Absolute Value Transformations

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Note: For Parent Functions and general transformations, see the Parent Graphs and Transformations section.

This section covers:

Absolute Value Transformations can be tricky, since we have two different types of problems:

  1. Transformations of Absolute Value Functions
  2. Performing Absolute Value Transformations on other functions

Transformations of the Absolute Value Parent Function

Let’s first work with transformations on the absolute value parent function.

Since the vertex (the “point”) of an absolute value parent function \(y=\left| x \right|\) is \(\left( {0,\,0} \right)\), an absolute value equation with new vertex \(\left( {h,\,k} \right)\) is \(\displaystyle f\left( x \right)=a\left| {\frac{1}{b}\left( {x-h} \right)} \right|+k\), where \(a\) is the vertical stretch, \(b\) is the horizontal stretch, \(h\) is the horizontal shift to the right, and \(k\) is the vertical shift upwards. If \(a\) is negative, the graph points up instead of down.

Here is an example with a t-chart:




\(\displaystyle \begin{array}{l}y=-3\left| {2x+4} \right|+1\\y=-3\left| {2(x+2)} \right|+1\end{array}\)


(have to take out a 2 to make \(x\) by itself)


Parent function:

\(\displaystyle y=\left| x \right|\)

\(\frac{1}{2}x-2\)    x y   3y + 1
    –3      –2 2       –5
 –2.5      –1 1       –2
    –2        0 0         1
 –1.5        1 1       2
    –1        2 2       –5


Domain:  \(\left( {-\infty ,\infty } \right)\)  Range:  \(\left( {-\infty ,1} \right]\)



Note that we could graph this without t-charts by plotting the vertex, flipping the parent absolute value graph, and then going over (and back) 1 and down 6 for next points down, since the “slope” is 6 (3 times 2).

Here’s an example of writing an absolute value function from a graph:

Graph Getting Equation
We see that this is an absolute value graph (parent graph \(y=\left| x \right|\)) since it is “pointy”, but flipped around \(x\)-axis, since it is facing down.

We are taking the absolute value of the whole function, since it “bounces” up from the \(x\) axis (only positive \(y\) values). This is weird, but it’s an absolute value of an absolute value function!

Therefore, the equation will be in the form \(y=\left| {a\left| {x-h} \right|+k} \right|\) with vertex \(\left( {h,\,\,k} \right)\), and \(a\) should be negative. Since the vertex of the graph is \(\left( {-1,\,\,10} \right)\), one equation of the graph could be \(y=\left| {a\left| {x+1} \right|+10} \right|\).

We need to find \(a\); use the point \(\left( {4,\,0} \right)\):

\(\displaystyle \begin{align}y&=\left| {a\left| {x+1} \right|+10} \right|\\0&=\left| {a\left| {4+1} \right|+10} \right|\\0&=\left| {a\left| 5 \right|+10} \right|\\0&=5a+10,\,\,\text{since}\,\,\left| 0 \right|\text{ =0}\\-5a&=10;\,\,\,\,\,\,a=-2\end{align}\)   \(\begin{array}{c}\text{The equation of the graph then is:}\\y=\left| {-2\left| {x+1} \right|+10} \right|\end{array}\)

(We could have also found \(a\) by noticing that the graph goes over/back 1 and down 2), so it’s “slope” is –2.

Be sure to check your answer by graphing or plugging in more points! √


Absolute Value Transformations of other Parent Functions

Now let’s look at taking the absolute value of functions, both on the outside (affecting the \(y\)’s) and the inside (affecting the \(x\)’s). These are a little trickier.

Let’s look at a function of points, and see what happens when we take the absolute value of the function “on the outside” and then “on the inside”.  Then we’ll show absolute value transformations using parent functions.

Note that with the absolute value on the outside (affecting the \(\boldsymbol{y}\)’s), we just take all negative \(\boldsymbol{y}\) values and make them positive, and with absolute value on the inside (affecting the \(\boldsymbol{x}\)’s), we take all the 1st and 4th quadrant points and reflect them over the \(\boldsymbol{y}\)-axis, so that the new graph is symmetric to the \(\boldsymbol{y}\)-axis.




Original Function


  x      y
–6    –4
–4      0
–3      2
  0      0
  3    –2
  4      0
  6      1

\(y=\left| {f\left( x \right)} \right|\)


Replace all negative \(y\) values with their absolute value (make them positive). Make sure that all (negative \(y\)) points on the graph are reflected across the \(x\)-axis to be positive.


Example Function: \(y=\left| {{{x}^{3}}+4} \right|\)

  x     y    |y|
–6  4     4
–4    0     0
–3    2     2
  0    0     0
  3  –2     2
  4    0     0
  6    1     1

Let’s try:


\(y=\left| {2f\left( x \right)-4} \right|\)


Reflect negative \(y\) values across the \(x\)-axis.

  x  y   |2y – 4|
–6  4     12
–4    0      4
–3    2      0
  0    0      4
  3  2      8
  4    0      4
  6    1      2

\(y=f\left( {\left| x \right|} \right)\)


Make a symmetrical graph from the positive \(x\)’s across the \(y\) axis. “Throw away” the left-hand side of the graph (negative \(x\)’s), and replace the left side of the graph with the reflection of the right-hand side.


For any negative \(x\)’s, replace the \(y\) value with the \(y\) value corresponding to the positive value (absolute value) of the negative \(x\)’s. For example, when \(x\) is –6, replace the \(y\) with a 1, since the \(y\) value for positive 6 is 1.


Example Function: \(y=4{{\left| x \right|}^{3}}-2\)

  x y    New y
–6 –4     1
–4   0     0
–3   2   2
 0   0
 3 2
 4   0
 6   1

Let’s try:


\(y=3f\left( {\left| x \right|} \right)+2\)


(The absolute value is directly around the \(x\).)


After performing the transformation on the \(y\), for any negative \(x\)’s, replace the \(y\) value with the \(y\) value corresponding to the positive value (absolute value) of the negative \(x\)’s, For example, when \(x\) is  –6, replace the \(y\) with a 5, since the \(y\) value for positive 6 is 5.

  x y   3y+2  New  y
–6 –4    –10      5
–4   0       2       2
–3   2     –7     –4
 0   0       2
 3 –2     –4
 4   0       2
 6   1       5


Here are examples of mixed absolute value transformations to show what happens when the inside absolute value is not just around the \(x\), versus just around the \(x\); you can see that this can get complicated.




\(y=\sqrt{{\left| {2\left( {x+3} \right)} \right|}}+4\)


With this mixed transformation, we need to perform the inner absolute value first:


For any original negative \(x\)’s, replace the \(y\) value with the \(y\) value corresponding to the positive value (absolute value) of the negative \(x\)’s. (See pink arrows)


Then with the new values, we can perform the shift for \(y\) (add 4) and the shift for \(x\) (divide by 2 and then subtract 3).


The best way to check your work is to put the graph in your calculator and check the table values.

Parent function:


(\(x\) must be \(\ge 0\) for original function, but not for transformed function)



Note: The boxed \(y\) is the \(y\) value associated with the absolute value of that \(x\) value.


If the absolute value sign was just around the \(x\), such as \(y=\sqrt{{2\left( {\left| x \right|+3} \right)}}+4\) (see next problem), we would have replaced the \(y\) values with those of the positive \(x\)’s after doing the \(x\) transformation, instead of before. Thus, the graph would be symmetrical around the \(y\)-axis. Tricky!


\(y=\sqrt{{2\left( {\left| x \right|+3} \right)}}+4\)


The best way to do this problem is to perform the transformations of a horizontal compression by \(\frac{1}{2}\), shift left 3, and up 4. You will first get a graph that is like the right-hand part of the graph above.


Then, “throw away” all the \(y\) values where \(x\) is negative and make the graph symmetrical to the \(y\)-axis.

Parent function:



A t-chart is just too messy, since the \(y\) values for all the negative \(x\) values (after the \(\tfrac{1}{2}x-3\) computation) would have to be replaced by the positive \(x\) values after the \(\tfrac{1}{2}x-3\) computation.

Here are more absolute value examples with parent functions:

Absolute Value Transformations T-chart Graph

\(y=\left| {{{x}^{3}}+4} \right|\)


Parent function:



Reflect all values below the \(y\)-axis to above the \(y\)-axis.

x y    |y + 4|
–2 –8       4
–1 –1       3
0  0        4
1  1        5
2  8      12

\(2{{\left| x \right|}^{3}}-1\)


Parent function:



For the two value of \(x\) that are negative (–2 and –1), replace the \(y\)’s with the \(y\) from the absolute value (2 and 1, respectively) for those points.


Note that this is like “erasing” the part of the graph to the left of the \(y\)-axis and reflecting the points from the right of the \(y\)-axis over to the left.

     x  y     2y – 1   New y
–2 –8    –17       15
–1 –1     –3        1
 0   0     –1
 1   1       1
 2   8     15

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


Parent function:



For the negative \(x\) value, just use the \(y\) values of the absolute value of these \(x\) values! Note that we pick up these new \(y\) values after we do the translation of the \(x\) values.


Note that this is like “erasing” the part of the graph to the left of the \(y\)-axis and reflecting the points from the right of the \(y\)-axis over to the left.

x + 3   x  y    New y
–2     –5 \(\frac{1}{{32}}\)   \(\color{#800000}{{\frac{1}{2}}}\)
–1      –4 \(\frac{1}{{16}}\)   \(\color{blue}{{\frac{1}{4}}}\)
  1       –2 \(\color{blue}{{\frac{1}{4}}}\)
  2      –1 \(\color{#800000}{{\frac{1}{2}}}\)
 3        0 1
 4        1 2
 5        2 4

\(\displaystyle y=\left| {\frac{3}{x}+3} \right|\)


Parent function:

\(\displaystyle y=\frac{1}{x}\)


Since the absolute value is on the “outside”, we can just perform the transformations on the \(y\), doing the absolute value last


Reflect all values below the \(y\)-axis to above the \(y\)-axis.

 x  y    |3y + 3|
–3   1/3       2
–2 –1/2     1.5
–1 –1        0
  1   1        6
  2  1/2     4.5
  3 1/3      4

\(y=\left| {{{{\log }}_{3}}\left( {x+4} \right)} \right|\)


Parent function:

\(y={{\log }_{3}}\left( x \right)\)


(Asymptote: \(x=0\) )


Reflect all values below the \(y\)-axis to above the \(y\)-axis.

x – 4    x  y     |y|
\(-3\frac{8}{9}\) –1/9 –2     2
\(-3\frac{2}{3}\) –1/3 –1     1
   –3       1  0      0
  –1       3  1      1
    5       9  2     2

New Asymptote:  \(x=-4\)



Note: These mixed transformations with absolute value are very tricky; it’s really difficult to know what order to use to perform them. The general rule of thumb is to perform the absolute value first for the absolute values on the inside, and the absolute value last for absolute values on the outside (work from the inside out). The best thing to do is to play around with them on your graphing calculator to see what’s going on.

For example, with something like \(y=\left| {{{2}^{x}}} \right|-3\), you perform the \(y\) absolute value function first (before the shift); with something like \(y=\left| {{{2}^{x}}-3} \right|\), you perform the \(y\) absolute value last (after the shift). (These two make sense, when you look at where the absolute value functions are.) But we saw that with \(y={{2}^{{\left| x \right|-3}}}\), we performed the \(x\) absolute value function last (after the shift). I also noticed that with \(y={{2}^{{\left| {x-3} \right|}}}\), you perform the \(x\) absolute value transformation first (before the shift).

I don’t think you’ll get this detailed with your transformations, but you can see how complicated this can get!

Here’s an example where we’re using what we know about the absolute value transformation, but we’re using it on an absolute value parent function! Pretty crazy, huh?

Transformation T-chart Graph

\(y=\left| {3\left| {x-1} \right|-2} \right|\)


Parent function:

\(y=\left| x \right|\)


Since we’re using the absolute value parent function, we only have to take the absolute value on the outside (\(y\)). We can do this, since the absolute value on the inside is a linear function (thus we can use the parent function).


x + 1    x y   |3y 2|
–1     –2 2       4
 0      –1 1       1
 1        0 0       2
 2       1 1       1
 3       2 2       4



More Absolute Value Transformations

What about \(\left| {f\left( {\left| x \right|} \right)} \right|\)?  Play around with this in your calculator with \(y=\left| {{{2}^{{\left| x \right|}}}-5} \right|\), for example. You’ll see that it shouldn’t matter which absolute value function you apply first, but it certainly doesn’t hurt to work from the inside out. And with \(-\left| {f\left( {\left| x \right|} \right)} \right|\), it’s a good idea to perform the inside absolute value first, then the outside, and then the flip across the \(x\) axis. So the rule of thumb with these absolute value functions and reflections is to move from the inside out.

Let’s do more complicated examples with absolute value and flipping – sorry that this stuff is so complicated! Just be careful about the order by trying real functions in your calculator to see what happens. These are for the more advanced Pre-Calculus classes!

Original Function Transformation Explanation

\(-f\left( {-x} \right)\)


Flip the function around the \(x\)-axis, and then around the \(y\)-axis.


It actually doesn’t matter which flip you perform first.

\(\left| {f\left( {-x} \right)} \right|\)



Flip the function around the \(x\)-axis, and then reflect everything below the \(x\)-axis to make it above the \(x\)-axis; this takes the absolute value (all positive \(y\) values).


We actually could have done this in the other order, and it would have worked!

\(\left| {f\left( {\left| x \right|} \right)} \right|\)



For the absolute value on the inside, throw away the negative \(x\) values, and replace them with the \(y\) values for the absolute value of the \(x\).


Then reflect everything below the \(x\)-axis to make it above the \(x\)-axis; this takes the absolute value (all positive \(y\) values).


We could have done this in any order.

\(-\left| {f\left( {\left| x \right|} \right)} \right|\)



Do everything we did in the transformation above, and then flip the function around the \(x\)-axis, because of the negative sign.


For this one, I noticed that we needed to do the flip around the \(x\)-axis last (we need to work “inside out”).

Learn these rules, and practice, practice, practice!

On to Piecewise Functions – you are ready!

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