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.
video by Prof Leonard 

video by Prof Leonard 

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The idea of transformations is that you can take the equation of a basic function and adjust it in very straightforward 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 xdirection (or negative xdirection), it makes sense that we will do something with the xvariable, 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 xdirection) \(a\) units, we make a new function \(g(x)=f(xa)\). So, for example, if we have \(f(x)=x^3\), we can shift this three units to the right by writing \(g(x)=f(x3)\). 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 xaxis 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 \(h(x) = f(x)2 = x^32\). 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 xdirection and the ydirection. Let's start with shrinking. We will continue working with \(f(x)=x^3\).
Figure 5  Shrinking yDirection [built with GeoGebra]
Shrinking yDirection
To shrink a graph in the ydirection 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 ydirection. 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 ydirection, not the xdirection.
\(f(1) = 1\) 
\(g(1) = 1/2\) 
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 yvalue has shrunk by two. So that's what is meant by shrinking in the ydirection.
Figure 6  Shrinking xDirection [built with GeoGebra]
Shrinking xDirection
The equation for shrinking in the xdirection 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 xdirection.
\(f(1) = 1\) 
\(h(2) = 1\) 
In order to get the same yvalue 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 onehalf instead of 2? This parallels what we experienced with shifting in the xdirection. 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 yDirection [built with GeoGebra]
Stretching yDirection
So now we are going to stretch a graph in the ydirection. Stretching is the opposite of shrinking. So to stretch \(f(x)=x^3\) by two in the ydirection, 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 xDirection [built with GeoGebra]
Stretching xDirection
Similarly, for stretching \(a\) units in the xdirection we replace \(x\) with \(a\cdot x\). In our current example, stretching two units in the xdirection 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 yAxis [built with GeoGebra]
yaxis Reflection
To reflect a graph about the yaxis, we replace \(x\) with \(x\). For reflecting, we will be using the function \(f(x)=\sqrt{x}\). Reflecting about (or across) the yaxis, we need \(g(x)=f(x) = \sqrt{x}\). The result is shown in Figure 9.
Figure 10  Reflecting Across the xAxis [built with GeoGebra]
xaxis Reflection
To reflect a graph about the xaxis, we just negate the yvalues. Reflecting about (or across) the xaxis, 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 xaxis and then about the yaxis. We can also rotate about the yaxis first and then the xaxis. 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 xaxis.
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 yaxis.
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 xaxis 
\(b\) affects horizontal shrinking/stretching and reflection about yaxis 
\(c\) affects horizontal shifting 
\(d\) affects vertical shifting 
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.
video by MIP4U 

UnTransforming 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.
\( f(x) = (x3)^3 \)
Problem Statement 

Use transformations to graph the function \( f(x) = (x3)^3 \)
Final Answer 

Problem Statement
Use transformations to graph the function \( f(x) = (x3)^3 \)
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
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
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
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
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
video by MIP4U 

Final Answer
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\(\displaystyle{ f(x) = \frac{1}{x3} + 1 }\)
Problem Statement 

Use transformations to graph the function \(\displaystyle{ f(x) = \frac{1}{x3} + 1 }\)
Final Answer 

Problem Statement
Use transformations to graph the function \(\displaystyle{ f(x) = \frac{1}{x3} + 1 }\)
Solution
video by MIP4U 

Final Answer
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\( y=2^{x3} \)
Problem Statement 

Use transformations to graph the function \( y=2^{x3} \)
Final Answer 

Problem Statement
Use transformations to graph the function \( y=2^{x3} \)
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 xaxis 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 xaxis 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 xaxis and shifts it to the left two units.
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
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
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
video by Brian McLogan 

Final Answer
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\( y=(x2)^2 + 3 \)
Problem Statement 

Use transformations to graph \( y=(x2)^2 + 3 \)
Final Answer 

Problem Statement
Use transformations to graph \( y=(x2)^2 + 3 \)
Solution
video by The Organic Chemistry Tutor 

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.
video by The Organic Chemistry Tutor 

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
video by The Organic Chemistry Tutor 

Final Answer
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\(\displaystyle{ f(x) = \frac{2}{x4} + 2 }\)
Problem Statement 

Use transformations to graph \(\displaystyle{ f(x) = \frac{2}{x4} + 2 }\)
Final Answer 

Problem Statement
Use transformations to graph \(\displaystyle{ f(x) = \frac{2}{x4} + 2 }\)
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
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
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
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
video by Brian McLogan 

Final Answer
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\(\displaystyle{ f(x) = \frac{1}{2}(x4)^2  5 }\)
Problem Statement 

Use transformations to graph \(\displaystyle{ f(x) = \frac{1}{2}(x4)^2  5 }\)
Final Answer 

Problem Statement
Use transformations to graph \(\displaystyle{ f(x) = \frac{1}{2}(x4)^2  5 }\)
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
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
video by Brian McLogan 

Final Answer
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