### 6.3 - Other types of graphs

Literally dozens of different types of graphs (or charts or plots) have been invented. We will describe the three types that the Algebra Coach uses. We will assume that the variables of interest are called x and y and that we want to plot the graphs on the cartesian plane.

### Graphs of functions

A function can be described as a machine that takes a number x as input, and applies some rule to it that gives a corresponding number y as output: This means that we can make a table of values of the function. This is a two column table whose first column is a list of x values of our choosing, and whose second column is the corresponding y values calculated using the function rule. (The x values should be uniformly spaced and they should be in order from smallest to biggest. Click here to see an example.)

We can then plot the table of values. Each row of the table becomes a point on the graph. If the function is not varying too rapidly then we can connect each point to the next one with a straight line segment. If the x values are chosen close enough together then the sequence of straight line segments will appear to be one smooth curve.

Notes:
• The graph of a function can have only one y value for any given x value.
• The Algebra Coach uses this table of values technique to plot functions.
• Click here to see graphs of various types of functions.

### Graphs of relations

A relation is simply an equation containing the two variables x and y, but that has not necessarily been solved for y. An example is the general form of the equation for the straight line
x + 3 y + 4 = 0. If a relation can be solved for y, then we again have a function which can be plotted as described above. For example the relation 2 x + 3 y = 4 can easily be solved for y. However, many relations can’t. An example is the relation
x 2 − 12 y + y 3 = 30,
which is graphed to the right. The problem is that if we can’t solve for y then we can’t make a table of values.

Thus we use a totally different approach; we systematically test every point of the cartesian plane (or every pixel on the computer screen) to see if the x and y values at that point satisfy the relation. If they do then we plot a point there; otherwise we don’t. The set of all points that satisfies the relation is called a locus of points and for this example the locus of points is a curve. Obviously testing every point in the plane would take forever but there are approximate methods that allow it to be done quickly.

Notes:
• The Algebra Coach uses such an approximate method to plot relations.

• Unlike the graph of a function, the graph of a relation can have more than one y value for a given x value.

• You may be familiar with topographical maps. These are maps that show the elevations of mountains, valleys and other geographical features by using contours, which are lines of constant elevation. The graph of a relation can be though of as a contour on a contour map. Here is an example showing why:

• Create a function of two variables from the left-hand-side of the above relation, like this:
f (xy) = x 2 − 12 y + y 3.
This function takes two numbers, x and y, as input; evaluates x 2 − 12 y + y 3; and returns that number as output. For example if x = 5 and y = 2, then the function calculates f (5, 2) = 5 2 −12 · 2 + 2 3 = 9, and returns the number 9.

• Now imagine that x and y are the longitude and latitude of a location and that the output of the function is the elevation at that location. For example at x = 5 and y = 2 the elevation is 9.

• Then finding all x and y that satisfy the relation
x 2 − 12 y + y 3 = 30 can be interpreted as finding the contour whose elevation is 30.

The graph to the right shows four contour lines at elevations of 10, 20, 30 and 40 for our function of two variables.

### Parametric plots

A parametric plot is a completely different way to draw a curve. Let (xy) denote a point along the curve. The idea is to let both x and y be functions of a new variable or parameter, usually called t. You can arrange it so that when t = 0 you are at one endpoint of the curve, when t = 1 you are at the other endpoint of the curve, and as t ranges from 0 to 1 you move along the curve from one endpoint to the other. You could imagine that t represents time and that as time progresses you move to different points along the curve.

Example: Let

x (t) = t · cos (4 π t),
and
y (t) = t · sin (4 π t),

be the x and y coordinates of points along a curve, expressed as functions of a parameter t. The 3-column table of values below shows values of x and y as the parameter t ranges uniformly from 0 to 1 in 21 steps. The points in the parametric plot to the right are the 21 values from the table. They have been connected with straight line segments. (Using smaller increments of t would have resulted in a smoother curve.)

If t denotes time then the curve could describe the path of an object that starts at the origin and spirals outward, slowly at first, and then ever more quickly (as evidenced by the fact that the dots are getting farther apart). ### Using graphs to solve equations

Many equations cannot be solved using algebra. To solve these we must resort to approximate methods involving graphs and computers. Here is the process:
• Prepare a function corresponding to the equation:
Let the unknown in the equation must be solved for be called x. Use algebra to move everything over to one side of the equation. This puts the equation into the form 0 = f (x), where f (x) denotes some expression involving x. Now replace the 0 on the left-hand-side by the variable y. The result is y = f (x), ie. y is some function of x.

• Graph the function and estimate the roots:
Make a graph of the function y = f (x), plotting y vertically versus x horizontally. Find the points where the function crosses the x axis. These are the values of x for which y = 0 or f (x) = 0. In other words, these are the required roots of the equation. Thus from the graph we can find the number of roots and their approximate values.

Example: Consider the equation . Find the number of solutions and their approximate values.

Solution: The first step is to prepare the function. Bring everything to one side of the equation. There are many ways that this can be done. One way is to simply subtract x from both sides of the equation to get:  Then replace the 0 on the left-hand-side by y: The second step is to graph this function, as we have done to the right. The roots are the points where y = 0 (i.e. the x intercepts). The graph shows that there are three such points, approximately at
x = −1, x = 1 and x = 5.2

Note: Another way to prepare the function is by first clearing denominators in the equation by multiplying through by x (x − 5): then distributing
x 3 − 5 x 2 = (x − 5) + x,
and then bringing everything to one side of the equation:
x 3 − 5 x 2 − 2 x + 5 = 0. Now replace 0 by y and graph the function:
y = x 3 − 5 x 2 − 2 x + 5,
as we have done to the right. This cubic function looks completely different from the previous function but crosses the x axis at exactly the same points. Thus we find the same three roots for the equation; namely x = −1, x = 1 and x = 5.2 (approximately).

One problem with multiplying through by x (x − 5) is that this can introduce extraneous roots. Therefore we must check our solutions by substituting them back into the original equation.

 Algebra Coach Exercises

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