Horizontal Stretches and Compressions

4.11.5 Horizontal Stretches and Compressions

4.11.5 • Horizontal Stretches and Compressions

Horizontal Stretches and Compressions

Access the Desmos guide PDF for tips on solving problems with the Desmos graphing calculator.

1. On the same coordinate grid, graph the following linear functions. You do not need to label the lines in the graphing tool.

Line Label	Function
Line $i$	$f (x) = x$
Line $j$	$f (2 x)$
Line $k$	$f (\frac{1}{2} x)$

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Compare your work:

A graph compares three functions: f of x equals x is the red dashed line, f of 2 times x is the blue solid line, and f of one-half times x is the green dotted line, each plotted on a coordinate plane with labeled axes. Each line goes through the orgin and has a positive slope.

2. What is the difference between what you entered for the functions of lines $j$ and $k$ in comparison to the function for line $j$ ?

Compare your answer: Line $i$ is the parent linear function. Lines $j$ and $k$ are the result of function f where the input is different from the original $x$ -values.

3. Now, graph lines $m$ and $n$ on the same graph as $i - k$ .

Line Label	Function
Line $m$	$m (x) = 2 x$
Line $n$	$n (x) = \frac{1}{2} x$

Compare your work:

A graph shows lines for functions: f of x equals x is the orange dashed line, f of 2 times x is the blue solid line, f of one-half times x is the green solid line, m of x equals 2 times x is the red dotted line, and n of x equals one-half times x is the black dashed line. All lines go through the origin with a positive slope.

4. How does line $j$ , representing $f (2 x)$ , compare to line $m$ , representing $m (x) = 2 x$ ?

Compare your answer: The same points are in both functions. The lines appear to be exactly the same.

5. How does line $k$ , representing $f (\frac{1}{2} x)$ , compare to line $n$ , representing $n (x) = \frac{1}{2} x$ ?

Compare your answer: The same points are in both functions. The lines appear to be exactly the same.

Horizontal dilations that stretch and compress linear functions are difficult to identify from graphs because they can appear as if they are vertical dilations. However, they are different because:

The changes to the function are applied to the input values of a function when the transformation is a horizontal dilation.
The changes to the function are applied to the output values of a function when the transformation is a vertical dilation.

6. Now graph the lines given below on the same coordinate grid.

Line Label	Function
Line $i$	$f (x) = x$
Line $j$	$f (2 x)$
Line $p$	$f (3 x)$
Line $q$	$f (4 x)$

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Compare your work:

Line graph comparing four functions: f of x equals x is the dashed orange line, f of 2 times x is the solid blue line, f of 3 times x is the dotted green line, and f of 4 times x is the solid red line, each with increasing steepness. All lines go through the origin with a positive slope.

7. As the coefficient on the input increases, what is happening to the graph?

Compare your answer: The line gets closer and closer to the $y$ -axis because the function is being horizontally compressed.

8. Graph the lines given below on the same coordinate grid.

Line Label	Function
Line $i$	$f (x) = x$
Line $k$	$f (\frac{1}{2} x)$
Line $r$	$f (\frac{1}{4} x)$
Line $s$	$f (\frac{1}{10} x)$

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Compare your work:

A graph shows four lines representing functions: f of x equals x is the dashed orange line, f of one-half times x is the solid blue line, f of one-fourth times x is the green dotted line, and f of one-tenth times x is the solid red line. All lines go through the origin with a positive slope.

9. As the coefficient on the input approaches 0, what is happening to the graph?

Compare your answer: The line gets farther from the $y$ -axis because the function is being horizontally stretched.

10. Graph the lines given below on the same coordinate grid.

Line Label	Function
Line $t$	$f (\frac{1}{2} x)$
Line $u$	$f (- \frac{1}{2} x)$

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Compare your work:

A graph shows two lines: a solid blue line for f of one-half times x, increasing from bottom left to top right, and a dotted green line for f of one-half times x, decreasing from top left to bottom right.

In the function $f (b x)$ , the $b$ is acting as the horizontal stretch or horizontal compression of the parent function. It dilates the graph.

When $b$ is negative, there is also a horizontal reflection of the graph over the $y$ -axis.
Multiplying the function's input by $b$ horizontally compresses the graph of $f$ by a factor of $b$ units if $b > 1$ .
The $b$ horizontally expands the graph of $f$ by a factor of $b$ units if $0 < b < 1$ .

Graph with eight colorful lines showing transformations of f of x equals a times x. Each line, labeled in matching colors, represents different functions: f of one-tenth times x, f of one-fourth times x, f of one-half times x, f of 2 times x, f of 3 times x, f of negative x, f of negative one-half times x.

Horizontal Dilations of a Function

A horizontal stretch or compression “transforms” the parent function into another function by horizontally dilating the graph by “b” units.

Horizontal Dilation → $f (b x)$

If the $b$ value is negative, the graph is reflected over the $y$ -axis. If the $b$ value is greater than 1, the graph is horizontally compressed. If the $b$ value is between 0 and 1, the graph is horizontally stretched.

Self Check

Which function represents $g(x)$ , a function transformed from $f(x)$ by a horizontal compression?

$g(x) = f(x + 5)$
$g(x) = f(x) + 5$
$g(x) = f(5x)$
$g(x) = 5f(x)$

Additional Resources

Horizontal Transformations of Linear Functions

Horizontal transformations are the result of altering the input values before applying the function rule. Vertical transformations result from altering the output values after the function rule has been applied.

Example 1

Describe how $f (x)$ is transformed into $f (x - 3)$ .

Solution

First, graph the given function, $f (x)$ , the parent linear function.

A thick, red dashed line with a positive slope passes through the origin on an x-y grid, extending from the lower left to the upper right. Both axes are labeled and ticked from negative 4 to 5 for x and negative 4 to 7 for y.

Now, let’s examine an input value of $x = 2$ using $f (x)$ .

For the parent function $f (x)$ , all points on the line are represented by $(x, f (x))$ .
So, when $x = 2$ , then $f (x)$ becomes $f (2)$ .
That means this point is located at $(2, f (2))$ or $(2, 2)$ .

In order to get the same output value for $f (x - 3)$ as we had for $f (x)$ , we need to figure out how to change the input value.

Because our transformed equation is subtracting three from what we input, we have to increase the prior input value by 3 units to yield the same output as $f (x)$ .
The input value that is 3 units larger than $x = 2$ is 5. In the transformed equation, we must use $x = 5$ .

So, in the original equation where $x = 2$ resulted in $f (2) = 2$ , in the transformed equation, we have to use $x = 5$ in order for $f (x - 3) = f (5 - 3) = f (2)$ .

A graph shows two lines: a red dashed line with open circles at (2,2), and a blue solid line passing through the points (2, negative 1) and (5,2) with an open circle.

On the graph, $x = 5$ is located 3 units to the right of $x = 2$ . So, the transformed equation of $f (x - 3)$ is 3 units to the right of $f (x)$ .

Horizontal transformations are represented by the rule $f (x - c)$ where $c$ indicates the number of units to the right or left that the graph is shifted.

Example 2

Graph $f (x + 2)$ when $f (x) = \frac{1}{2} x - 3$ . Then, determine the equation of the transformed function.

Solution

Step 1 - Graph the given function, $f (x) = \frac{1}{2} x - 3$ .

The $y$ -intercept is (0, -3) and the slope is $\frac{1}{2}$ .

For horizontal shifts, it is helpful to identify the $x$ -intercept. On our graph, the $x$ -intercept or zero is located at (6,0).

A graph with a dashed red line with a positive slope. There is a white open circle on the line at the point (6, 0).

Step 2 - Apply the designated transformation. Remember that the rule for a horizontal shift is $f (x - c)$ . If we rewrite $f (x + 2)$ in an equivalent form, we can make it is easier to see $x - c$ .

$f (x + 2) = f (x - (- 2))$ That means $c = - 2$ .

So, $f (x + 2)$ represents a horizontal shift of two units to the left. On our graph, we move the $x$ -intercept two units left from (6, 0) to (4, 0).

A graph with a solid blue line and a dashed red line are parallel on a grid. The blue line passes through points (4, 0), and (6,0), which are marked with open circles. Axes range from -2 to 8 on x and -5 to 6 on y.

The orange dashed line is the graph of the original function and the blue solid line is the transformed function.

Step 3 - Determine the equation of the transformed function.

The $y$ -intercept of the transformed line is (0, -2) and the slope of the line is $\frac{2}{4}$ or $\frac{1}{2}$ .

That means the equation $f (x + 2) = \frac{1}{2} x - 2$

Want to check your work? We can check our equation algebraically by substituting $x + 2$ into $f (x)$ .

$f (x) = \frac{1}{2} x - 3$

$f (x + 2) = \frac{1}{2} (x + 2) - 3 = \frac{1}{2} x + 1 - 3 = \frac{1}{2} x - 2$

Example 3

Graph $f (4 x)$ when $f (x) = \frac{1}{2} x - 3$ . Then, determine the equation of the transformed function.

Solution

Step 1 - Graph the given function, $f (x) = \frac{1}{2} x - 3$ .

The $y$ -intercept is (0, -3) and the slope is $\frac{1}{2}$ .

For horizontal shifts, it is helpful to identify the $x$ -intercept. On our graph, the x-intercept or zero is located at (6,0).

Step 2 - Apply the designated transformation. Remember that the rule for a horizontal shift is $f (b x)$ . And, when the value of $b$ is greater than 1, the function experiences a horizontal compression.

As a result of the horizontal compression, the linear function should look steeper because it is being compressed by a scale factor of 4.

On our graph, this means the $x$ -intercept is compressed from (6, 0) to (1.5, 0).

A graph showing a blue line with positive slope crossing the y-axis at -3, and a red dashed line with positive slope crossing the y-axis at negative 3. The lines intersect at point (0, negative 3) with an open white circle. There are also open white circles at (1.5, 0) and (6, 0).

The orange dashed line is the graph of the original function and the blue solid line is the transformed function.

Step 3 - Determine the equation of the transformed function.

The $y$ -intercept of the transformed line is (0, -3) and the slope of the line is 3/1.5 or 2.

That means the equation $f (4 x) = 2 x - 3$ .

Want to check your work? We can check our equation algebraically by substituting 4x into $f (x)$ .

$f (x) = \frac{1}{2} x - 3$

$f (4 x) = \frac{1}{2} (4 x) - 3 = 2 x - 3$ ✔

Try it

Horizontal Transformations of Linear Functions

Name the transformation to go from the parent function, $f (x) = x$ , to $f (x - 1)$ .

Here is how to determine the transformation:

Because $f (x - c)$ defines a horizontal shift, the transformation $f (x - 1)$ moves the parent function 1 unit to the right.