# 1.7: Horizontal Transformations - Mathematics

In the previous section, we introduced the concept of transformations. We made a change to the basic equation y = f(x), such as y = af(x), y = −f(x), y = f(x) − c, or y = f(x) + c, then studied how these changes affected the shape of the graph of y = f(x). In that section, we concentrated strictly on transformations that applied in th vertical direction. In this section, we will study transformations that will affect the shape of the graph in the horizontal direction.

We begin our task with an example that requires that we read the graph of a function to capture several key points that lie on the graph of the function.

Example (PageIndex{1})

Consider the graph of f presented in Figure (PageIndex{1})(a). Use the graph of f to complete the table in Figure (PageIndex{1})(b).

Solution

To compute f(−2), for example, we would first locate −2 on the x-axis, draw a vertical arrow to the graph of f, then a horizontal arrow to the y-axis, as shown in Figure (PageIndex{2})(a). The y-value of this final destination is the value of f(−2). That is, f(−2) = −4. This allows us to complete one entry in the table, as shown in Figure (PageIndex{2})(b). Continue in this manner to complete all of the entries in the table. The result is shown in Figure (PageIndex{2})(c).

## Horizontal Scaling

In the narrative that follows, we will have repeated need of the graph in Figure (PageIndex{2})(a) and the table in Figure (PageIndex{2})(c). They characterize the basic function that will be the starting point for the concepts of scaling, reflection, and translation that we develop in this section. Consequently, let’s place them side-by-side for emphasis in Figure (PageIndex{3}).

We are now going to scale the graph of f in the horizontal direction.

Example (PageIndex{2})

If y = f(x) has the graph shown in Figure (PageIndex{3})(a), sketch the graph of y = f(2x).

Solution

In the previous section, we investigated the graph of y = 2f(x). The number 2 was outside the function notation and as a result we stretched the graph of y = f(x) vertically by a factor of 2. However, note that the 2 is now inside the function notation y = f(2x). Intuition would demand that this might have something to do with scaling in the x-direction (horizontal direction), but how?

Again, when we’re unsure of the shape of the graph, we rely on plotting a table of points. We begin by picking these x-values: x = −2, −1, 0, 1, and 2. Note that these are precisely half of each of the x-values presented in the table in Figure (PageIndex{3})(b). We will now evaluate the function y = f(2x) at each of these x-values. For example, to compute y = f(2x) at x = −2, we first insert x = −2 for x to obtain [y=f(2(-2))=f(-4)]

To complete the computation, we must now evaluate f(−4). However, this result is recorded in the table in Figure (PageIndex{3})(b). There we find that f(−4) = 0, and we can complete the computation started above.

[y=f(2(-2))=f(-4)=0]

In similar fashion, to evaluate the function y = f(2x) at x = −1, first substitute x = −1 in y = f(2x) to obtain

[y=f(2(-1))=f(-2)]

Now, note that f(−2) is the next recorded value in the table in Figure (PageIndex{3})(b). There we find that f(−2) = −4, so we can complete the computation started above.

[y=f(2(-1))=f(-2)=-4]

At this point, you might see why we chose x-values: −2, −1, 0, 1, and 2. These are precisely half of the x-values in the table of original values for the function y = f(x) in Figure (PageIndex{3})(b). When the values −2, −1, 0, 1, and 2 are substituted into the function y = f(2x), they are first doubled before we go to look up the function value in the table in Figure (PageIndex{3})(b).

Continuing in this manner, we evaluate the function y = f(2x) at the remaining values of x, namely, 0, 1, and 2.

[egin{aligned} y &=f(2(0))=f(0)=0 y &=f(2(1))=f(2)=2 y &=f(2(2))=f(4)=0 end{aligned}]

We enter these values into the table in Figure 4(b) and plot them to determine the graph of y = f(2x) in Figure (PageIndex{4})(a).

At this point, there are a number of comparisons you can make.

1. Compare the data in the table in Figure (PageIndex{4})(b) with the original function data in the table in Figure (PageIndex{3})(b). Note that the y-values in each table are identical. However, note that each x-value in the table of Figure (PageIndex{4})(b) is precisely half of the corresponding x-value in the table of Figure (PageIndex{3})(b).

2. Compare the graph of y = f(2x) in Figure (PageIndex{4})(a) with the original graph of y = f(x) in Figure (PageIndex{3})(a). Note that each x-value at each point on the graph of y = f(2x) in

Figure (PageIndex{4})(a) is precisely half the x-value of the corresponding point on the graph of y = f(x) in Figure (PageIndex{3})(a).

Note the result. The graph of y = f(2x) is compressed horizontally (toward the y-axis), both positively and negatively, by a factor of 2. Note that this is exactly the opposite of what you might expect by intuition, but a careful examination of the data in the tables in Figures (PageIndex{3})(b) and (PageIndex{4})(b) will explain why.

Let’s look at another example.

Example (PageIndex{3})

If y = f(x) has the graph shown in Figure (PageIndex{3})(a), sketch the graph of y = f((1/2)x).

Solution

Rather than doubling each value of x at the start, this function first halves each value of x. Thus, we will want to evaluate the function y = f((1/2)x) at x = −8, −4, 0, 4, and 8. For example, to evaluate the function y = f((1/2)x) at x = −8, first substitute x = −8 to obtain [y=f((1 / 2)(-8))=f(-4)]

Now, look up this value in the table in Figure (PageIndex{3})(b) and note that f(−4) = 0. Thus, we can complete the computation as follows.

[y=f((1 / 2)(-8))=f(-4)=0]

Similarly, to evaluate the function y = f((1/2)x) at x = −4, first substitute x = −4 to obtain

[y=f((1 / 2)(-4))=f(-2)]

Now, look up this value in the table in Figure (PageIndex{3})(b) and note that f(−2) = −4. Thus, we can complete the computation as follows.

[y=f((1 / 2)(-4))=f(-2)=-4]

At this point, you will see why we chose x-values: −8, −4, 0, 4, and 8. These values are precisely double the x-values in the table of original values for the function y = f(x) in Figure (PageIndex{3})(b). When the values −8, −4, 0, 4, and 8 are substituted into the function y = f((1/2)x), they are first halved before we go to look up the function value in the table in Figure (PageIndex{3})(b). This halving leads to the values −4, −2, 0, 2, and 4, which are precisely the values available in the table in Figure (PageIndex{3})(b).

We make similar computations at the remaining values of x, namely x = 0, 4, and 8.

[egin{aligned} y &=f((1 / 2)(0))=f(0)=0 y &=f((1 / 2)(4))=f(2)=2 y &=f((1 / 2)(8))=f(4)=0 end{aligned}]

Hopefully, these computations explain our choice of x-values above. Each of these results is recorded in the table in Figure (PageIndex{5})(b) and plotted on the graph shown in Figure (PageIndex{5})(a).

Again, note that the y-values in the table in Figure (PageIndex{5})(b) are identical to the y-values in the table in Figure (PageIndex{3})(b). However, each x-value in the table in Figure (PageIndex{5})(b) is precisely double the corresponding x-value in the table in Figure (PageIndex{3})(b).

This doubling of the x-values is apparent in the graph of y = f((1/2)x) shown in Figure (PageIndex{5})(a), where the graph is stretched by a factor of 2 horizontally (away from the y-axis), both positively and negatively. Note that this is exactly the opposite of what you might expect by intuition, but a careful examination of the data in the tables in Figures (PageIndex{3})(b) and (PageIndex{5})(b) will explain why.

Let’s summarize our findings.

A Visual Summary — Horizontal Scaling

Consider the images in Figure (PageIndex{6}).

• In Figure (PageIndex{6})(a), we see pictured the graph of the original function y = f(x).
• In Figure (PageIndex{6})(b), note that each key point on the graph of y = f(2x) has an x-value that is precisely half the x-value of the corresponding point on the graph of y = f(x) in Figure (PageIndex{6})(a).
• In Figure (PageIndex{6})(c), note that each key point on the graph of y = f((1/2)x) has an x-value that is twice the x-value of the corresponding point on the graph of y = f(x) in Figure (PageIndex{6})(a).
• Note that the y-value of each transformed point remains the same.

The visual summary in Figure (PageIndex{6}) makes sketching the graphs of y = f(2x) and y = f(1/2)x) an easy task.

• Given the graph of y = f(x), to sketch the graph of y = f(2x), simply take each point on the graph of y = f(x) and cut its x-value in half, keeping the same y-value.
• Given the graph of y = f(x), to sketch the graph of y = f((1/2)x), simply take each point on the graph of y = f(x) and double its x-value, keeping the same y-value.

Follow the same procedures for other scaling factors. For example, in the case of y = f(3x), take each point on the graph of y = f(x) and divide its x-value by 3, keeping the same y-value. On the other hand, to draw the graph of y = f((1/3)x), take each point on the graph of f and multiply its x-value by 3, keeping the same y-value.

In general, we can state the following.

summary

Suppose we are given the graph of y = f(x).

• If a > 1, the graph of y = f(ax) compresses horizontally (toward the y-axis) , both positively and negatively, by a factor of a.
• If 0 < a < 1, the graph of y = f(ax) stretches horizontally (away from the y-axis), both positively and negatively, by a factor of 1/a.

In the case of the first item in Summary, when we compare the general form y = f(ax) with y = f(2x), we see that a = 2. In this case, note that a > 1 and the graph of y = f(2x) compresses horizontally by a factor of 2 when compared with the graph of y = f(x) (see Figure (PageIndex{6})(b)).

In the case of the second item in Summary, when we compare the general form y = f(ax) with the equation y = f((1/2)x), we see that a = 1/2, so

[frac{1}{a}=frac{1}{1 / 2}=2]

The second item in Summary 4 says that when 0 < a < 1, the graph of y = f(ax) stretches horizontally by a factor of 1/a. Indeed, this is exactly what happened in the case of y = f((1/2)x), which stretched in the horizontal direction by a factor of 1/(1/2), or 2 (see Figure (PageIndex{6})(c)).

## Horizontal Reflections

For convenience, we begin by repeating the original graph of y = f(x) and its accompanying data in Figure (PageIndex{7}). We are now going to reflect the graph of y = f(x) in the horizontal direction (across the y-axis).

Example (PageIndex{4})

If y = f(x) has the graph shown in Figure (PageIndex{7})(a), draw the graph of y = f(−x).

Solution

In the previous section, we were asked to draw the graph of y = −f(x). Note how the minus sign appears on the outside of the function. Clearly, the y-values of y = −f(x) must be opposite in sign to the y-values of y = f(x). That is why the graph of y = −f(x) was a reflection of the graph of y = f(x) across the x-axis.

However, in this example, the minus sign is inside the function, leaving one to intuit that it is the x-values, not the y-values, that are being negated. We will choose the following x-values: x = 4, 2, 0, −2, and −4. This is a bit deceptive, as it looks like we are choosing the same x-values, only in reverse order. This is not the case. We are choosing the negative of each x-value in the table in Figure (PageIndex{7})(b).

To evaluate y = f(−x) at our first x-value, namely x = 4, we perform the following calculation. First substitute x = 4 to obtain

[y=f(-(4))=f(-4)]

Now, look up this value in the table in Figure (PageIndex{7})(b) and note that f(−4) = 0. Thus, we can complete the computation as follows.

[y=f(-(4))=f(-4)=0]

Similarly, to evaluate the function y = f(−x) at x = 2, first substitute x = 2 to obtain

[y=f(-(2))=f(-2)]

Now, look up this value in the table in Figure (PageIndex{7})(b) and note that f(−2) = −4. Thus, we can complete the computation as follows.

[y=f(-(2))=f(-2)=-4]

At this point, you will see why we chose x-values: 4, 2, 0, −2, and −4. These values are the negatives of the x-values in the table of original values for the function y = f(x) in Figure (PageIndex{7})(b). When the values 4, 2, 0, −2, and −4 are substituted into the function y = f(−x), they are first negated before we go to look up the function value in the table in Figure (PageIndex{7})(b). This negating leads to the values −4, −2, 0, 2, and 4, which are precisely the values available in the table in Figure (PageIndex{7})(b).

We make similar computations at the remaining values of x, namely x = 0, −2, and −4.

[egin{array}{l}{y=f(-(0))=f(0)=0} {y=f(-(-2))=f(2)=2} {y=f(-(-4))=f(4)=0}end{array}]

We organize these points in the table in Figure (PageIndex{8})(b), then plot them in Figure (PageIndex{8})(a).

When you compare the entries in the table in Figure (PageIndex{8})(b) with those in the table in Figure (PageIndex{7})(b), note that the y-values appear in the same order, but the x-values of the table in Figure (PageIndex{7})(b) have been negated in the table in Figure (PageIndex{8})(b). This means that a former point such as (−2, −4) is transformed to the point (2, −4), which is a reflection of the point (−2, −4) across the y-axis.

Thus, to produce the graph of y = f(−x), simply reflect the graph of y = f(x) across the y-axis.

Let’s summarize what we’ve learned about horizontal reflections.

A Visual Summary — Horizontal Reflections

Consider the images in Figure (PageIndex{9}).

• In Figure (PageIndex{9})(a), we see pictured the original graph of y = f(x).
• In Figure (PageIndex{9})(b), the graph of y = f(−x) is a reflection of the graph of y = f(x) across the y-axis.

Thus, given the graph of y = f(x), it is a simple task to draw the graph of y = f(−x).

• To draw the graph of y = f(−x), take each point on the graph of y = f(x) and reflect it across the y-axis, keeping the y-value the same, but negating the x-value.

## Horizontal Translations

In the previous section, we saw that the graphs of y = f(x) + c and y = f(x) − c were vertical translations of the graph of y = f(x). If c is a positive number, then the graph of y = f(x) + c shifts c units upward while the graph of y = f(x) − c shifts c units downward.

In this section, we will study horizontal translations. For convenience, we begin by repeating the original graph of y = f(x) and its accompanying data in Figure 10.

Example (PageIndex{5})

If y = f(x) has the graph shown in Figure (PageIndex{10})(a), sketch the graph of y = f(x + 1).

Solution

In the previous section, we drew the graph of y = f(x)+1. Note that in y = f(x)+1, the number 1 is outside the function. The result was a graph that was shifted 1 unit upwards in the y-direction.

In this case, y = f(x + 1) and the 1 is inside the function notation, leading one to intuit that the translation might be in the horizontal direction (x-direction). But how?

Again, we will set up a table of points that satisfy the equation y = f(x + 1), then plot them. Because this function requires that we first add 1 to each x-value before inserting it into the function, we will choose x-values appropriately, namely x = −5, −3, −1, 1, and 3. In a moment, it will be clear why we have chosen these particular values of x. Perhaps you already see why?

We need to evaluate the function y = f(x + 1) at each of these chosen values of x. To evaluate y = f(x+ 1) at the first value, namely x = −5, we insert x = −5 and make the calculation [y=f(-5+1)=f(-4)]

To complete the calculation, we must now evaluate f(−4). However, this result is recorded in the table in Figure (PageIndex{10})(b). There we find that f(−4) = 0, and we can complete the calculation started above.

[y=f(-5+1)=f(-4)=0]

In similar fashion, we can evaluate the function y = f(x+1) at x = −3. First, substitute x = −3 in y = f(x + 1) to obtain

[y=f(-3+1)=f(-2)]

To complete the calculation, we must now evaluate f(−2). There we find that f(−2) = −4, and we can complete the calculation started above.

[y=f(-3+1)=f(-2)=-4]

At this point, you might see why we chose x-values: −5, −3, −1, 1, and 3. These are precisely one less than the x-values in the table of original values for the function y = f(x) in Figure (PageIndex{10})(b). When the values −5, −3, −1, 1, and 3 are substituted into the function y = f(x + 1), we first add 1 to each value before we go to look up the function value in the table in Figure (PageIndex{10})(b). This adding of 1 leads to the values −4, −2, 0, 2, and 4, which are precisely the values available in the table in Figure (PageIndex{10})(b).

Continuing in this manner, we evaluate the function y = f(x + 1) at the remaining values of x, namely, −1, 1, and 3.

[egin{aligned} y &=f(-1+1)=f(0)=0 y &=f(1+1)=f(2)=2 y &=f(3+1)=f(4)=0 end{aligned}]

We assemble these results in the table in Figure (PageIndex{11})(b) and plot them in Figure (PageIndex{11})(a).

When you compare the points on the graph of y = f(x+1) in the table in Figure (PageIndex{11})(b) with the original points on the graph of y = f(x) in the table in Figure (PageIndex{10})(b), note that the y-values are identical, but the x-values in the table in Figure (PageIndex{11})(b) are all 1 unit less than the corresponding x-values in the table in Figure (PageIndex{10})(b). It is no wonder that the graph of y = f(x + 1) in Figure (PageIndex{11})(a) is shifted 1 unit to the left of the original graph of y = f(x) in Figure (PageIndex{10})(a).

Note that this is somewhat counterintuitive, because we’re seemingly adding 1 to each x-value in y = f(x + 1). Why doesn’t the graph move one unit to the right? Well, a careful comparison of the x-values in the tables in Figures (PageIndex{10})(b) and (PageIndex{11})(b) reveals the answer. In order to use the data in the table in Figure (PageIndex{10})(b), we must first subtract 1 from each x-value to produce the x-values in the table in Figure (PageIndex{11})(b). This is why the graph of y = f(x + 1) moves 1 unit to the left instead of 1 unit to the right.

You might also recall that the function y = f(2x) compressed by a factor of 2, which is also the opposite of what intuition might dictate. Similarly, the function y = f((1/2)x) stretches by a factor of 2, which also goes counter to intuition. With these thoughts in mind, it is not surprising that y = f(x+ 1) shifts one unit to the left. Still, a comparison of the x-values in the tables in Figures (PageIndex{10})(b) and (PageIndex{11})(b) provide irrefutable evidence that the shift is 1 unit to the left.

Let’s look at another example.

Example (PageIndex{6})

If y = f(x) has the graph shown in Figure (PageIndex{10})(a), sketch the graph of y = f(x − 2).

Solution

Again, we will set up a table of points that satisfy the equation y = f(x − 2), then plot them. Because this function requires that we first subtract 2 from each x-value before inserting it into the function, we will choose x-values: −2, 0, 2, 4, and 6. We need to evaluate the function y = f(x − 2) at each of these values of x.

To evaluate y = f(x − 2) at the first value, namely x = −2, insert x = −2 into the function y = f(x − 2) to obtain [y=f(-2-2)=f(-4)]

In the table in Figure (PageIndex{10})(b), we find that f(−4) = 0, which allows us to complete the calculation above.

[y=f(-2-2)=f(-4)=0]

In similar fashion, we evaluate y = f(x − 2) at x = 0 to obtain

[y=f(0-2)=f(-2)]

In the table in Figure (PageIndex{10})(b), we find that f(−2) = −4, which allows us to complete the calculation above.

[y=f(0-2)=f(-2)=-4]

Hopefully, you see why we chose the x-values: −2, 0, 2, 4, and 6. These values are 2 larger than the x-values in the table of original values for the function y = f(x) in Figure (PageIndex{10})(b). When the values −2, 0, 2, 4, and 6 are substituted into the function y = f(x − 2), we first subtract 2 from each value before we go to look up the function value in the table in Figure (PageIndex{10})(b). This subtracting of 2 leads to −4, −2, 0, 2, and 4, precisely the values that are available in the table in Figure (PageIndex{10})(b).

Continuing in this manner, we evaluate y = f(x − 2) at the remaining values of x, namely, x = 2, 4, and 6.

[egin{array}{l}{y=f(2-2)=f(0)=0} {y=f(4-2)=f(2)=2} {y=f(6-2)=f(4)=0}end{array}]

We assemble these results in the table in Figure (PageIndex{12})(b) and plot them in Figure (PageIndex{12})(a).

When you compare the points on the graph of y = f(x−2) in the table in Figure (PageIndex{12})(b) with the original points on the graph of y = f(x) in the table in Figure (PageIndex{10})(b), note that the y-values are identical, but the x-values in the table in Figure (PageIndex{12})(b) are all 2 larger than the corresponding x-values in the table in Figure (PageIndex{10})(b). It is no wonder that the graph of y = f(x − 2) in Figure (PageIndex{12})(a) is shifted 2 units to the right of the original graph of y = f(x) in Figure (PageIndex{10})(a).

Again, this runs counterintuitive (why doesn’t the graph of y = f(x − 2) shift 2 units to the left?), but a comparison of the x-values in the tables in Figures (PageIndex{10})(b) and (PageIndex{12})(b) clearly indicates a shift to the right.

Let’s summarize what we’ve learned about horizontal translations.

Visual Summary — Horizontal Translations (Shifts)

Consider the images in Figure (PageIndex{13}).

• In Figure (PageIndex{13})(a), we see pictured the graph of the original function y = f(x).
• In Figure (PageIndex{13})(b), note that each point on the graph of y = f(x + 1) has an x-value that is 1 unit less than the x-value of the corresponding point on the graph of y = f(x) in Figure (PageIndex{13})(a).
• In Figure (PageIndex{13})(c), note that each point on the graph of y = f(x − 2) has an x-value that is 2 units greater than the x-value of the corresponding point on the graph of y = f(x) in Figure (PageIndex{13})(a).
• Note that the y-value of each transformed point remains the same.

The visual summary in Figure (PageIndex{13}) makes sketching the graphs of y = f(x + 1) and y = f(x − 2) an easy task.

• Given the graph of y = f(x), to sketch the graph of y = f(x + 1), simply take each point on the graph of y = f(x) and shift it 1 unit to the left, keeping the same y-value.
• Given the graph of y = f(x), to sketch the graph of y = f(x − 2), simply take each point on the graph of y = f(x) and shift it 2 units to the right, keeping the same y-value.

In general, we can state the following.

Summary

Suppose that we are given the graph of y = f(x) and suppose that c is any positive real number.

• The graph of y = f(x+c) is shifted c units to the left of the graph of y = f(x).
• The graph of y = f(x−c) is shifted c units to the right of the graph of y = f(x).

When we looked at vertical translations in the previous section, a translation was described by first imagining a graph on a sheet of transparent plastic, then sliding the transparency (without rotating it) over a coordinate system on a sheet of graph paper. Horizontal translations can be thought of in the same way, as sliding the graph on the transparency c units to the left, or c units to the right.

In this section, let’s take the concepts from the Visual Summaries and put them to work on some final examples.

Example (PageIndex{7})

Consider the graph of f in Figure (PageIndex{14}).

Use the concepts from the Visual Summaries (scaling, reflection, and translation) to sketch the graphs of y = f(2x), y = f(−x), and y = f(x + 2) without creating and referring to tables.

Solution

To sketch the graph of y = f(2x), simply take each point on the graph of y = f(x) in Figure (PageIndex{15})(a) and divide its x-value by 2, keeping the same y-value. The result is shown in Figure (PageIndex{15})(b).

To sketch the graph of y = f(−x), simply take each point on the graph of y = f(x) in Figure (PageIndex{16})(a) and negate its x-value, keeping the same y-value. The result is shown in Figure (PageIndex{16})(b).

To sketch the graph of y = f(x + 2), simply take each point on the graph of y = f(x) in Figure (PageIndex{17})(a) and subtract 2 from its x-value, keeping the same y-value. The result is shown in Figure (PageIndex{17})(b).

## Summary

In this section we’ve seen how a handful of transformations greatly enhance our graphing capability. We end this section by listing the transformations presented in this section and their effects on the graph of a function.

Vertical Transformations

Suppose we are given the graph of y = f(x).

• If a > 1, the graph of y = f(ax) compresses horizontally (toward the y-axis) , both positively and negatively, by a factor of a.
• If 0 < a < 1, the graph of y = f(ax) stretches horizontally (away from the y-axis), both positively and negatively, by a factor of 1/a.
• The graph of y = f(−x) is a reflection of the graph of y = f(x) across the y-axis.
• If c > 0, then the graph of y = f(x + c) is shifted c units to the left of the graph of y = f(x).
• If c > 0, then the graph of y = f(x − c) is shifted c units to the right of the graph of y = f(x).

## Missing Factors 1 to 7 (Horizontal Questions - Full Page)

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## 7.2: Sides and Angles (15 minutes)

### Activity

The purpose of this activity is for students to see that translations, rotations, and reflections preserve lengths and angle measures. Students can use tracing paper to help them draw the figures and make observations about the preservation of side lengths and angle measures under transformations. While the grid helps measure lengths of horizontal and vertical segments, the students may need more guidance when asked to measure diagonal lengths. It is important in the launch to demonstrate for students how to either use the tracing paper or an index card to mark off unit lengths using the grid (MP5).

Since students are creating their own measuring tool, they can only give an estimate, and some flexibility should be allowed in the response. During the discussion, highlight different reasonable answers that students find for the lengths which are not whole numbers.

As students work individually, monitor and ask them to explain how they are performing their transformations and finding the side lengths and angle measures. During the discussion, select students who mention corresponding sides and angles, which they learned in grade 7 when making scaled copies, to share. Also select students who estimated the side lengths for Figure C correctly using either the tracing paper or index card.

### Launch

Tell students, “In this activity you will be performing transformations. You can use tracing paper to help you draw the images of the figures or to check your work.”

Point students to Figure C and tell them, “When you are asked to measure side lengths here, you will need to make a ruler on either tracing paper or on a blank edge of an index card.” This reinforces the strategies and estimates students made in the warm-up.

Give students 3 minutes of quiet think time. Be sure to save at least 5 minutes for the discussion.

For classrooms using the digital version of the activity, the applets contain tools for the three rigid transformations students need. They have to choose which tool to use in each problem. Caution students that a quick click is all that is needed to select a figure. If they move the cursor away, and the image does not seem “highlighted,” it is likely they selected and de-selected the figure.

1. Translate Polygon (A) so point (P) goes to point (P') . In the image, write in the length of each side, in grid units, next to the side using the draw tool.
1. In the image, write the length of each side, in grid units, next to the side.
2. In the image, write the measure of each angle in the interior.

### Student Response

For access, consult one of our IM Certified Partners.

### Launch

Tell students, “In this activity you will be performing transformations. You can use tracing paper to help you draw the images of the figures or to check your work.”

Point students to Figure C and tell them, “When you are asked to measure side lengths here, you will need to make a ruler on either tracing paper or on a blank edge of an index card.” This reinforces the strategies and estimates students made in the warm-up.

Give students 3 minutes of quiet think time. Be sure to save at least 5 minutes for the discussion.

For classrooms using the digital version of the activity, the applets contain tools for the three rigid transformations students need. They have to choose which tool to use in each problem. Caution students that a quick click is all that is needed to select a figure. If they move the cursor away, and the image does not seem “highlighted,” it is likely they selected and de-selected the figure.

1. Translate Polygon (A) so point (P) goes to point (Q) . In the image, write the length of each side, in grid units, next to the side.

Expand Image

Expand Image

In the image, write the length of each side, in grid units, next to the side. You may need to make your own ruler with tracing paper or a blank index card.

In the image, write the measure of each angle in the interior.

Expand Image

### Student Response

For access, consult one of our IM Certified Partners.

### Anticipated Misconceptions

Students may try to count the grid squares on the diagonal side lengths. Remind students to measure these lengths with their tracing paper or index card. Students may also struggle estimating the diagonal side lengths on their self-marked index card or tracing paper. Remind students of how they estimated the lengths for the questions in the warm-up where the ruler was not marked.

### Activity Synthesis

Ask selected students to share how they performed the given transformation for each question. After each explanation, ask the class if they agree or disagree. Introduce students to the idea of corresponding sides and corresponding angles. Ask students to identify the corresponding angles in the first question and the corresponding side lengths in the second since they were not asked about these attributes the first time. The point here is not to find the actual values but to note that the corresponding measurements are equal. Since it is sometimes not possible to measure angles or side lengths exactly, student estimates for these values (both corresponding sides and corresponding angles) may be slightly different.

Point out that for each of the transformations in this activity, the lengths of the sides of the original figure equal the lengths of the corresponding sides in the image, and the measures of the angles in the original figure equal the measures of the corresponding angles in the image. For this reason, we call these transformations rigid transformations: they behave as if we are moving the shapes around without stretching, bending, or breaking them. An example of a non-rigid transformation is one that compresses a figure vertically, like this:

Expand Image

Tell them that a rigid transformation is a transformation where all pairs of corresponding distances and angle measures in the figure and its image are equal. It turns out that translations, reflections, and rotations are the building blocks for all rigid transformations, and we will explore that next.

## How to horizontally stretch a function?

Now that we’ve learned about horizontal stretches and how they affect a base function, it’s time that we apply these on functions’ graphs. Before we start stretching functions horizontally by a certain factor, remember these pointers to stretch graphs faster horizontally:

• Only stretch the base of the graph horizontally so that the y-coordinates would remain in the same position.
• Since the y-coordinates will remain the same, the y-intercept stays the same as well.
• Make sure to double-check critical points on the graph, such as its intercepts, maximum points, and more.
• See if the coordinate points are scaled correctly as well.

Let’s use the table of values shown in the previous section to graph both y = |x| and y = |x/3| to visualize the effects of a horizontal stretch on a graph.

As we have discussed, we’re expecting the graphs to stretch along the base with the values for y remaining constant.

To attain y = |x/3|, we stretch the parent function y = |x| by a factor of 3. The graph shown above confirms this, and we can apply the same process when horizontally stretching the graphs of other functions.

Ready to graph more functions and apply horizontal stretches? Let’s summarize what we have learned so far first before we try out more questions.

### Summary of horizontal stretch definition and properties

Here are some important pointers to remember when answering problems and graphing functions that involve horizontal stretches:

• We can only horizontally stretch a graph by a factor of 1/a when the input value is also increased by a.
• When f(x) is stretched horizontally to f(ax), multiply the x-coordinates by a.
• Retain the y-intercepts’ position.
• The resulting function will have the same range but may have a different domain.
• Given a point (m, n), it becomes (am, n) when stretched horizontally.

Let’s always go back to these five-pointers when in doubt. Being able to master the technique of horizontally stretching graphs can help us graph functions faster and understand their behavior.

Are you ready to test your knowledge? Let’s go ahead and try out some of these problems!

The function, g(x), is obtained by horizontally stretching f(x) = 8x by a scale factor of 1/4. Which of the following is the correct expression for g(x)?

a. g(x) = 32x
b. g(x) =16x
c. g(x) = 2x
d. g(x) =1/2 x

Remember that when we horizontally stretch a function by 1/a, we divide the input value by a. Why don’t we apply this to the problem? We replace x with x/4 to find the expression for g(x).

Simplifying the expression will lead to g(x) = 2x.

Example 2

Write the expressions for g(x) and h(x) in terms of f(x) given the following conditions:

a. The function g(x) is the result of f(x) being stretched horizontally by a factor of 1/5.
b. When we horizontally stretch g(x) by a scale factor of 1/2, we obtain h(x).

Let’s start with g(x). We can horizontally stretch f(x) to obtain g(x), so we divide the input value of f(x) by 5 to obtain g(x)’s expression: f(x/5).

Now that we have g(x), we can find the expression for h(x). We divide g(x)’s input value by 2. Hence, we have:

This means that in terms of f(x), g(x) = f(x/5) and h(x)= f(x/10).

Example 3

The function, f(x), passes through the point (6, 4). If f(x) is horizontally stretched by a scale factor of 1/2, what would be the new x-coordinate of the point?

When we stretch a graph horizontally, we multiply the base function’s x-coordinate by the given scale factor’s denominator to find the new point ly ing along the same y-coordinate.

Hence, we have (6, 4) → (2 ∙ 6, 4). The new x-coordinate of the point will be (12, 4).

The table of values for f(x) is shown below. If h(x) is the result of f(x) being horizontally stretched by a scale factor of 1/4, construct its table of values and retain the current output values.

 x -3 -2 -1 0 1 2 3 f(x) -6 -4 -2 0 2 4 6

When the function is stretched horizontally, we multiply the input values (for this case, it’s the values of x) by the given scale factor’s denominator.

This means that (-2, -4) will be transformed to (4∙ -2, -4) = (-8, -4). We apply the same process for the rest of the values and have h(x)’s table of values as shown below.

 x -12 -8 -4 0 4 8 12 h(x) -6 -4 -2 0 2 4 6

Observe the functions shown below. What is the relationship between f(x) and g(x)?

Just by looking at the graph, we can see that g(x) results from f(x) being stretched horizontally. Why don’t we inspect some points from both graphs?

This is what happens to the points:

• (-3, 9) → (-9, 9)
• (-2, 4) → (-6, 4)
• (-1, 1) → (-3, 1)
• (0, 0) → (0, 0)
• (1, 1) → (3, 1)
• (2, 4) → (6, 4)
• (3, 9) → (9, 9)

Notice how for each case: the x-coordinates of g(x) are all three times larger than f(x). This means that the scale factor used to stretch f(x) is 1/3.

The function g(x) is the result of f(x) being horizontally stretched by a scale factor of 1/3.

The image below shows the graph of f(x). Graph g(x) using the fact that it is the result of f(x) being stretched horizontally by a factor of 1/2. Make sure to include the new critical points for g(x).

Let’s go ahead and find the new critical points of g(x) first. Since f(x) will be stretched horizontally, we’ll multiply the x-coordinates by 2. Hence, we have the following critical points:

Plot these points and stretch the graph of f(x) by 3. Make sure that the y-coordinates and y-intercept remain the same.

This is how the resulting graph should look like. Since g(x) is the result of f(x) being stretched horizontally, we stretch the graph of f(x) by a scale of 3.

Describe the transformations done on the following functions shown below.

b. m(x) = √x → n(x) = 3√(x/4)
c. p(x) = 3x – 1→ q(x) = 3x/4 – 2

If you might have noticed that some may not only exhibit horizontal stretches, you are right!

Keep in mind that we sometimes have to apply different transformation techniques for us to obtain a particular function’s expression.

To check the scale factor applied in the input value of f(x) to attain g(x), let’s express g(x) as a perfect square: g(x) = (x/3) 2 = (1/3 · x) 2 .

Expressing g(x) in terms of f(x), we have g(x) = f(x/3). Hence, g(x) is the result of f(x) being horizontally stretched by a scale factor of 1/3.

We can see two scale factors applied to n(x): 3 on the output value and 1/4 for the input value. Applying what we know on vertical and horizontal stretches, we have n(x) = 3·m(1/4 · x). Meaning, n(x) is the result of m(x) being vertically stretched by a scale factor of 3 and horizontally stretched by a scale factor of 1/4.

Lastly, let’s observe the translations done on p(x).

From this, we can see that q(x) results from p(x) being stretched horizontally by a scale factor of 1/4 and translated one unit downward.

What are the transformations done on f(x) so that it results to g(x) = 2|x/3| – 1? Use the graph of f(x) shown below to guide you. Apply the transformations to graph g(x).

Let’s go ahead and express g(x) in terms of f(x).

g(x) = 2 · |1/3 · x| – 1

This means that the translations on f(x) to obtain g(x) are:

• Horizontally stretched by a scale factor of 1/3.
• Vertically stretched by a scale factor of 2.
• Translated downwards by 1 unit.

Let’s slowly apply these transformations on f(x), starting with horizontally stretching f(x).

Let’s now stretch the resulting graph vertically by a scale factor of 2.

Lastly, let’s translate the graph one unit downward.

Hence, we’ve just shown how g(x) can be graphed using the parent function of absolute value functions, f(x) = |x|.

### Practice Questions

1. The function, g(x), is obtained by horizontally stretching f(x) = 16x 2 by a scale factor of 2. Which of the following is the correct expression for g(x)?

2. Write the expressions for g(x) and h(x) in terms of f(x) given the following conditions:

a. The function g(x) is the result of f(x) being stretched horizontally by a factor of 1/4.

b. When we horizontally stretch g(x) by a scale factor of 1/3, we obtain h(x).

3. The function, f(x), passes through the point (10, 8). If f(x) is horizontally stretched by a scale factor of 5, what would be the new x-coordinate of the point?

4. The table of values for f(x) is shown below. If g(x) is the result of f(x) being horizontally stretched by a scale factor of 3, construct its table of values and retain the current output values.

 x -9 -6 -3 0 3 6 9 f(x) -17 -11 -5 1 7 13 19

5. Observe the functions shown below. What is the relationship between f(x) and g(x)?

6. The image below shows the graph of f(x). Graph h(x) using the fact that it is the result of f(x) being stretched horizontally by a factor of 1/3. Make sure to include the new critical points for g(x).

7. Describe the transformations done on the following functions shown below.

8. What are the transformations done on f(x) so that it results in g(x) = 3√(x/2)? Use the graph of f(x) shown below to guide you. Apply the transformations to graph g(x).

## Horizontal Compression – Properties, Graph, & Examples

Is it possible for us to shrink or compress graphs horizontally? When do we compress graphs along the x-axis, and how does it affect its expression? These are some of the questions you’ll be able to answer once we learn about this unique transformation technique: horizontal compression.

Horizontal compressions occur when the function’s base graphis shrunk along the x-axis and, consequent, away from the y-axis.

Isn’t it interesting that by inspecting coefficients, we can either stretch or compress a function’s graph? Mastering the different types of transformations will save us time and help us better understand functions and graphs.

This is why we’ve written about this topic extensively. Think you need a refresher in any of these topics below? Feel free to click the links!

What is a horizontal compression?

Given that y = f(x) is the function that we want to transform, f(x) will undergo a horizontal compression when the scale factor, a ( where a > 1), is multiplied to the input value or x for this case.

This means that when we multiply x by a scale factor greater than 1, we expect its graph to shrink by the same scale factor.

Why don’t we compress f(x) = x 2 by scale factors of 2 and 4?

As can be observed from the graph, when we multiply x by 2, the new graph is a compressed version of the original graph. We also expect a similar effect when x is multiplied by 4, but this time, the graph compresses by a scale factor of 4.

Note that despite being compressed horizontally, the y-coordinates and intercepts remain the same. What do we expect from the transformed functions’ coordinates?

If the base function passes through (m, n), the horizontally compressed graph will pass through (m/a, n).

How to horizontally compress a function?

We’ve now seen how horizontal compressions affect the graph and points of a function’s graph. Why don’t we start applying our knowledge to compress and graph functions? Before we do, here are some important reminders to remember:

• Make sure to compress the base graph horizontally by checking the reference points’ new positions.
• The y-coordinates would remain in the same position. Consequently, make sure that the y-intercepts remain in the same position.
• Make sure that the graph and its points are scaled correctly.

Why don’t we graph the parent function, f(x) = x 2 , and plot some reference points as well? Our goal is to graph the function g(x) = (4x) 2 .

Based on what we’ve just learned, make sure that the coordinate points’ x-coordinates are reduced by 1/4. The current points will now be:

Let’s go ahead and plot these points and the graph of g(x) to compare the two graphs.

As we have expected, since g(x)’s input value is scaled by a factor of 4, we expect the graph of f(x) to shrink by the same scale factor.

The graph shown above confirms this and shows how g(x) is the result of f(x) being horizontally compressed by a scale factor of 4.

Summary of horizontal compression definition and properties

Are you ready to answer more problems involving horizontal compressions? Why don’t we recap what we learned so far first before we do:

• Horizontal compression depends on the scale factor multiplied by the input value (usually x).
• When f(x) is compressed horizontally to f(ax), divide the x-coordinates by a.
• Retain the y-intercepts and coordinates’ positions.
• The transformed function will have the same range but may have a different domain.
• When a function is compressed horizontally, its point transforms from (m, n) to (m/a, n).

When stuck on a problem or unsure what to do next when compressing a function horizontally, don’t hesitate to go back to these five pointers.

Let’s go ahead and try out problems that involve horizontal compression!

Given that f(x) is equal to 12|x|, what is the expression for g(x) given that is the result of f(x) being compressed horizontally by a scale factor of 4?

When we compress a function horizontally, we multiply the input value by the scale factor. Hence, we have

Write the expressions for g(x) and h(x) in terms of f(x) given the following conditions:

a. When we horizontally compress f(x) by a scale factor of 4, we obtain g(x).
b. The function h(x) is the result of g(x) being compressed horizontally by a factor of 2.

Let’s first find the expression for g(x) in terms of f(x). Since, g(x) is the result of horizontally compressing f(x), we multiply the input value of f(x) by 4.

Now that we have g(x) in terms of f(x), we can use this to find the expression for h(x).

Hence, we have g(x) = f(2x) and h(x) = f(4x).

Determine the relationship shared by f(x) and g(x) based on the graphs shown below.

From the graph alone, we can see that g(x) is the result of f(x) being horizontally compressed. Let’s go ahead and note some reference points to find the scale factor needed to compress f(x) to g(x) horizontally.

We can see that the input values decreased by 1/3, so the scale factor applied on f(x) is 3, and so, g(x) = f(3x).

This means that the function g(x) is the result of f(x) being horizontally compressed by a scale factor of 3.

Apply what you know so far and describe the transformations performed for each pair of functions.

b. m(x) = √x → n(x) = 2√(3x)
c. p(x) = |x| → q(x) = |2x| + 4

By reviewing each pair of graphs, we can see that some may require other transformations that we have learned from the past.

Express g(x) as a perfect square to determine the changes applied to f(x). We have g(x) = (4x) 2 . From this, we can see that g(x) is the result of f(x) being compressed horizontally by a scale factor of 4.

From the two expressions, we can see that n(x) is the result if m(x) being scaled vertically and horizontally. Let’s inspect the coefficients and see what they represent:

n(x) = 2 ∙ m(3x)

We can see that n(x) is the result of m(x) being stretched vertically by a scale factor of 2 and compressed horizontally by a scale factor of 3.

Lastly, let’s see the transformations done on p(x) to reach q(x).

q(x) = p(2x) + 4

Hence, q(x) results from p(x) being compressed horizontally by a scale factor of 2 and translated 4 units downward.

What are the transformations done on f(x) = sin x so that it results to g(x) = 3 sin (2x) + 1? Use the graph of f(x) (from -π to π) shown below then apply the transformations to graph g(x).

First, let’s observe the transformations applied on f(x) to attain g(x).

g(x) = 3 ∙ sin (2 ∙ x) + 1

From this, we can see that to attain g(x), we need to compress f(x) horizontally by a factor of 2, vertically stretch f(x) by a scale factor of 3, and translate the resulting function 1 unit upward.

Let’s slowly graph g(x) by applying these transformations. Let’s start by horizontally compressing f(x) by a scale factor of 2.

Next, let’s vertically stretch the resulting function by a scale factor of 3.

After performing the horizontal compression and vertical stretch on f(x), let’s move the graph one unit upward.

Hence, we have the g(x) graph just by transforming its parent function, y = sin x.

Practice Questions

1. Given that f(x) is equal to 2x 2, what is the expression for g(x), given that is the result of f(x) being compressed horizontally by a scale factor of 3?

a. g(x) = 6x 2
b. g(x) =12x 2
c. g(x) = 18x 2
d. g(x) = 36x 2

2. Write the expressions for m(x) and n(x) in terms of f(x) given the following conditions:

a. When we horizontally compress f(x) by a scale factor of 3, we obtain m(x).
b. The function n(x) is the result of m(x) being compressed horizontally by a factor of 4.

3. Determine the relationship shared by f(x) and g(x) based on the graphs shown below.

4. Apply what you know so far and describe the transformations performed for each pair of functions.

5. What are the transformations done on f(x) = sin x so that it results to g(x) = 4 cos (3x) – 1? Use the graph of f(x) (from -π to π) shown below then apply the transformations to graph g(x).

## Transformation of Functions

We all know that a flat mirror enables us to see an accurate image of ourselves and whatever is behind us. When we tilt the mirror, the images we see may shift horizontally or vertically. But what happens when we bend a flexible mirror? Like a carnival funhouse mirror, it presents us with a distorted image of ourselves, stretched or compressed horizontally or vertically. In a similar way, we can distort or transform mathematical functions to better adapt them to describing objects or processes in the real world. In this section, we will take a look at several kinds of transformations.

### Graphing Functions Using Vertical and Horizontal Shifts

Often when given a problem, we try to model the scenario using mathematics in the form of words, tables, graphs, and equations. One method we can employ is to adapt the basic graphs of the toolkit functions to build new models for a given scenario. There are systematic ways to alter functions to construct appropriate models for the problems we are trying to solve.

#### Identifying Vertical Shifts

One simple kind of transformation involves shifting the entire graph of a function up, down, right, or left. The simplest shift is a vertical shift, moving the graph up or down, because this transformation involves adding a positive or negative constant to the function. In other words, we add the same constant to the output value of the function regardless of the input. For a function g ( x ) = f ( x ) + k ,

units. See [link] for an example.

To help you visualize the concept of a vertical shift, consider that y = f ( x ) .

value increases or decreases depending on the value of k .

The result is a shift upward or downward.

a new function g ( x ) = f ( x ) + k ,

is a constant, is a vertical shift of the function f ( x ) .

All the output values change by k

is positive, the graph will shift up. If k

is negative, the graph will shift down.

To regulate temperature in a green building, airflow vents near the roof open and close throughout the day. [link] shows the area of open vents V

(in square feet) throughout the day in hours after midnight, t .

During the summer, the facilities manager decides to try to better regulate temperature by increasing the amount of open vents by 20 square feet throughout the day and night. Sketch a graph of this new function.

We can sketch a graph of this new function by adding 20 to each of the output values of the original function. This will have the effect of shifting the graph vertically up, as shown in [link].

Notice that in [link], for each input value, the output value has increased by 20, so if we call the new function S ( t ) ,

This notation tells us that, for any value of t , S ( t )

can be found by evaluating the function V

at the same input and then adding 20 to the result. This defines S

as a transformation of the function V ,

in this case a vertical shift up 20 units. Notice that, with a vertical shift, the input values stay the same and only the output values change. See [link].

 t 0 8 10 17 19 24 V ( t ) 0 0 220 220 0 0 S ( t ) 20 20 240 240 20 20

Given a tabular function, create a new row to represent a vertical shift.

1. Identify the output row or column.
2. Determine the magnitude of the shift.
3. Add the shift to the value in each output cell. Add a positive value for up or a negative value for down.

is given in [link]. Create a table for the function g ( x ) = f ( x ) − 3.

 x 2 4 6 8 f ( x ) 1 3 7 11

The formula g ( x ) = f ( x ) − 3

tells us that we can find the output values of g

by subtracting 3 from the output values of f .

Subtracting 3 from each f ( x )

value, we can complete a table of values for g ( x )

 x 2 4 6 8 f ( x ) 1 3 7 11 g ( x ) −2 0 4 8

As with the earlier vertical shift, notice the input values stay the same and only the output values change.

The function h ( t ) = − 4.9 t 2 + 30 t

of a ball (in meters) thrown upward from the ground after t

seconds. Suppose the ball was instead thrown from the top of a 10-m building. Relate this new height function b ( t )

and then find a formula for b ( t ) .

#### Identifying Horizontal Shifts

We just saw that the vertical shift is a change to the output, or outside, of the function. We will now look at how changes to input, on the inside of the function, change its graph and meaning. A shift to the input results in a movement of the graph of the function left or right in what is known as a horizontal shift, shown in [link].

For example, if f ( x ) = x 2 ,

is a new function. Each input is reduced by 2 prior to squaring the function. The result is that the graph is shifted 2 units to the right, because we would need to increase the prior input by 2 units to yield the same output value as given in f .

a new function g ( x ) = f ( x − h ) ,

is a constant, is a horizontal shift of the function f .

is positive, the graph will shift right. If h

is negative, the graph will shift left.

Returning to our building airflow example from [link], suppose that in autumn the facilities manager decides that the original venting plan starts too late, and wants to begin the entire venting program 2 hours earlier. Sketch a graph of the new function.

to be the original program and F ( t )

to be the revised program.

In the new graph, at each time, the airflow is the same as the original function V

was 2 hours later. For example, in the original function V ,

the airflow starts to change at 8 a.m., whereas for the function F ,

the airflow starts to change at 6 a.m. The comparable function values are V ( 8 ) = F ( 6 ) .

See [link]. Notice also that the vents first opened to 220 ft 2

at 10 a.m. under the original plan, while under the new plan the vents reach 220 ft 2

In both cases, we see that, because F ( t )

starts 2 hours sooner, h = − 2.

That means that the same output values are reached when F ( t ) = V ( t − ( − 2 ) ) = V ( t + 2 ) .

has the effect of shifting the graph to the left.

Horizontal changes or “inside changes” affect the domain of a function (the input) instead of the range and often seem counterintuitive. The new function F ( t )

uses the same outputs as V ( t ) ,

but matches those outputs to inputs 2 hours earlier than those of V ( t ) .

Said another way, we must add 2 hours to the input of V

to find the corresponding output for F : F ( t ) = V ( t + 2 ) .

Given a tabular function, create a new row to represent a horizontal shift.

1. Identify the input row or column.
2. Determine the magnitude of the shift.
3. Add the shift to the value in each input cell.

is given in [link]. Create a table for the function g ( x ) = f ( x − 3 ) .

 x 2 4 6 8 f ( x ) 1 3 7 11

The formula g ( x ) = f ( x − 3 )

tells us that the output values of g

are the same as the output value of f

when the input value is 3 less than the original value. For example, we know that f ( 2 ) = 1.

To get the same output from the function g ,

we will need an input value that is 3 larger. We input a value that is 3 larger for g ( x )

because the function takes 3 away before evaluating the function f .

We continue with the other values to create [link].

 x 5 7 9 11 x − 3 2 4 6 8 f ( x − 3 ) 1 3 7 11 g ( x ) 1 3 7 11

The result is that the function g ( x )

has been shifted to the right by 3. Notice the output values for g ( x )

remain the same as the output values for f ( x ) ,

but the corresponding input values, x ,

have shifted to the right by 3. Specifically, 2 shifted to 5, 4 shifted to 7, 6 shifted to 9, and 8 shifted to 11.

[link] represents both of the functions. We can see the horizontal shift in each point.

[link] represents a transformation of the toolkit function f ( x ) = x 2 .

Relate this new function g ( x )

and then find a formula for g ( x ) .

Notice that the graph is identical in shape to the f ( x ) = x 2

function, but the x-values are shifted to the right 2 units. The vertex used to be at (0,0), but now the vertex is at (2,0). The graph is the basic quadratic function shifted 2 units to the right, so

Notice how we must input the value x = 2

to get the output value y = 0

the x-values must be 2 units larger because of the shift to the right by 2 units. We can then use the definition of the f ( x )

function to write a formula for g ( x )

To determine whether the shift is + 2

, consider a single reference point on the graph. For a quadratic, looking at the vertex point is convenient. In the original function, f ( 0 ) = 0.

In our shifted function, g ( 2 ) = 0.

To obtain the output value of 0 from the function f ,

we need to decide whether a plus or a minus sign will work to satisfy g ( 2 ) = f ( x − 2 ) = f ( 0 ) = 0.

For this to work, we will need to subtract 2 units from our input values.

gives the number of gallons of gas required to drive m

miles. Interpret G ( m ) + 10

can be interpreted as adding 10 to the output, gallons. This is the gas required to drive m

miles, plus another 10 gallons of gas. The graph would indicate a vertical shift.

can be interpreted as adding 10 to the input, miles. So this is the number of gallons of gas required to drive 10 miles more than m

miles. The graph would indicate a horizontal shift.

Given the function f ( x ) = x ,

graph the original function f ( x )

and the transformation g ( x ) = f ( x + 2 )

on the same axes. Is this a horizontal or a vertical shift? Which way is the graph shifted and by how many units?

are shown below. The transformation is a horizontal shift. The function is shifted to the left by 2 units.

#### Combining Vertical and Horizontal Shifts

Now that we have two transformations, we can combine them together. Vertical shifts are outside changes that affect the output ( y -

) axis values and shift the function up or down. Horizontal shifts are inside changes that affect the input ( x -

) axis values and shift the function left or right. Combining the two types of shifts will cause the graph of a function to shift up or down and right or left.

Given a function and both a vertical and a horizontal shift, sketch the graph.

1. Identify the vertical and horizontal shifts from the formula.
2. The vertical shift results from a constant added to the output. Move the graph up for a positive constant and down for a negative constant.
3. The horizontal shift results from a constant added to the input. Move the graph left for a positive constant and right for a negative constant.
4. Apply the shifts to the graph in either order.

sketch a graph of h ( x ) = f ( x + 1 ) − 3.

is our toolkit absolute value function. We know that this graph has a V shape, with the point at the origin. The graph of h

is a change on the inside of the function, giving a horizontal shift left by 1, and the subtraction by 3 in f ( x + 1 ) − 3

is a change to the outside of the function, giving a vertical shift down by 3. The transformation of the graph is illustrated in [link].

Let us follow one point of the graph of f ( x ) = | x | .

is transformed first by shifting left 1 unit:

is transformed next by shifting down 3 units:

sketch a graph of h ( x ) = f ( x − 2 ) + 4.

Write a formula for the graph shown in [link], which is a transformation of the toolkit square root function.

The graph of the toolkit function starts at the origin, so this graph has been shifted 1 to the right and up 2. In function notation, we could write that as

Using the formula for the square root function, we can write

Note that this transformation has changed the domain and range of the function. This new graph has domain [ 1 , ∞ )

Write a formula for a transformation of the toolkit reciprocal function f ( x ) = 1 x

that shifts the function’s graph one unit to the right and one unit up.

### Graphing Functions Using Reflections about the Axes

Another transformation that can be applied to a function is a reflection over the x- or y-axis. A vertical reflection reflects a graph vertically across the x-axis, while a horizontal reflection reflects a graph horizontally across the y-axis. The reflections are shown in [link].

Notice that the vertical reflection produces a new graph that is a mirror image of the base or original graph about the x-axis. The horizontal reflection produces a new graph that is a mirror image of the base or original graph about the y-axis.

a new function g ( x ) = − f ( x )

is a vertical reflection of the function f ( x ) ,

sometimes called a reflection about (or over, or through) the x-axis.

a new function g ( x ) = f ( − x )

is a horizontal reflection of the function f ( x ) ,

sometimes called a reflection about the y-axis.

Given a function, reflect the graph both vertically and horizontally.

1. Multiply all outputs by –1 for a vertical reflection. The new graph is a reflection of the original graph about the x-axis.
2. Multiply all inputs by –1 for a horizontal reflection. The new graph is a reflection of the original graph about the y-axis.

Reflect the graph of s ( t ) = t

(a) vertically and (b) horizontally.

Reflecting the graph vertically means that each output value will be reflected over the horizontal t-axis as shown in [link].

Because each output value is the opposite of the original output value, we can write

Notice that this is an outside change, or vertical shift, that affects the output s ( t )

values, so the negative sign belongs outside of the function.

Reflecting horizontally means that each input value will be reflected over the vertical axis as shown in [link].

Because each input value is the opposite of the original input value, we can write

Notice that this is an inside change or horizontal change that affects the input values, so the negative sign is on the inside of the function.

Note that these transformations can affect the domain and range of the functions. While the original square root function has domain [ 0 , ∞ )

the vertical reflection gives the V ( t )

and the horizontal reflection gives the H ( t )

function the domain ( − ∞ , 0 ] .

Reflect the graph of f ( x ) = | x − 1 |

(a) vertically and (b) horizontally.

is given as [link]. Create a table for the functions below.

the negative sign outside the function indicates a vertical reflection, so the x-values stay the same and each output value will be the opposite of the original output value. See [link].

the negative sign inside the function indicates a horizontal reflection, so each input value will be the opposite of the original input value and the h ( x )

values stay the same as the f ( x )

is given as [link]. Create a table for the functions below.

A common model for learning has an equation similar to k ( t ) = − 2 − t + 1 ,

is the percentage of mastery that can be achieved after t

practice sessions. This is a transformation of the function f ( t ) = 2 t

shown in [link]. Sketch a graph of k ( t ) .

This equation combines three transformations into one equation.

• A horizontal reflection: f ( − t ) = 2 − t
• A vertical reflection: − f ( − t ) = − 2 − t
• A vertical shift: − f ( − t ) + 1 = − 2 − t + 1

We can sketch a graph by applying these transformations one at a time to the original function. Let us follow two points through each of the three transformations. We will choose the points (0, 1) and (1, 2).

1. First, we apply a horizontal reflection: (0, 1) (–1, 2).
2. Then, we apply a vertical reflection: (0, −1) (-1, –2).
3. Finally, we apply a vertical shift: (0, 0) (-1, -1).

This means that the original points, (0,1) and (1,2) become (0,0) and (-1,-1) after we apply the transformations.

In [link], the first graph results from a horizontal reflection. The second results from a vertical reflection. The third results from a vertical shift up 1 unit.

As a model for learning, this function would be limited to a domain of t ≥ 0 ,

with corresponding range [ 0 , 1 ) .

Given the toolkit function f ( x ) = x 2 ,

Take note of any surprising behavior for these functions.

Notice: g ( x ) = f ( − x )

### Determining Even and Odd Functions

Some functions exhibit symmetry so that reflections result in the original graph. For example, horizontally reflecting the toolkit functions f ( x ) = x 2

will result in the original graph. We say that these types of graphs are symmetric about the y-axis. Functions whose graphs are symmetric about the y-axis are called even functions.

If the graphs of f ( x ) = x 3

were reflected over both axes, the result would be the original graph, as shown in [link].

We say that these graphs are symmetric about the origin. A function with a graph that is symmetric about the origin is called an odd function.

Note: A function can be neither even nor odd if it does not exhibit either symmetry. For example, f ( x ) = 2 x

is neither even nor odd. Also, the only function that is both even and odd is the constant function f ( x ) = 0.

A function is called an even function if for every input x

The graph of an even function is symmetric about the y -

A function is called an odd function if for every input x

The graph of an odd function is symmetric about the origin.

Given the formula for a function, determine if the function is even, odd, or neither.

Determine whether the function satisfies f ( x ) = f ( − x ) .

Is the function f ( x ) = x 3 + 2 x

Without looking at a graph, we can determine whether the function is even or odd by finding formulas for the reflections and determining if they return us to the original function. Let’s begin with the rule for even functions.

This does not return us to the original function, so this function is not even. We can now test the rule for odd functions.

in [link]. Notice that the graph is symmetric about the origin. For every point ( x , y )

on the graph, the corresponding point ( − x , − y )

is also on the graph. For example, (1, 3) is on the graph of f ,

and the corresponding point ( −1 , −3 )

Is the function f ( s ) = s 4 + 3 s 2 + 7

### Graphing Functions Using Stretches and Compressions

Adding a constant to the inputs or outputs of a function changed the position of a graph with respect to the axes, but it did not affect the shape of a graph. We now explore the effects of multiplying the inputs or outputs by some quantity.

We can transform the inside (input values) of a function or we can transform the outside (output values) of a function. Each change has a specific effect that can be seen graphically.

#### Vertical Stretches and Compressions

When we multiply a function by a positive constant, we get a function whose graph is stretched or compressed vertically in relation to the graph of the original function. If the constant is greater than 1, we get a vertical stretch if the constant is between 0 and 1, we get a vertical compression. [link] shows a function multiplied by constant factors 2 and 0.5 and the resulting vertical stretch and compression.

a new function g ( x ) = a f ( x ) ,

is a constant, is a vertical stretch or vertical compression of the function f ( x ) .

then the graph will be stretched.

then the graph will be compressed.

then there will be combination of a vertical stretch or compression with a vertical reflection.

Given a function, graph its vertical stretch.

the graph is stretched by a factor of a .

the graph is compressed by a factor of a .

the graph is either stretched or compressed and also reflected about the x-axis.

models the population of fruit flies. The graph is shown in [link].

A scientist is comparing this population to another population, Q ,

whose growth follows the same pattern, but is twice as large. Sketch a graph of this population.

Because the population is always twice as large, the new population’s output values are always twice the original function’s output values. Graphically, this is shown in [link].

If we choose four reference points, (0, 1), (3, 3), (6, 2) and (7, 0) we will multiply all of the outputs by 2.

The following shows where the new points for the new graph will be located.

Symbolically, the relationship is written as

This means that for any input t ,

the value of the function Q

is twice the value of the function P .

Notice that the effect on the graph is a vertical stretching of the graph, where every point doubles its distance from the horizontal axis. The input values, t ,

stay the same while the output values are twice as large as before.

Given a tabular function and assuming that the transformation is a vertical stretch or compression, create a table for a vertical compression.

is given as [link]. Create a table for the function g ( x ) = 1 2 f ( x ) .

 x 2 4 6 8 f ( x ) 1 3 7 11

The formula g ( x ) = 1 2 f ( x )

tells us that the output values of g

are half of the output values of f

with the same inputs. For example, we know that f ( 4 ) = 3.

We do the same for the other values to produce [link].

 x 2 4 6 8 g ( x ) 1 2 3 2 7 2 11 2

The result is that the function g ( x )

has been compressed vertically by 1 2 .

Each output value is divided in half, so the graph is half the original height.

is given as [link]. Create a table for the function g ( x ) = 3 4 f ( x ) .

 x 2 4 6 8 f ( x ) 12 16 20 0

The graph in [link] is a transformation of the toolkit function f ( x ) = x 3 .

Relate this new function g ( x )

and then find a formula for g ( x ) .

When trying to determine a vertical stretch or shift, it is helpful to look for a point on the graph that is relatively clear. In this graph, it appears that g ( 2 ) = 2.

With the basic cubic function at the same input, f ( 2 ) = 2 3 = 8.

Based on that, it appears that the outputs of g

the outputs of the function f

From this we can fairly safely conclude that g ( x ) = 1 4 f ( x ) .

We can write a formula for g

by using the definition of the function f .

Write the formula for the function that we get when we stretch the identity toolkit function by a factor of 3, and then shift it down by 2 units.

#### Horizontal Stretches and Compressions

Now we consider changes to the inside of a function. When we multiply a function’s input by a positive constant, we get a function whose graph is stretched or compressed horizontally in relation to the graph of the original function. If the constant is between 0 and 1, we get a horizontal stretch if the constant is greater than 1, we get a horizontal compression of the function.

results in a horizontal stretch or compression. Consider the function y = x 2 .

Observe [link]. The graph of y = ( 0.5 x ) 2

is a horizontal stretch of the graph of the function y = x 2

by a factor of 2. The graph of y = ( 2 x ) 2

is a horizontal compression of the graph of the function y = x 2

a new function g ( x ) = f ( b x ) ,

is a constant, is a horizontal stretch or horizontal compression of the function f ( x ) .

then the graph will be compressed by

then the graph will be stretched by

then there will be combination of a horizontal stretch or compression with a horizontal reflection.

Given a description of a function, sketch a horizontal compression or stretch.

1. Write a formula to represent the function.
2. Set g ( x ) = f ( b x )

Suppose a scientist is comparing a population of fruit flies to a population that progresses through its lifespan twice as fast as the original population. In other words, this new population, R ,

will progress in 1 hour the same amount as the original population does in 2 hours, and in 2 hours, it will progress as much as the original population does in 4 hours. Sketch a graph of this population.

Symbolically, we could write

See [link] for a graphical comparison of the original population and the compressed population.

Note that the effect on the graph is a horizontal compression where all input values are half of their original distance from the vertical axis.

is given as [link]. Create a table for the function g ( x ) = f ( 1 2 x ) .

 x 2 4 6 8 f ( x ) 1 3 7 11

The formula g ( x ) = f ( 1 2 x )

tells us that the output values for g

are the same as the output values for the function f

at an input half the size. Notice that we do not have enough information to determine g ( 2 )

because g ( 2 ) = f ( 1 2 ⋅ 2 ) = f ( 1 ) ,

and we do not have a value for f ( 1 )

in our table. Our input values to g

will need to be twice as large to get inputs for f

that we can evaluate. For example, we can determine g ( 4 ) .

We do the same for the other values to produce [link].

 x 4 8 12 16 g ( x ) 1 3 7 11

[link] shows the graphs of both of these sets of points.

Because each input value has been doubled, the result is that the function g ( x )

has been stretched horizontally by a factor of 2.

looks like the graph of f ( x )

horizontally compressed. Because f ( x )

values have been compressed by 1 3 ,

We might also notice that g ( 2 ) = f ( 6 )

Either way, we can describe this relationship as g ( x ) = f ( 3 x ) .

This is a horizontal compression by 1 3 .

Notice that the coefficient needed for a horizontal stretch or compression is the reciprocal of the stretch or compression. So to stretch the graph horizontally by a scale factor of 4, we need a coefficient of 1 4

This means that the input values must be four times larger to produce the same result, requiring the input to be larger, causing the horizontal stretching.

Write a formula for the toolkit square root function horizontally stretched by a factor of 3.

so using the square root function we get g ( x ) = 1 3 x

### Performing a Sequence of Transformations

When combining transformations, it is very important to consider the order of the transformations. For example, vertically shifting by 3 and then vertically stretching by 2 does not create the same graph as vertically stretching by 2 and then vertically shifting by 3, because when we shift first, both the original function and the shift get stretched, while only the original function gets stretched when we stretch first.

When we see an expression such as 2 f ( x ) + 3 ,

which transformation should we start with? The answer here follows nicely from the order of operations. Given the output value of f ( x ) ,

we first multiply by 2, causing the vertical stretch, and then add 3, causing the vertical shift. In other words, multiplication before addition.

Horizontal transformations are a little trickier to think about. When we write g ( x ) = f ( 2 x + 3 ) ,

for example, we have to think about how the inputs to the function g

relate to the inputs to the function f .

Suppose we know f ( 7 ) = 12.

would produce that output? In other words, what value of x

will allow g ( x ) = f ( 2 x + 3 ) = 12 ?

we would first subtract 3, resulting in a horizontal shift, and then divide by 2, causing a horizontal compression.

This format ends up being very difficult to work with, because it is usually much easier to horizontally stretch a graph before shifting. We can work around this by factoring inside the function.

Let’s work through an example.

Now we can more clearly observe a horizontal shift to the left 2 units and a horizontal compression. Factoring in this way allows us to horizontally stretch first and then shift horizontally.

When combining vertical transformations written in the form a f ( x ) + k ,

first vertically stretch by a

and then vertically shift by k .

When combining horizontal transformations written in the form f ( b x - h ) ,

first horizontally shift by h

and then horizontally stretch by 1 b .

When combining horizontal transformations written in the form f ( b ( x - h ) ) ,

first horizontally stretch by 1 b

and then horizontally shift by h .

Horizontal and vertical transformations are independent. It does not matter whether horizontal or vertical transformations are performed first.

Given [link] for the function f ( x ) ,

create a table of values for the function g ( x ) = 2 f ( 3 x ) + 1.

 x 6 12 18 24 f ( x ) 10 14 15 17

There are three steps to this transformation, and we will work from the inside out. Starting with the horizontal transformations, f ( 3 x )

is a horizontal compression by 1 3 ,

which means we multiply each x -

 x 2 4 6 8 f ( 3 x ) 10 14 15 17

Looking now to the vertical transformations, we start with the vertical stretch, which will multiply the output values by 2. We apply this to the previous transformation. See [link].

 x 2 4 6 8 2 f ( 3 x ) 20 28 30 34

Finally, we can apply the vertical shift, which will add 1 to all the output values. See [link].

 x 2 4 6 8 g ( x ) = 2 f ( 3 x ) + 1 21 29 31 35

in [link] to sketch a graph of k ( x ) = f ( 1 2 x + 1 ) − 3.

To simplify, let’s start by factoring out the inside of the function.

By factoring the inside, we can first horizontally stretch by 2, as indicated by the 1 2

on the inside of the function. Remember that twice the size of 0 is still 0, so the point (0,2) remains at (0,2) while the point (2,0) will stretch to (4,0). See [link].

Next, we horizontally shift left by 2 units, as indicated by x + 2.

Last, we vertically shift down by 3 to complete our sketch, as indicated by the − 3

on the outside of the function. See [link].

Access this online resource for additional instruction and practice with transformation of functions.

### Key Equations

 Vertical shift g ( x ) = f ( x ) + k (up for k > 0 ) Horizontal shift g ( x ) = f ( x − h ) (right for h > 0 ) Vertical reflection g ( x ) = − f ( x ) Horizontal reflection g ( x ) = f ( − x ) Vertical stretch g ( x ) = a f ( x ) ( a > 0 ) Vertical compression g ( x ) = a f ( x ) ( 0 < a < 1 ) Horizontal stretch g ( x ) = f ( b x ) ( 0 < b < 1 ) Horizontal compression g ( x ) = f ( b x ) ( b > 1 )

### Key Concepts

• A function can be shifted vertically by adding a constant to the output. See [link] and [link].
• Relating the shift to the context of a problem makes it possible to compare and interpret vertical and horizontal shifts. See [link].
• Vertical and horizontal shifts are often combined. See [link] and [link].
• A vertical reflection reflects a graph about the x -

axis. A graph can be reflected vertically by multiplying the output by –1.

axis. A graph can be reflected horizontally by multiplying the input by –1.

axis, whereas odd functions are symmetric about the origin.

### Section Exercises

#### Verbal

When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal shift from a vertical shift?

A horizontal shift results when a constant is added to or subtracted from the input. A vertical shifts results when a constant is added to or subtracted from the output.

When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal stretch from a vertical stretch?

When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal compression from a vertical compression?

A horizontal compression results when a constant greater than 1 is multiplied by the input. A vertical compression results when a constant between 0 and 1 is multiplied by the output.

When examining the formula of a function that is the result of multiple transformations, how can you tell a reflection with respect to the x-axis from a reflection with respect to the y-axis?

How can you determine whether a function is odd or even from the formula of the function?

Simplify. If the resulting function is the same as the original function, f ( − x ) = f ( x ) ,

then the function is even. If the resulting function is the opposite of the original function, f ( − x ) = − f ( x ) ,

then the original function is odd. If the function is not the same or the opposite, then the function is neither odd nor even.

#### Algebraic

Write a formula for the function obtained when the graph of f ( x ) = x

is shifted up 1 unit and to the left 2 units.

Write a formula for the function obtained when the graph of f ( x ) = | x |

is shifted down 3 units and to the right 1 unit.

Write a formula for the function obtained when the graph of f ( x ) = 1 x

is shifted down 4 units and to the right 3 units.

Write a formula for the function obtained when the graph of f ( x ) = 1 x 2

is shifted up 2 units and to the left 4 units.

For the following exercises, describe how the graph of the function is a transformation of the graph of the original function f .

is a horizontal shift to the left 43 units of the graph of f .

is a horizontal shift to the right 4 units of the graph of f .

is a vertical shift up 8 units of the graph of f .

is a vertical shift down 7 units of the graph of f .

is a horizontal shift to the left 4 units and a vertical shift down 1 unit of the graph of f .

For the following exercises, determine the interval(s) on which the function is increasing and decreasing.

#### Graphical

For the following exercises, use the graph of f ( x ) = 2 x

shown in [link] to sketch a graph of each transformation of f ( x ) .

For the following exercises, sketch a graph of the function as a transformation of the graph of one of the toolkit functions.

#### Numeric

Tabular representations for the functions f , g ,

are given below. Write g ( x )

as transformations of f ( x ) .

 x −2 −1 0 1 2 f ( x ) −2 −1 −3 1 2
 x −1 0 1 2 3 g ( x ) −2 −1 −3 1 2
 x −2 −1 0 1 2 h ( x ) −1 0 −2 2 3

Tabular representations for the functions f , g ,

are given below. Write g ( x )

as transformations of f ( x ) .

 x −2 −1 0 1 2 f ( x ) −1 −3 4 2 1
 x −3 −2 −1 0 1 g ( x ) −1 −3 4 2 1
 x −2 −1 0 1 2 h ( x ) −2 −4 3 1 0

For the following exercises, write an equation for each graphed function by using transformations of the graphs of one of the toolkit functions.

For the following exercises, use the graphs of transformations of the square root function to find a formula for each of the functions.

For the following exercises, use the graphs of the transformed toolkit functions to write a formula for each of the resulting functions.

For the following exercises, determine whether the function is odd, even, or neither.

For the following exercises, describe how the graph of each function is a transformation of the graph of the original function f .

is a vertical reflection (across the x

is a vertical stretch by a factor of 4 of the graph of f .

is a horizontal compression by a factor of 1 5

is a horizontal stretch by a factor of 3 of the graph of f .

is a horizontal reflection across the y

-axis and a vertical stretch by a factor of 3 of the graph of f .

For the following exercises, write a formula for the function g

that results when the graph of a given toolkit function is transformed as described.

-axis and horizontally compressed by a factor of 1 4

-axis and horizontally stretched by a factor of 2.

The graph of f ( x ) = 1 x 2

is vertically compressed by a factor of 1 3 ,

then shifted to the left 2 units and down 3 units.

is vertically stretched by a factor of 8, then shifted to the right 4 units and up 2 units.

is vertically compressed by a factor of 1 2 ,

then shifted to the right 5 units and up 1 unit.

is horizontally stretched by a factor of 3, then shifted to the left 4 units and down 3 units.

For the following exercises, describe how the formula is a transformation of a toolkit function. Then sketch a graph of the transformation.

The graph of the function f ( x ) = x 2

is shifted to the left 1 unit, stretched vertically by a factor of 4, and shifted down 5 units.

is stretched vertically by a factor of 2, shifted horizontally 4 units to the right, reflected across the horizontal axis, and then shifted vertically 3 units up.

The graph of the function f ( x ) = x 3

is compressed vertically by a factor of 1 2 .

The graph of the function is stretched horizontally by a factor of 3 and then shifted vertically downward by 3 units.

is shifted right 4 units and then reflected across the vertical line x = 4.

For the following exercises, use the graph in [link] to sketch the given transformations.

### Glossary

horizontal compression a transformation that compresses a function’s graph horizontally, by multiplying the input by a constant b > 1 horizontal reflection a transformation that reflects a function’s graph across the y-axis by multiplying the input by −1 horizontal shift a transformation that shifts a function’s graph left or right by adding a positive or negative constant to the input horizontal stretch a transformation that stretches a function’s graph horizontally by multiplying the input by a constant 0 < b < 1 odd function a function whose graph is unchanged by combined horizontal and vertical reflection, f ( x ) = − f ( − x ) ,

and is symmetric about the origin

vertical compression a function transformation that compresses the function’s graph vertically by multiplying the output by a constant 0 < a < 1 vertical reflection a transformation that reflects a function’s graph across the x-axis by multiplying the output by −1 vertical shift a transformation that shifts a function’s graph up or down by adding a positive or negative constant to the output vertical stretch a transformation that stretches a function’s graph vertically by multiplying the output by a constant a > 1

## 1.7: Horizontal Transformations - Mathematics

Fundamentals of Graph Transformations.

We would like to begin looking at the transformations of the graphs of functions. In particular, we will use our familiarity with quadratic equations

with a≠0, where we will be concerned with three general types of transformations in the variables x and y. Let us first look specifically at the basic monic quadratic equation for a parabola with vertex at the origin, (0,0): y = x². Its graph is given below.

We will consider horizontal translations, horizontal scaling, vertical translations and vertical scaling first.

Horizontal translations affect the domain on the function we are graphing. Recall that the domain is the set of all values that we can put in for x in the function without breaking a rule of algebra, such as division by 0, or taking the logarithm of a negative number. If we translate by some positive real number a, then our parabolas equation is changed "in the parentheses". When we say "in the parentheses" in this context we are referring to the notation: f(x) = x² we will mean that we are changing the value in the parentheses. We denote a horizontal translation as follows: y = (x-a)². Let's look at how this affects our graph for a=2:

Notice how the graph moved to the right by 2 units. These are the rules we will follow: We will always write our equations in the form "x-something" and resulting horizontal translation is in the direction of the sign of the something and a number of units equal to its absolute value. The two possibilities are that we have either y = (x-a)², or y = (x+a)² = (x-(-a))². Now on to horizontal scaling.

Horizontal scaling involves multiplying inside the parentheses by a nonzero constant b ≠ 0. If b>1, the values of y associated with a given x in the domain occur in "1/b time" in the absence of a horizontal translation, and if b<1, the values of y associated to a given x in the domain occur in "b time". Let's consider a horizontal scaling by b = 2 > 1 of our basic parabola. Our function will take on the form: y = (2x)² = 4x². Understanding what we mean by 1/b time. Consider our basic parabola without the scaling: y = x², here, if we set x = 4, then y = 16. On the other hand, if we put x = 2 into our scaled equation, we get 16. For our scaled equation we see that the value y =16 is achieved at x=2, 1/2 = 1/b of that needed in the unscaled equation. Let's look at the graph of y = (2x)².

We will now consider vertical translations and scaling. Such transformations affect the range of the function and happen "outside" the parentheses. Let's begin by considering a vertical translation by d ≠ 0, where when d > 0 our translation move the graph vertically up d units, and when d < 0 it the graph is moved vertically down d units. Let's look again at our basic parabola after such a transformation. Let d = 2, then our equation takes the form y = x² + 2. Notice that the 2 is not inside the parentheses as before, and is not squared as a result. Here is a picture of the graph of this equation.

As the last of our basic transformations we will give a vertical scaling. This differs from the horizontal scaling as follows: In the horizontal scaling the x-value needed to produce a given y-value was scaled. However, in the case of a vertical scaling, the y-value resulting from a given x-value is scaled. A vertical scaling by a nonzero constant d > 0 will "grow" the y-value associated to some x-value in the domain by d if d > 1, and "shrink" it if d < 1. Let's look at the vertical scaling of our basic parabola by c = 2.

Now, how do we insure we get the correct graph? Consider the following possible graphs for the equation y = (5x-3)², where we have translated first in the top graph and scaled first in the bottom graph.

How do we know which is correct. One simple manner to do so is to plug some different values of x into the equation y = (5x-3)² and compare the resulting y-values to each of the graphs. Doing so rules out the top graph, pointing us to the correct graph. How could we have known this without checking different x-values? The correct order is a consequence of the order of operations we learned when we were first learning about multiplication and addition we will always handle scaling first, then translations.

What of the transformations I alluded to earlier not already covered? The remaining transformations are horizontal and vertical reflections. The former affects the domain, taking the x-value which produces a specific y-value and sending it to its negative. In our equation it is manifested by allowing our b-values from the scaling above to take on negative value. Because it is a type of scaling, it is handled before translations. Let's look at a graph of the horizontal reflection of the parabola with equation: y = (x+2)². It is given by the equation: y = (-x+2)². The purple graph is associated to the former, and the red to the latter. Notice that the horizontal reflection of a graph is across the y-axis.

Similarly the vertical reflection of a graph send the y-value associated to a given x-value to its negative, reflecting the graph across the x-axis. Consider the graphs of y = x² and y=(-x)² below.

Looking at the Graph of a more complicated Quadratic Equation .

In Exercise 4 of Assignment 2 we are asked to begin by graphing the parabola with equation y = 2x² + 3x -4. Without further adieu I present to you the graph of the aforementioned:

We can see more clearly here by one, or both, of the following means: 1. determining the vertex using the formula for the coordinates of the vertex of a parabola, or 2. completing the square and placing the equation in vertex form. The latter encompasses the former and allows us to see the transformations that yielded this graph. Doing so yields the following two equivalent equations represented by the graph:

The basic equation as given:

The equation after being put into vertex form using the method of completing the square:

Notice that the coordinates of the vertex are:

These are clearly indicated in the vertex form. This is not a coincidence. In our first tasking following the initial graphing is to overlay the graph after making the horizontal translations by a = 4. The graphs are given below. The original is rendered in red, the translation in purple.

We are then asked to make a change to our equation which will move the vertex to the second quadrant. If we assume the graph to be moved is the graph obtained by our prior translation then we will be moving the vertex of our graph from the fourth quadrant to the second. A horizontal reflection followed by a vertical translation will accomplish the desired goal. We will first indicate the horizontal reflection with the translated graph from the prior step, then the graph of the horizontal reflection together with the vertical translation below.

The equations of the horizontal translation of the basic equation from exercise 4, and the subsequent horizontal reflection are given below in their respective orders.

We would now like to move the vertex of the parabola to second quadrant by a vertical translation. In the step following this it will be of great advantage if we shift the vertex to a coordinate with integer value for the y-coordinate. Recalling our observations from before that horizontal translations and reflections leave the range unaffected, we know that the y-coordinate of the vertex is still -41/8, we therefore translate by 49/8 vertically up. Our resulting equation for the parabola and the associated graph are given below. We have graphed the graph resulting from our horizontal reflection together with the graph from this transformation.

The following graph depicts out graph from the completion of step 1 together with our graph from completion of this step. The graph from the completion of step 1 is depicted in red.

Step 3 ask us to make one more change to the graph for the explicit quadratic equation given. After the completion of this step we will attempt to generalize the result. We would like to transform the graph in such a manner that we preserve its vertex coordinates and the new graph is concave down. Recall that the property of being concave down is determined by the sign second derivative. The graph will be concave down if the second derivation of the equation describing it is a negative constant. This can only result from the leading coefficient of the equation being negative, or equivalently, by applying a vertical reflection to the equation. However, the vertical reflection of our graph in the absence of an additional vertical translation results in the following equation and graph.

In order to insure the same vertex we translate by twice the negative of the y-coordinate for the vertex of the parabola we reflected. To obtain the equations and associated graph:

We offer the follow simplifications of the standard form of the quadratic equation after our work and the vertex form of our equation, respectively.

We can then recall the original equation in standard form and vertex to synthesize the changes that have taken place.

Looking at the vertex form we see that we have, in aggregate moved our vertex from

Given the aggregate change to the equation of exercise 4 upon completion of step 3, we might ask what can we say for a general quadratic, y = ax²+bx+c, with a ≠ 0? We must first reformulate the changes in steps 1 through 3 to see what would have happened to the general quadratic and its parabolic graph. We will first consider a, b, and c such that the vertex of the parabola lies in the third quadrant, as are original equation was.

Let y= ax²+bx+c is a non-degenerate quadratic with vertex in the third quadrant and a>0, then the discriminant is greater than zero b²-4ac >0, and c is less than, or equal to, the ratio of b² to 4a. We make the following transformations to the equation given and then show that the vertex of the resulting equation lies in the second quadrant and its graph is concave down on its domain.

Here is a setting where the students will have experience, having nothing to do with skills in algebra: daylight savings time (since you're in the USA). When we advance clocks ahead by an hour we lose an hour of sleep. Thus replacing $t$ with $t+1$ shifts your schedule for the day back by one hour.

Here, $t$ is the old "function" and $t + 1$ is the new "function". $t$ is one hour behind $t + 1$, but after this transformation, $t + 1$ shortens the day by one hour. So in the end, we lose an hour. Point out to the students that it also works the other way when we turn the clocks back ($t$ is replaced with $t-1$) and then gain an hour in our schedule.

As described in the link, to plot a graph, simply imagine the $x$-axis covered in coconuts, one for every $x$ value, like this:

with "$color<[cdot]>$" indicating the coconut at the origin. The Function Monkey strolls along the axis, picks up each $x$ coconut, evaluates the corresponding $y$ value (as is his wont), and throws the coconut to the appropriate height (or depth). Graph plotted!

But, wait! No one ever said that the values of the coconuts were required to match the values of the $x$ coordinates which have been coloured for convenience. Hmmmm .

If the coconut at (the coloured) location $x$ has value $x+1$, then the row of coconuts looks like this: $cdots quad (-2) quad color<#9509A5> <(-1)>quad color <()>quad color <[1]>quad color< #2B87CD> <(2)>color<#D86907> quad (4) quad cdots$

Importantly, the coconut locations in colour have not changed. The coconuts still sit on the axis at locations $x = cdots, -3, color<#9509A5><-2>,color<-1>, color<0>,color<#2B87CD><1>, color<#D86907><2>, 3, cdots$ still indicating the coconut at the origin.

So far as the Function Monkey is concerned, adding $1$ to each coconut value has effectively shifted the row of coconuts to the left. (Likewise, adding $-1$ to each coconut value effectively shifts the row of coconuts to the right.) Since the Function Monkey throws coconuts vertically, the plotted graph shifts the same way.

In a similar manner, multiplying each (original) coconut value by $2$ yields this row of coconuts:

Again, the locations of the coconuts haven't changed, but we see that the span of coconut values from $-6$ to $6$ has been compressed into the space between locations $x=-3$ and $x=3$. The graph will exhibit the same, horizontal, compression.

A plot-less way of thinking about this, which is close to your "head start" interpretation, is that "$f(x+1)$" fools Function Monkey $f$ into thinking that each input value is one unit bigger than it really is. Similarly "$f(2x)$" fools him into thinking that each input is twice its actual size. The effective changes in output values correspond to the Function Monkey's altered perceptions.

## Constructions and Rigid Motions

Students construct (with both compass and straight-edge and technology, rather than the physical models and transparencies used in eighth grade) perpendicular lines, parallel lines, and regular polygons, and develop formal definitions of these objects. Students then develop more precise definitions for translations, rotations, and reflections, and use these to describe symmetries - single rigid transformations that carry objects to themselves. Careful attention is given to properties of figures that are preserved (for example, as the result of a translation, a line segment is both parallel and congruent to the pre-image), as they will be important to following work with transformational proofs. Additionally, coordinates are used to represent rigid motions as functions that map points in the plane to points in the plane.

Students have worked with geometric shapes since kindergarten. In grade 4, they classified two-dimensional shapes by properties of their sides and angles, and in grade 5, they classified these shapes in a hierarchy based on properties. In grade 8, they were introduced to the concepts of rotation, reflection, and translation via physical models, transparencies, or geometry software. They defined congruence of two-dimensional figures in terms of these rigid motions, understanding that two figures were congruent if one could be obtained from the other via a sequence of rigid motions. Given a two-dimensional figure, students described the effect of a rotation, reflection, or translation on the figure in terms of coordinates. Students also worked with determining distances in the coordinate plane using the Pythagorean Theorem.

## Review

1. Graph the function f(x)=2|x&minus1|&minus3 without a calculator.
2. What is the vertex of the graph and how do you know?
3. Does it open up or down and how do you know?
4. For the function: f(x)=|x|+c if c is positive, the graph shifts in what direction?
5. For the function: f(x)=|x|+c if c is negative, the graph shifts in what direction?
6. The function g(x)=|x&minusa| represents a shift to the right or the left?
7. The function h(x)=|x+a| represents a shift to the right or the left?
8. If a graph is in the form a&sdotf(x). What is the effect of changing the a?

Describe the transformation that has taken place for the parent function f(x)=|x|.

Write an equation that reflects the transformation that has taken place for the parent function ( g(x)=frac<1>), for it to move in the following ways:

1. Move two spaces up
2. Move four spaces to the right
3. Stretch it by 2 in the y-direction

Write an equation for each described transformation.

1. a V-shape shifted down 4 units.
2. a V-shape shifted left 6 units
3. a V-shape shifted right 2 units and up 1 unit.

The following graphs are transformations of the parent function f(x)=|x| in the form of f(x)=a|x&minush|=k. Graph or sketch each to observe the type of transformation.

1. f(x)=|x|+2. What happens to the graph when you add a number to the function? (i.e. f(x) + k).
2. f(x)=|x|&minus4. What happens to the graph when you subtract a number from the function? (i.e. f(x) - k).
3. f(x)=|x&minus4|. What happens to the graph when you subtract a number in the function? (i.e. f(x - h)).
4. f(x)=|x+2|. What happens to the graph when you add a number in the function? (i.e. f(x + h)).

Practice: Graph the following.

1. f(x)=2|x|
2. ( f(x)=frac<5><2>|x|)
3. ( f(x)=frac<1><2>|x|)
4. ( f(x)=frac<2><5>|x|)
5. Let f(x)=x 2 . Let g(x) be the function obtained by shifting the graph of f(x) two units to the right and then up three units. Find a formula for g(x) and then draw its graph

Suppose H(t) gives the height of high tide in Hawaii(H) on a Tuesday, (t) of the year. Use shifts of the function H(t) to find formulas of each of the following functions: