Now that we have two transformations, we can combine them together. Vertical shifts are outside changes that affect the output ( [latex]y\text{-}[/latex] ) axis values and shift the function up or down. Horizontal shifts are inside changes that affect the input ( [latex]x\text{-}[/latex] ) 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.

Example: Graphing Combined Vertical and Horizontal Shifts

Given [latex]f\left(x\right)=|x|[/latex], sketch a graph of [latex]h\left(x\right)=f\left(x+1\right)-3[/latex].

The function [latex]f[/latex] is our toolkit absolute value function. We know that this graph has a V shape, with the point at the origin. The graph of [latex]h[/latex] has transformed [latex]f[/latex] in two ways: [latex]f\left(x+1\right)[/latex] is a change on the inside of the function, giving a horizontal shift left by 1, and the subtraction by 3 in [latex]f\left(x+1\right)-3[/latex] is a change to the outside of the function, giving a vertical shift down by 3. The transformation of the graph is illustrated below.

Let us follow one point of the graph of [latex]f\left(x\right)=|x|[/latex].

• The point [latex]\left(0,0\right)[/latex] is transformed first by shifting left 1 unit: [latex]\left(0,0\right)\to \left(-1,0\right)[/latex]
• The point [latex]\left(-1,0\right)[/latex] is transformed next by shifting down 3 units: [latex]\left(-1,0\right)\to \left(-1,-3\right)[/latex]

Below is the graph of [latex]h[/latex].

Try It

Given [latex]f\left(x\right)=|x|[/latex], sketch a graph of [latex]h\left(x\right)=f\left(x - 2\right)+4[/latex].
Check your work with Desmos.

Solution

Example: Identifying Combined Vertical and Horizontal Shifts

Write a formula for the graph shown below, which is a transformation of the toolkit square root function.

Solution

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

[latex]h\left(x\right)=f\left(x - 1\right)+2[/latex]

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

[latex]h\left(x\right)=\sqrt{x - 1}+2[/latex]

Analysis of the Solution

Note that this transformation has changed the domain and range of the function. This new graph has domain [latex]\left[1,\infty \right)[/latex] and range [latex]\left[2,\infty \right)[/latex].

Example: Applying a Learning Model Equation

A common model for learning has an equation similar to [latex]k\left(t\right)=-{2}^{-t}+1[/latex], where [latex]k[/latex] is the percentage of mastery that can be achieved after [latex]t[/latex] practice sessions. This is a transformation of the function [latex]f\left(t\right)={2}^{t}[/latex] shown below. Sketch a graph of [latex]k\left(t\right)[/latex].

Solution

This equation combines three transformations into one equation.

• A horizontal reflection: [latex]f\left(-t\right)={2}^{-t}[/latex]
• A vertical reflection: [latex]-f\left(-t\right)=-{2}^{-t}[/latex]
• A vertical shift: [latex]-f\left(-t\right)+1=-{2}^{-t}+1[/latex]

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 the graphs below, 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.

Analysis of the Solution

As a model for learning, this function would be limited to a domain of [latex]t\ge 0[/latex], with corresponding range [latex]\left[0,1\right)[/latex].

Try It

Given the toolkit function [latex]f\left(x\right)={x}^{2}[/latex], graph [latex]g\left(x\right)=-f\left(x\right)[/latex] and [latex]h\left(x\right)=f\left(-x\right)[/latex]. Take note of any surprising behavior for these functions.

Solution

Notice: [latex]g(x)=f(−x)[/latex] looks the same as [latex]f(x)[/latex].

Combine Shifts and Stretches

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 [latex]2f\left(x\right)+3[/latex], which transformation should we start with? The answer here follows nicely from the order of operations. Given the output value of [latex]f\left(x\right)[/latex], 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 [latex]g\left(x\right)=f\left(2x+3\right)[/latex], for example, we have to think about how the inputs to the function [latex]g[/latex] relate to the inputs to the function [latex]f[/latex]. Suppose we know [latex]f\left(7\right)=12[/latex]. What input to [latex]g[/latex] would produce that output? In other words, what value of [latex]x[/latex] will allow [latex]g\left(x\right)=f\left(2x+3\right)=12[/latex]? We would need [latex]2x+3=7[/latex]. To solve for [latex]x[/latex], 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.

[latex]f\left(bx+p\right)=f\left(b\left(x+\frac{p}{b}\right)\right)[/latex]

Let’s work through an example.

[latex]f\left(x\right)={\left(2x+4\right)}^{2}[/latex]

We can factor out a 2.

[latex]f\left(x\right)={\left(2\left(x+2\right)\right)}^{2}[/latex]

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.

Example: Finding a Triple Transformation of a Graph

Use the graph of [latex]f\left(x\right)[/latex] to sketch a graph of [latex]k\left(x\right)=f\left(\frac{1}{2}x+1\right)-3[/latex].

Solution

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

[latex]f\left(\frac{1}{2}x+1\right)-3=f\left(\frac{1}{2}\left(x+2\right)\right)-3[/latex]

By factoring the inside, we can first horizontally stretch by 2, as indicated by the [latex]\frac{1}{2}[/latex] 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).

Next, we horizontally shift left by 2 units, as indicated by [latex]x+2[/latex].

Last, we vertically shift down by 3 to complete our sketch, as indicated by the [latex]-3[/latex] on the outside of the function.