})+2}\) \({}=3\)	\((0,3)\)
\(1\)	\(\frac{6}{(\substitute{1})^2-2(\substitute{1})+2}\) \({}=6\)	\((1,6)\)
\(2\)	\(\frac{6}{(\substitute{2})^2-2(\substitute{2})+2}\) \({}=3\)	\((2,3)\)
\(3\)	\(\frac{6}{(\substitute{3})^2-2(\substitute{3})+2}\) \({}=1.2\)	\((3,1.2)\)

epspdfpngsvgtex

Subsection 9.1.3 Graphing Lines Using Intercepts

As noted earlier, sometimes the form of an equation suggests an alternative way that we could graph it. In the case of a linear equation, an alternative to making a table is to find the line's “intercepts”. These are the locations where the line crosses either the \(x\)-axis or the \(y\)-axis. In the case of a straight line, that is theoretically all you need to graph the complete line. Here, we review this approach. We also hope this example simply serves as a reminder of what intercepts are.

Recall that the standard form (3.7.1) of a line equation is \(Ax+By=C\) where where \(A\text{,}\) \(B\text{,}\) and \(C\) are three numbers (each of which might be \(0\text{,}\) although at least one of \(A\) and \(B\) must be nonzero). If a linear equation is given in standard form, we can relative easily find the line's \(x\)- and \(y\)-intercepts by substituting in \(y=0\) and \(x=0\text{,}\) respectively.

Example 9.1.14.

Find the intercepts of \(3x+5y=60\text{,}\) and then graph the equation given those intercepts.

To find the \(x\)-intercept, set \(y=0\) and solve for \(x\text{.}\)

\begin{align*} 3x+5\substitute{(0)}\amp=60\\ 3x\amp=60\\ x\amp=20 \end{align*}

So the \(x\)-intercept is the point \((\firsthighlight{20},\secondhighlight{0})\text{.}\)

To find the \(y\)-intercept, set \(x=0\) and solve for \(y\text{.}\)

\begin{align*} 3\substitute{(0)}+5y\amp=60\\ 5y\amp=60\\ y\amp=12 \end{align*}

So, the \(y\)-intercept is the point \((\firsthighlight{0},\secondhighlight{12})\text{.}\)

Next, we just plot these points and draw the line that runs through them.

epspdfpngsvgtex