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Section 3.2 Graphing Equations

permalinkWe have graphed points in a coordinate system, and now we will graph lines and curves.

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Figure 3.2.1. Alternative Video Lesson

permalinkA graph of an equation is a picture of that equation's solution set. For example, the graph of y=βˆ’2x+3 is shown in Figure 3.2.3.(c). The graph plots the ordered pairs whose coordinates make y=βˆ’2x+3 true. Figure 3.2.2 shows a few points that make the equation true.

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y=βˆ’2x+3 (x,y)
5=βœ“βˆ’2(βˆ’1)+3 (βˆ’1,5)
3=βœ“βˆ’2(0)+3 (0,3)
1=βœ“βˆ’2(1)+3 (1,1)
βˆ’1=βœ“βˆ’2(2)+3 (2,βˆ’1)
βˆ’3=βœ“βˆ’2(3)+3 (3,βˆ’3)
βˆ’5=βœ“βˆ’2(4)+3 (4,βˆ’5)
Figure 3.2.2.

Figure 3.2.2 tells us that the points (βˆ’1,5), (0,3), (1,1), (2,βˆ’1), (3,βˆ’3), and (4,βˆ’5) are all solutions to the equation y=βˆ’2x+3, and so they should all be shaded as part of that equation's graph. You can see them in Figure 3.2.3.(a). But there are many more points that make the equation true. More points are plotted in Figure 3.2.3.(b). Even more points are plotted in Figure 3.2.3.(c)β€”so many, that together the points look like a straight line.

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permalinka Cartesian graph of the points listed in the text that are solutions to the equation y=-2x+3;
(a) A few points…
permalinka second Cartesian graph showing all of the points listed in the first graph and additional solutions to y=-2x+3 between those points
(b) More points…
permalinka third Cartesian graph showing so many points that are solutions of the line y=-2x+3 that they overlap and form a solid line
(c) So many points it looks like a straight line
Figure 3.2.3. Graphs of the Equation y=βˆ’2x+3

permalinkThe graph of an equation shades all the points (x,y) that make the equation true once the x- and y-values are substituted in. Typically, there are so many points shaded, that the final graph appears to be a continuous line or curve that you could draw with one stroke of a pen.

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Checkpoint 3.2.4.
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Checkpoint 3.2.5.

permalinkTo make our own graph of an equation with two variables x and y, we can choose some reasonable x-values, then calculate the corresponding y-values, and then plot the (x,y)-pairs as points. For many algebraic equations, connecting those points with a smooth curve will produce an excellent graph.

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Example 3.2.6.

Let's plot a graph for the equation y=βˆ’2x+5. We use a table to organize our work:

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x y=βˆ’2x+5 Point
βˆ’2 βˆ’2(βˆ’2)+5=9 (βˆ’2,9)
βˆ’1 βˆ’2(βˆ’1)+5=7 (βˆ’1,7)
0 βˆ’2(0)+5=5 (0,5)
1 βˆ’2(1)+5=3 (1,3)
2 βˆ’2(2)+5=1 (2,1)
(a) Set up the table
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x y=βˆ’2x+5 Point
βˆ’2 βˆ’2(βˆ’2)+5=9 (βˆ’2,9)
βˆ’1 βˆ’2(βˆ’1)+5=7 (βˆ’1,7)
0 βˆ’2(0)+5=5 (0,5)
1 βˆ’2(1)+5=3 (1,3)
2 βˆ’2(2)+5=1 (2,1)
(b) Complete the table
Figure 3.2.7. Making a table for y=βˆ’2x+5

We use points from the table to graph the equation in Figure 3.2.8. First, we need a coordinate system to draw on. We will eventually need to see the x-values βˆ’2, βˆ’1, 0, 1, and 2. So drawing an x-axis that runs from βˆ’5 to 5 will be good enough. We will eventually need to see the y-values 9, 7, 5, 3, and 1. So drawing a y-axis that runs from βˆ’1 to 10 will be good enough.

Axes should always be labeled using the variable names they represent. In this case, with β€œx” and β€œy”. The axes should have tick marks that are spaced evenly. In this case it is fine to place the tick marks one unit apart (on both axes). Labeling at least some of the tick marks is necessary for a reader to understand the scale. Here we label the even-numbered ticks.

Then, connect the points with a smooth curve. Here, the curve is a straight line. Lastly, we can communicate that the graph extends further by sketching arrows on both ends of the line.

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All that we have decided so far is drawn in Figure 3.2.(a).

Next we carefully plot each point from the table in Figure 3.2.(b). It's not necessary to label each point's coordinates, but we do so here.

Last we connect the points with a smooth curve. Here, the β€œcurve” is a straight line. We communicate that the line extends further by putting arrowheads on both ends.

permalinka blank Cartesian grid
(a) Staging the coordinate system
permalinka Cartesian grid with points (-2,9),(-1,7),(0,5),(1,3),(2,1)
(b) Use points from the table
permalinka Cartesian grid with points (-2,9),(-1,7),(0,5),(1,3),(2,1) connected with a solid line
(c) Connect the points in whatever pattern is apparent
Figure 3.2.8. Graphing the Equation y=βˆ’2x+5
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Remark 3.2.9.

Note that our choice of x-values in the table was arbitrary. As long as we determine the coordinates of enough points to indicate the behavior of the graph, we may choose whichever x-values we like. Having a few negative x-values will be good. For simpler calculations, it's fine to choose βˆ’2, βˆ’1, 0, 1, and 2. However sometimes the equation has context that suggests using other x-values, as in the next examples.

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Example 3.2.10.

The gas tank in Sofia's car holds 14 gal of fuel. Over the course of a long road trip, her car uses fuel at an average rate of 0.032 gal⁄mi. If Sofia fills the tank at the beginning of a long trip, then the amount of fuel remaining in the tank, y, after driving x miles is given by the equation y=14βˆ’0.032x. Make a suitable table of values and graph this equation.

Explanation

Choosing \(x\)-values from \(-2\) to \(2\text{,}\) as in our previous example, wouldn't make sense here. Sofia cannot drive a negative number of miles, and any long road trip is longer than \(2\) miles. So in this context, choose \(x\)-values that reflect the number of miles Sofia might drive in a day.

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\(x\) \(y=14-0.032x\) Point
\(20\) \(13.36\) \((20,13.36)\)
\(50\) \(12.4\) \((50,12.4)\)
\(80\) \(11.44\) \((80,11.44)\)
\(100\) \(10.8\) \((100,10.8)\)
\(200\) \(7.6\) \((200,7.6)\)
Figure 3.2.11. Make the table
permalinka Cartesian graph of the points listed in the table; the points are connected by a solid line with an arrow to the right
Figure 3.2.12. Make the graph

In Figure 3.2.12, notice how both axes are also labeled with units (β€œgallons” and β€œmiles”). When the equation you plot has context like this example, including the units in the axes labels is very important to help anyone who reads your graph to understand it better.

In the table, \(x\)-values ran from \(20\) to \(200\text{,}\) while \(y\)-values ran from \(13.36\) down to \(7.6\text{.}\) So it was appropriate to make the \(x\)-axis cover something like from \(0\) to \(250\text{,}\) and make the \(y\)-axis cover something like from \(0\) to \(20\text{.}\) The scales on the two axes ended up being different.

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Example 3.2.13.

Plot a graph for the equation y=43xβˆ’4.

Explanation

This equation doesn't have any context to help us choose \(x\)-values for a table. We could use \(x\)-values like \(-2\text{,}\) \(-1\text{,}\) and so on. But note the fraction in the equation. If we use an \(x\)-value like \(-2\text{,}\) we will have to multiply by the fraction \(\frac{4}{3}\) which will leave us still holding a fraction. And then we will have to subtract \(4\) from that fraction. Since we know that everyone can make mistakes with that kind of arithmetic, maybe we can avoid it with a more wise selection of \(x\)-values.

If we use only multiples of \(3\) for the \(x\)-values, then multiplying by \(\frac{4}{3}\) will leave us with an integer, which will be easy to subtract \(4\) from. So we decide to use \(-6\text{,}\) \(-3\text{,}\) \(0\text{,}\) \(3\text{,}\) and \(6\) for \(x\text{.}\)

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\(x\) \(y=\frac{4}{3}x-4\) Point
\(-6\) \(\phantom{\frac{4}{3}(\substitute{-6})-4=\highlight{-12}}\) \(\phantom{(-6,-12)}\)
\(-3\) \(\phantom{\frac{4}{3}(\substitute{-3})-4=\highlight{-8}}\) \(\phantom{(-3,-8)}\)
\(0\) \(\phantom{\frac{4}{3}(\substitute{0})-4=\highlight{-4}}\) \(\phantom{(0,-4)}\)
\(3\) \(\phantom{\frac{4}{3}(\substitute{3})-4=\highlight{0}}\) \(\phantom{(3,0)}\)
\(6\) \(\phantom{\frac{4}{3}(\substitute{6})-4=\highlight{4}}\) \(\phantom{(6,4)}\)
(a) Set up the table
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\(x\) \(y=\frac{4}{3}x-4\) Point
\(-6\) \(\frac{4}{3}(\substitute{-6})-2=\highlight{-12}\) \((-6,-12)\)
\(-3\) \(\frac{4}{3}(\substitute{-3})-2=\highlight{-8}\) \((-3,-8)\)
\(0\) \(\frac{4}{3}(\substitute{0})-2=\highlight{-4}\) \((0,-4)\)
\(3\) \(\frac{4}{3}(\substitute{3})-2=\highlight{0}\) \((3,0)\)
\(6\) \(\frac{4}{3}(\substitute{6})-2=\highlight{4}\) \((6,4)\)
(b) Complete the table
Figure 3.2.14. Making a table for \(y=\frac{4}{3}x-4\)

We use points from the table to graph the equation. First, plot each point carefully. Then, connect the points with a smooth curve. Here, the curve is a straight line. Lastly, we can communicate that the graph extends further by sketching arrows on both ends of the line.

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permalinka Cartesian grid with points (-6,-12),(-3,-8),(0,-4),(3,0),(6,4)
(a) Use points from the table
permalinka Cartesian grid with points (-6,-12),(-3,-8),(0,-4),(3,0),(6,4) connected with a solid line
(b) Connect the points in whatever pattern is apparent
Figure 3.2.15. Graphing the Equation \(y=\frac{4}{3}x-4\)

permalinkNot all equations make a straight line once they are plotted.

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Example 3.2.16.

Build a table and graph the equation y=x2. Use x-values from βˆ’3 to 3.

Explanation
\(x\) \(y=x^2\) Point
\(-3\) \((-3)^2=9\) \((-3,9)\)
\(-2\) \((-2)^2=4\) \((-2,4)\)
\(-1\) \((-1)^2=1\) \((-1,1)\)
\(0\) \((0)^2=0\) \((0,0)\)
\(1\) \((1)^2=1\) \((0,1)\)
\(2\) \((2)^2=4\) \((2,4)\)
\(3\) \((3)^2=9\) \((3,9)\)

In this example, the points do not fall on a straight line. Many algebraic equations have graphs that are non-linear, where the points do not fall on a straight line. We connected the points with a smooth curve, sketching from left to right.

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Reading Questions Reading Questions

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1.

When a point like (5,8) is on the graph of an equation, where the equation has variables x and y, what happens when you substitute in 5 for x and 8 for y?

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2.

What are all the things to label when you set up a Cartesian coordinate system?

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3.

When you start making a table for some equation, you have to choose some x-values. Explain three different ways to choose those x-values that were demonstrated in this section.

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4.

What is an example of an equation that does not make a straight line once you make a graph of it?

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Exercises Exercises

Testing Points as Solutions
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1.

Consider the equation

y=7x+10

Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer.

  • (βˆ’5,βˆ’25)

  • (βˆ’4,βˆ’18)

  • (2,25)

  • (0,12)

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2.

Consider the equation

y=8x+6

Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer.

  • (βˆ’4,βˆ’26)

  • (0,11)

  • (5,50)

  • (βˆ’3,βˆ’18)

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3.

Consider the equation

y=βˆ’3xβˆ’8

Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer.

  • (9,βˆ’35)

  • (0,βˆ’8)

  • (βˆ’4,6)

  • (βˆ’6,10)

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4.

Consider the equation

y=βˆ’2xβˆ’2

Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer.

  • (4,βˆ’10)

  • (βˆ’9,16)

  • (0,βˆ’2)

  • (βˆ’4,11)

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5.

Consider the equation

y=23xβˆ’4

Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer.

  • (0,0)

  • (βˆ’6,βˆ’6)

  • (βˆ’15,βˆ’14)

  • (15,6)

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6.

Consider the equation

y=23xβˆ’1

Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer.

  • (12,7)

  • (0,0)

  • (βˆ’6,0)

  • (βˆ’15,βˆ’11)

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7.

Consider the equation

y=βˆ’34xβˆ’3

Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer.

  • (0,βˆ’3)

  • (8,βˆ’5)

  • (βˆ’20,14)

  • (βˆ’16,9)

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8.

Consider the equation

y=βˆ’34xβˆ’5

Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer.

  • (βˆ’16,10)

  • (βˆ’4,βˆ’2)

  • (4,βˆ’4)

  • (0,βˆ’5)

Tables for Equations
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9.

Make a table for the equation.

The first row is an example.

x y=βˆ’x+6 Points
βˆ’3 9 (βˆ’3,9)
βˆ’2
βˆ’1
0
1
2
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10.

Make a table for the equation.

The first row is an example.

x y=βˆ’x+7 Points
βˆ’3 10 (βˆ’3,10)
βˆ’2
βˆ’1
0
1
2
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11.

Make a table for the equation.

The first row is an example.

x y=5x+5 Points
βˆ’3 βˆ’10 (βˆ’3,βˆ’10)
βˆ’2
βˆ’1
0
1
2
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12.

Make a table for the equation.

The first row is an example.

x y=6x+1 Points
βˆ’3 βˆ’17 (βˆ’3,βˆ’17)
βˆ’2
βˆ’1
0
1
2
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13.

Make a table for the equation.

The first row is an example.

x y=βˆ’2x+8 Points
βˆ’3 14 (βˆ’3,14)
βˆ’2
βˆ’1
0
1
2
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14.

Make a table for the equation.

The first row is an example.

x y=βˆ’5x+4 Points
βˆ’3 19 (βˆ’3,19)
βˆ’2
βˆ’1
0
1
2
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15.

Make a table for the equation.

The first row is an example.

x y=32x+7 Points
βˆ’6 βˆ’2 (βˆ’6,βˆ’2)
βˆ’4
βˆ’2
0
2
4
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16.

Make a table for the equation.

The first row is an example.

x y=38xβˆ’5 Points
βˆ’24 βˆ’14 (βˆ’24,βˆ’14)
βˆ’16
βˆ’8
0
8
16
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17.

Make a table for the equation.

The first row is an example.

x y=βˆ’54x+4 Points
βˆ’12 19 (βˆ’12,19)
βˆ’8
βˆ’4
0
4
8
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18.

Make a table for the equation.

The first row is an example.

x y=βˆ’56xβˆ’4 Points
βˆ’18 11 (βˆ’18,11)
βˆ’12
βˆ’6
0
6
12
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19.

Make a table for the equation.

x y=7x
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20.

Make a table for the equation.

x y=12x
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21.

Make a table for the equation.

x y=8x+8
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22.

Make a table for the equation.

x y=10x+2
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23.

Make a table for the equation.

x y=53xβˆ’8
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24.

Make a table for the equation.

x y=187xβˆ’6
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25.

Make a table for the equation.

x y=βˆ’59xβˆ’4
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26.

Make a table for the equation.

x y=βˆ’65xβˆ’1
Cartesian Plots in Context
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27.

A certain water heater will cost you $900 to buy and have installed. This water heater claims that its operating expense (money spent on electricity or gas) will be about $31 per month. According to this information, the equation y=900+31x models the total cost of the water heater after x months, where y is in dollars. Make a table of at least five values and plot a graph of this equation.

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28.

You bought a new Toyota Corolla for $18,600 with a zero interest loan over a five-year period. That means you'll have to pay $310 each month for the next five years (sixty months) to pay it off. According to this information, the equation y=18600βˆ’310x models the loan balance after x months, where y is in dollars. Make a table of at least five values and plot a graph of this equation. Make sure to include a data point representing when you will have paid off the loan.

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29.

The pressure inside a full propane tank will rise and fall if the ambient temperature rises and falls. The equation P=0.1963(T+459.67) models this relationship, where the temperature T is measured in Β°F and the pressure and the pressure P is measured in lb⁄in2. Make a table of at least five values and plot a graph of this equation. Make sure to use T-values that make sense in context.

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30.

A beloved coworker is retiring and you want to give her a gift of week-long vacation rental at the coast that costs $1400 for the week. You might end up paying for it yourself, but you ask around to see if the other 29 office coworkers want to split the cost evenly. The equation y=1400x models this situation, where x people contribute to the gift, and y is the dollar amount everyone contributes. Make a table of at least five values and plot a graph of this equation. Make sure to use x-values that make sense in context.

Graphs of Equations
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31.

Create a table of ordered pairs and then make a plot of the equation y=2x+3.

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32.

Create a table of ordered pairs and then make a plot of the equation y=3x+5.

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33.

Create a table of ordered pairs and then make a plot of the equation y=βˆ’4x+1.

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34.

Create a table of ordered pairs and then make a plot of the equation y=βˆ’xβˆ’4.

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35.

Create a table of ordered pairs and then make a plot of the equation y=52x.

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36.

Create a table of ordered pairs and then make a plot of the equation y=43x.

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37.

Create a table of ordered pairs and then make a plot of the equation y=βˆ’25xβˆ’3.

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38.

Create a table of ordered pairs and then make a plot of the equation y=βˆ’34x+2.

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39.

Create a table of ordered pairs and then make a plot of the equation y=x2+1.

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40.

Create a table of ordered pairs and then make a plot of the equation y=(xβˆ’2)2. Use x-values from 0 to 4.

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41.

Create a table of ordered pairs and then make a plot of the equation y=βˆ’3x2.

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42.

Create a table of ordered pairs and then make a plot of the equation y=βˆ’x2βˆ’2xβˆ’3.