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17Calculus Precalculus - Graph Transformations

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On this page, we give you an overview of the main types of basic graph transformations: shifting, scaling and reflecting. Then on separate pages, we discuss the specifics and how to transform graphs, as well as how to determine if an equation represents a basic function that has been transformed. This will aid you in quickly graphing a function.

If you want complete lectures on this topic, we recommend these videos.

Prof Leonard - Introduction to Graph Transformations [48mins]

video by Prof Leonard

Prof Leonard - How to Graph with Transformations [33mins-9secs]

video by Prof Leonard

The idea of transformations is that you can take the equation of a basic function and adjust it in very straight-forward ways to either move it with shifts, shrink or stretch it, or reflect it about a line to get a different function.
Similarly, we can do the opposite. We can be given a function and, in order to easily graph it, we can extract a basic function if we can tell how the equation was manipulated using transformations. Now, all that sounds complicated but it's not really. First, let's talk about the various kinds of transformations.

Horizontal Shifts

If we have a graph, we can shift it either left or right. Since we are talking about the x-direction (or negative x-direction), it makes sense that we will do something with the x-variable, right? And that's exactly what we do. If we are given a function \(f(x)\), we shift it horizontally as follows.

Figure 1 - Shift Right [built with GeoGebra]

Shift Right

Shifting to the right (positive x-direction) \(a\) units, we make a new function \(g(x)=f(x-a)\). So, for example, if we have \(f(x)=x^3\), we can shift this three units to the right by writing \(g(x)=f(x-3)\). Figure 1 shows both \(f(x)\) and \(g(x)\) on the same set of axes.

Figure 2 - Shift Left [built with GeoGebra]

Shift Left

Similarly, we shift to the left \(b\) units by adding \(b\) to \(x\) to get a new function. Using the same example as above, given that \(f(x)=x^3\) and we want to shift to the left two units, we get \(h(x) = f(x+2) = (x+2)^3\). Figure 2 shows both \(f(x)\) and \(h(x)\) on the same set of axes.

More on Horizontal Shifts

Another way to think about horizontal shifts is to think about the sign of \(a\). We can use one equation to represent horizontal shifts.

For \(f(x+a)\),
if \(a \gt 0\), we shift left,
if \( a \lt 0 \), we shift right.

Now, if you are like me you might be thinking, wait, going right on the x-axis is in the positive direction, so shouldn't we ADD 3 instead of subtract 3? There are several ways to think about this but the one that helps me the most is that if we think about the graph \(f(x)\), we move the AXES to the left 3 units. We do not really move the graph. So, when we shift to the right, we shift the axes to the left, which is the negative direction. If you are not convinced, check a few points on the graphs and then you should see that the equations are correct.

Vertical Shifts

Vertical shifts make a bit more intuitive sense than horizontal shifts, as you will see. We will continue to work with the cubic function, \(f(x)=x^3\).

Figure 3 - Shift Up [built with GeoGebra]

Shift Up

To shift up we just ADD a number to the function. To shift \(f(x)\) up \(a\) units, we write \(g(x) = f(x) + a\). For example, to shift \(f(x)=x^3\) up one unit, we get \(g(x) = f(x) + 1 = x^3 +1\). This is shown in Figure 3.

Shifting up makes sense in that we actually shift the function (not the axes) up. That means to shift in the positive direction, we ADD the amount we want to shift.

Figure 4 - Shift Down [built with GeoGebra]

Shift Down

Similarly, to shift down we subtract the required distance from the function. In the example where we have \(f(x)=x^3\), to shift down two units we write \(h(x) = f(x)-2 = x^3-2\). This is shown in Figure 4.

More on Vertical Shifts

Another way to think about vertical shifts is to think about the sign of \(a\). We can use one equation to represent shifts, \(f(x)+a\). If \(a \gt 0\), we shift up. If \( a \lt 0 \), we shift down.

Shrinking

Shrinking and stretching a graph involving shrinking and stretching in the x-direction and the y-direction. Let's start with shrinking. We will continue working with \(f(x)=x^3\).

Figure 5 - Shrinking y-Direction [built with GeoGebra]

Shrinking y-Direction

To shrink a graph in the y-direction by a value, just divide the entire function by that value. In this example, we want to shrink \(f(x)=x^3\) by two in the y-direction. Dividing the entire function by two gives us \(g(x) = f(x)/2 = (x^3)/2\). This is shown in Figure 5.

The word 'shrinking' may give you the impression that the graph will shrink. However, looking at Figure 5, it may not look like anything shrank. In fact, it looks like the function actually expanded outward. However, let's look at specific values. Pay special attention to what is going on the y-direction, not the x-direction.

\(f(1) = 1\)

   

\(g(1) = 1/2\)

Notice that for a given \(x\), one in this case, the value of the 'shrunk' function, \(g(x)\) is half the value of the original function. So the y-value has shrunk by two. So that's what is meant by shrinking in the y-direction.

Figure 6 - Shrinking x-Direction [built with GeoGebra]

Shrinking x-Direction

The equation for shrinking in the x-direction works as you would expect. To shrink by a value \(a\), just write \(h(x) = f(x/a)\). Here is a graph of \(f(x)=x^3\) and \(h(x)=f(x/2)=(x/2)^3\).

Again, the word shrinking doesn't seem to describe what is going here. But let's look at individual values and focus on the x-direction.

\(f(1) = 1\)

   

\(h(2) = 1\)

In order to get the same y-value of one, we have to go out to \(x=2\) in the shrunk function. This seems backwards, doesn't it? If it shrank, shouldn't we get one-half instead of 2? This parallels what we experienced with shifting in the x-direction. It's the opposite of what we think of intuitively. We can think of the axes as shrinking instead of the graph. Strange, I know!

Stretching

Figure 7 - Stretching y-Direction [built with GeoGebra]

Stretching y-Direction

So now we are going to stretch a graph in the y-direction. Stretching is the opposite of shrinking. So to stretch \(f(x)=x^3\) by two in the y-direction, we multiply the entire function by two, i.e. \(g(x) = 2f(x) = 2x^3\). The graphs of \(f(x)\) and \(g(x)\) are shown in Figure 7, plotted on the same set of axes.

The result in the figure may not be what you intuitively thought might happen. However, if you think about it, the graph of \(g(x)\) is going up twice as fast as \(f(x)\). Let's look at the single point \(x=1\) to see if this makes sense.

\(f(1) = 1\)

   

\(g(1) = 2\)

Figure 8 - Stretching x-Direction [built with GeoGebra]

Stretching x-Direction

Similarly, for stretching \(a\) units in the x-direction we replace \(x\) with \(a\cdot x\). In our current example, stretching two units in the x-direction gives us \(h(x) = f(2x) = (2x)^3\). The result is shown in Figure 8. Similar to what we have discussed previously about how to understand what is happening with the graph, think about it and look at several points to make sure you understand the graphs.

Reflecting

Figure 9 - Reflecting Across the y-Axis [built with GeoGebra]

y-axis Reflection

To reflect a graph about the y-axis, we replace \(x\) with \(-x\). For reflecting, we will be using the function \(f(x)=\sqrt{x}\). Reflecting about (or across) the y-axis, we need \(g(x)=f(-x) = \sqrt{-x}\). The result is shown in Figure 9.

Figure 10 - Reflecting Across the x-Axis [built with GeoGebra]

x-axis Reflection

To reflect a graph about the x-axis, we just negate the y-values. Reflecting about (or across) the x-axis, we need \(h(x)=-f(x) = -\sqrt{x}\). The graphs are shown in Figure 10.

Figure 11 - Rotating About the Origin [built with GeoGebra]

Origin Rotation

There is a third type of 'reflection' that is sometimes used in calculus, called origin rotation. Essentially, it is a double reflection, i.e. we reflect about the x-axis and then about the y-axis. We can also rotate about the y-axis first and then the x-axis. As you would expect, we multiply the function by negative one and also replace \(x\) with \(-x\). For our example, this looks like \(p(x) = -f(-x) = -\sqrt{-x}\). The result is shown in Figure 11.

More on Shrinking, Stretching and Reflecting

Another way to think about shrinking, stretching and reflecting is to think about the sign of \(a\). We can use one equation to represents all three, \(a \cdot f(x)\). If \(\abs{a} \gt 1\), the graph stretches. If \( \abs{a} \lt 1 \), the graph shrinks. And, if \(a \lt 0 \), the resulting graph is a reflection about the x-axis.

Similarly, for \( f(a \cdot x) \), the graph shrinks for \(\abs{a} \gt 1\). The graph stretches for \( \abs{a} \lt 1 \) and for \(a \lt 0 \), the resulting graph is a reflection about the y-axis.

Multiple Transformations

As you may have deduced from some of the above discussion, we can do multiple transformations in one equation. Here is a summary.

Multiple Transformations of \(f(x)\)

\( af( b(x+c) ) + d \)

\(a\) affects vertical shrinking/stretching and reflection about x-axis

\(b\) affects horizontal shrinking/stretching and reflection about y-axis

\(c\) affects horizontal shifting

\(d\) affects vertical shifting

See also Math Is Fun - Function Transformations

Caution - The order in which you perform the transformations is important. For example, if you write \(f(bx)\) and then want to shift \(c\) units and try to write \(f(bx+c)\), you will not get a shift of \(c\). You will get a shift of \(c/b\). So notice the parentheses in the multiple transformations equation and do not leave them out.

Here is a great video going through all these variables with plenty of graphs and examples.

MIP4U - Functions Transformations: A Summary [8mins-19secs]

video by MIP4U

Un-Transforming a Function

Sometimes you are given a function that you need to graph. If you can determine the basic function and how it was tranformed, you should be able to easily graph it. We recommend comparing the equation to \( af( b(x+c) ) + d \). This is easier said than done and it does take some experience to do it. So let's get some experience working the practice problems.

Practice

Unless otherwise instructed, use the concept of transformations to graph these functions.

\( f(x) = (x-3)^3 \)

Problem Statement

Use transformations to graph the function \( f(x) = (x-3)^3 \)

Final Answer

Problem Statement

Use transformations to graph the function \( f(x) = (x-3)^3 \)

Solution

Brian McLogan - 4221 video solution

video by Brian McLogan

Final Answer

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\( f(x) = x^3 + 7 \)

Problem Statement

Use transformations to graph the function \( f(x) = x^3 + 7 \)

Final Answer

Problem Statement

Use transformations to graph the function \( f(x) = x^3 + 7 \)

Solution

Brian McLogan - 4222 video solution

video by Brian McLogan

Final Answer

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\( f(x) = 3 \abs{x} \)

Problem Statement

Use transformations to graph the function \( f(x) = 3 \abs{x} \)

Final Answer

Problem Statement

Use transformations to graph the function \( f(x) = 3 \abs{x} \)

Solution

MIP4U - 4223 video solution

video by MIP4U

Final Answer

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\( f(x) = \sqrt{2x} \)

Problem Statement

Use transformations to graph the function \( f(x) = \sqrt{2x} \)

Final Answer

Problem Statement

Use transformations to graph the function \( f(x) = \sqrt{2x} \)

Solution

MIP4U - 4224 video solution

video by MIP4U

Final Answer

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\( f(x) = -\abs{x} \)

Problem Statement

Use transformations to graph the function \( f(x) = -\abs{x} \)

Final Answer

Problem Statement

Use transformations to graph the function \( f(x) = -\abs{x} \)

Solution

MIP4U - 4225 video solution

video by MIP4U

Final Answer

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\( f(x) = \sqrt[3]{-x} \)

Problem Statement

Use transformations to graph the function \( f(x) = \sqrt[3]{-x} \)

Final Answer

Problem Statement

Use transformations to graph the function \( f(x) = \sqrt[3]{-x} \)

Solution

MIP4U - 4226 video solution

video by MIP4U

Final Answer

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

Problem Statement

Use transformations to graph the function \(\displaystyle{ f(x) = \frac{1}{x-3} + 1 }\)

Final Answer

Problem Statement

Use transformations to graph the function \(\displaystyle{ f(x) = \frac{1}{x-3} + 1 }\)

Solution

MIP4U - 4227 video solution

video by MIP4U

Final Answer

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\( y=-2^{x-3} \)

Problem Statement

Use transformations to graph the function \( y=-2^{x-3} \)

Final Answer

Problem Statement

Use transformations to graph the function \( y=-2^{x-3} \)

Solution

Brian McLogan - 4228 video solution

video by Brian McLogan

Final Answer

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Find the function \(g(x)\) that takes \( f(x) = 1/x \), reflects it about the x-axis and shifts it to the left two units.

Problem Statement

Find the function \(g(x)\) that takes \( f(x) = 1/x \), reflects it about the x-axis and shifts it to the left two units.

Final Answer

\(\displaystyle{ g(x) = \frac{-1}{x+2} }\)

Problem Statement

Find the function \(g(x)\) that takes \( f(x) = 1/x \), reflects it about the x-axis and shifts it to the left two units.

Solution

Brian McLogan - 4229 video solution

video by Brian McLogan

Final Answer

\(\displaystyle{ g(x) = \frac{-1}{x+2} }\)

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\( f(x) = \log_2(x+5)+3 \)

Problem Statement

Use transformations to graph the function \( f(x) = \log_2(x+5)+3 \)

Final Answer

Problem Statement

Use transformations to graph the function \( f(x) = \log_2(x+5)+3 \)

Solution

Brian McLogan - 4230 video solution

video by Brian McLogan

Final Answer

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\( f(x) = \sqrt{x/2} - 4 \)

Problem Statement

Use transformations to graph the function \( f(x) = \sqrt{x/2} - 4 \)

Final Answer

Problem Statement

Use transformations to graph the function \( f(x) = \sqrt{x/2} - 4 \)

Solution

Brian McLogan - 4231 video solution

video by Brian McLogan

Final Answer

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\( f(x) = 3^{-x}+2 \)

Problem Statement

Use transformations to graph \( f(x) = 3^{-x}+2 \)

Final Answer

Problem Statement

Use transformations to graph \( f(x) = 3^{-x}+2 \)

Solution

Brian McLogan - 4232 video solution

video by Brian McLogan

Final Answer

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\( y=(x-2)^2 + 3 \)

Problem Statement

Use transformations to graph \( y=(x-2)^2 + 3 \)

Final Answer

Problem Statement

Use transformations to graph \( y=(x-2)^2 + 3 \)

Solution

The Organic Chemistry Tutor - 4233 video solution

Final Answer

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\( y = 3 - (x + 2)^2 \)

Problem Statement

Use transformations to graph \( y = 3 - (x + 2)^2 \)

Final Answer

Problem Statement

Use transformations to graph \( y = 3 - (x + 2)^2 \)

Solution

In this video solution, he demonstrates very well how to draw a more accurate sketch by plotting points.

The Organic Chemistry Tutor - 4234 video solution

Final Answer

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\( y = 4 - \sqrt{3 - x} \)

Problem Statement

Use transformations to graph \( y = 4 - \sqrt{3 - x} \)

Final Answer

Problem Statement

Use transformations to graph \( y = 4 - \sqrt{3 - x} \)

Solution

The Organic Chemistry Tutor - 4235 video solution

Final Answer

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\(\displaystyle{ f(x) = \frac{2}{x-4} + 2 }\)

Problem Statement

Use transformations to graph \(\displaystyle{ f(x) = \frac{2}{x-4} + 2 }\)

Final Answer

Problem Statement

Use transformations to graph \(\displaystyle{ f(x) = \frac{2}{x-4} + 2 }\)

Solution

Brian McLogan - 4236 video solution

video by Brian McLogan

Final Answer

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

Problem Statement

Use transformations to graph \(f(x) = -2\abs{x + 1} - 3 \)

Final Answer

Problem Statement

Use transformations to graph \(f(x) = -2\abs{x + 1} - 3 \)

Solution

Brian McLogan - 4237 video solution

video by Brian McLogan

Final Answer

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\( f(x) = 2(x+3)^2 + 4 \)

Problem Statement

Use transformations to graph \( f(x) = 2(x+3)^2 + 4 \)

Final Answer

Problem Statement

Use transformations to graph \( f(x) = 2(x+3)^2 + 4 \)

Solution

Brian McLogan - 4238 video solution

video by Brian McLogan

Final Answer

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\( f(x) = -3(x + 3)^2 + 4 \)

Problem Statement

Use transformations to graph \( f(x) = -3(x + 3)^2 + 4 \)

Final Answer

Problem Statement

Use transformations to graph \( f(x) = -3(x + 3)^2 + 4 \)

Solution

Brian McLogan - 4239 video solution

video by Brian McLogan

Final Answer

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\( f(x) = \sqrt{-2x} + 1 \)

Problem Statement

Use transformations to graph \( f(x) = \sqrt{-2x} + 1 \)

Final Answer

Problem Statement

Use transformations to graph \( f(x) = \sqrt{-2x} + 1 \)

Solution

Brian McLogan - 4240 video solution

video by Brian McLogan

Final Answer

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\(\displaystyle{ f(x) = \frac{1}{2}(x-4)^2 - 5 }\)

Problem Statement

Use transformations to graph \(\displaystyle{ f(x) = \frac{1}{2}(x-4)^2 - 5 }\)

Final Answer

Problem Statement

Use transformations to graph \(\displaystyle{ f(x) = \frac{1}{2}(x-4)^2 - 5 }\)

Solution

Brian McLogan - 4241 video solution

video by Brian McLogan

Final Answer

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\( f(x) = 2(x+4)^2 - 3 \)

Problem Statement

Use transformations to graph \( f(x) = 2(x+4)^2 - 3 \)

Final Answer

Problem Statement

Use transformations to graph \( f(x) = 2(x+4)^2 - 3 \)

Solution

MIP4U - 4242 video solution

video by MIP4U

Final Answer

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\( f(x) = -\abs{x+4} + 8 \)

Problem Statement

Use transformations to graph \( f(x) = -\abs{x+4} + 8 \)

Final Answer

Problem Statement

Use transformations to graph \( f(x) = -\abs{x+4} + 8 \)

Solution

Brian McLogan - 4243 video solution

video by Brian McLogan

Final Answer

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Practice Instructions

Unless otherwise instructed, use the concept of transformations to graph these functions.

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