Absolute Value Transformations

$ \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|$

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).

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! √

Original Function

(Points)

$ 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}}} \right|$

Let’s try:

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

Reflect negative $ y$-values across the $ x$-axis.

Example Function: $ \displaystyle y=\left| {2\left( {\frac{1}{x}} \right)-4} \right|$

$ 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.

OR

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={{\left| x \right|}^{3}}$

Let’s try:

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

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. (Do this after the transformation on the $ y$, since the $ \left| {\,\,} \right|$ is directly around the $ x$.) For example, when $ x$ is  –6, replace the $ y$ with a 5, since the $ y$-value for positive 6 is 5.

Example Function: $ y=3{{\left| x \right|}^{3}}+2$

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

With this mixed transformation, replace the $ y$-values right away for negative $ x$-values, even if they those $ y$-values aren’t defined in the original function!

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:

$ y=\sqrt{x}$

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

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

In this case,  replace 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!

Parent function:

$ y=\sqrt{x}$

A t-chart is messy, since we’d have to find the $ y$-values with those of the positive $ x$’s after doing the transformation, but you can see how the graph is different.

The only point that would work in the t-chart above is $ (1.5, 7)$, since $ 1.5$ is positive.

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

Parent function:

$ y={{x}^{3}}$

Reflect all values below the $ y$-axis to above the $ y$-axis.

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

Parent function:

$ y={{x}^{3}}$

For the two values 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.

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

Parent function:

$ y={{2}^{x}}$ (exponential)

For the negative $ x$-value, 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, since the $ \left| {\,\,} \right|$ is just around the $ x$. 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.

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

Parent function:

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

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.

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

Parent function:

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

(Asymptote: $ x=0$ )

Reflect all values below the $ y$-axis to above the $ y$-axis.

New Asymptote:  $ x=-4$

$ 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).

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

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

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

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

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

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|$

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