This section on stretching and shifting of graphs begins a summary of general properties of functions. Later sections will look at other properties, including domains and ranges as well as the various ways in which functions can be combined.

You have seen numerous examples of how graphs stretch and shift. You probably recall that

the transformation y = f(x-h) shifts the graph of y = f(x) by h units in the x direction,

the transformation y = f(x) + k shifts the graph of y = f(x) by k units in the y direction, and

the transformation y = A f(x) shifts the graph of y = f(x) by factor A from the x axis.

Note the following patterns in the table:

The values in the x = -3 row alternate between positive and negative values. The same pattern occurs for the x = -2 and x = -1 rows.

The x = .5 and x = -.5 powers are the largest (absolute) numbers shown on the table.

The x = .5 and x = -.5 powers get further and further from 0 as the power p goes from -2 to -5.

The values in the x = 0 row all indicate division by 0. Recall that this is usually associated with a vertical asymptote.

The values for x = 1, 2 and 3 are identical to those for x = -1, -2 and -3, except for the negative signs that occur for odd powers of x.

The x = 2 and x = 3 values get closer and closer to zero as the power p goes from -2 to -5.

The figure below shows the approximate graphs of the functions between x = -1 and x = 1. The basic points corresponding to x = .5 are clearly indicated. Note how the points mimic the x = .5 row of the table.

You will recall that replacing x by x-h will shift a graph h units to the right, while adding k to the y value will raise the graph k units. You will also recall that multiplying y values by a number A will move all graph points A times as far from the x axis.

We will investigate these ideas for the negative-power functions.

We now consider what happens to the graph of y = x^-2 when we replace x by (x-.7), and then when we add -1.5 to y.  Each of these processes is thought of as a transformation of the graph of the function on which it is performed.

Note that the x values change by increments of .7.

When we replace x by (x-.7), we obtain the function y = (x-.7)^-2, with values indicated in the third column. Note how the division by 0 occurs when x = .7, compared to x = 0 as with the original y = x^-2 function. This is easy enough to understand, since when x = .7, y = (x - .7) ^-2 will equal 0^-2, or 1 / 0^2.

We can see similarly that every y value in the (x-.7)^2 column is displaced 1 position downward (in the direction of positive x) from the corresponding value in the x^-2 column.

If we multiply the y values of a function by 2, we double all the y values. Since y values are represented on the graph as distances from the x axis, multiplying the values of a function by 2 has the effect of moving every point twice as far from the x axis.   If we multiply y = f(x) by 2 we get y = 2 f(x).

If we were to multiply all y values by .5, we would move every point twice as close to the x axis.  If we multiply y = f(x) by .5 we get y = .5 f(x).

If we multiply the y values by a negative number, then every point above the x axis, which must represent a positive y value, will move below the x axis, representing the resulting negative value that results when we multiply the original positive y value by a negative number. Similarly, since every point below the x axis represents a negative y value which will become positive when multiplied by our negative number, points below the x axis will move above the x axis. Thus the graph is inverted in addition to being stretched.

The table clearly shows in the x = .5 row, for example, how the values of 3 x^-3 and -5 x^-3 are respectively 3 and -5 times as great as the values of the y = x^-3 function. You should look at the other rows and verify that the same pattern occurs.

The x = 0 row clearly shows that in all cases, if x = 0 we have division by 0. Once again we see that as we divide by smaller and smaller numbers we get larger and larger numbers, resulting in a vertical asymptote at x = 0.

The original y = x^3 graph is shown by the solid black line. The graph shows how the y values are positive when x is positive and negative when x is negative. It also shows how the y values increase in magnitude as x approaches 0, forming the vertical asymptote at y = 0, and how the y values approach the

y = 0 asymptote (the x axis) when x becomes large.

The y = 3 x^3 graph is shown by the light (blue) dotted line. For any given x value, the point on the y = 3 x^3 will be three times as far from the x axis as the y = x^3 point. This is indicated for an x value of approximately -1.2. You should look at several other points and convince yourself that the same sort of statement could be made for those points.

3. For the function y = (x-1.7) ^ -4, show how division by 0 occurs at x = 1.7.

Show by evaluating the function at x = 1.5, 1.6, 1.8 and 1.9 we can see how a vertical asymptote forms at the line x = 1.7.

4. Construct a table showing how the y = x^-3 function can be transformed first into y = (x - .4) ^ -3, then into y = -2 (x - .4) ^ -3, and finally into y = -2 (x - .4) ^ -3 + .6. When constructing your table be sure you use an interval of .4 between your x values.

Identify the patterns that occur on your table.

Construct a graph of all the functions on your table. Explain how the patterns of the table appear on your graph.

5. Sketch on one set of coordinate axes the graphs of y = x ^ .5 and y = 3 x^.5, for x values between 0 and 4. For at least 3 randomly selected x values, show on your graph the ratios of y values between the two functions.

The original function is denoted by y = f(x). When multiplied by A, every point moves A times as far from the x axis. The resulting function is denoted y = A f(x).

The approximate values of the two functions y = f(x) and y = A f(x) are indicated at three arbitrary points by pairs of vertical bars.

The longer (blue) bar at each point represents the value of y = A f(x), while the shorter (red) bar represents the value of y = f(x).

At each point we see that the (blue) bar representing y = A f(x) is a little more than twice as long as the (red) bar representing y = f(x).

The y = A f(x) graph is then shifted k units higher. Each point on the graph is raised by k units to the graph of y = A f(x) + k.  This is indicated by the (green) vertical displacement lines of length k.

It is important to understand that had we done the vertical shift by k units before doing the vertical stretch, the result would have been somewhat different. This is because we would be vertically stretching not only the original function but the vertical shift itself. In the former example, the vertical shift had no effect on the vertical shift factor A.

Using function notation this is somewhat easier to understand:

If we begin with y = f(x) and shift it k units vertically we obtain the function y = f(x) + k.

If we then stretch by factor A, we multiply every y value by A, so we multiply the entire expression y = f(x) + k by A.

If we replace x by (x-h), the function y = f(x) becomes y = f(x-h).

The table of y = f(x-h) will be the same as that of y = f(x), but shifted h units in the direction of x (look again at the table of y = (x-.7)^-2 vs. x^-2). The graph correspondingly shifts h units in the x direction.

The horizontal shift h is shown for three points. The original function y = f(x) and the shifted function y = f(x-h) are connected by these shifts.

6. In Exercise 4 you constructed the graph of y = A f(x-h) + k for f(x) = x^-3, A = -2, h = .4 and k = .6. If you have not yet successfully completed that exercise, do so now.

Now for the same functions, construct the graph of y = A [ f(x-h) + k ], and compare the resulting graph to that of exercise 4.

Write out the function in the form y = A [ f(x-h) + k ] and simplify your expression using the distributive law. Use this expression to explain the difference in your graphs for this exercise and exercise 4.