# Parent Functions and Transformations

This section covers:

Before we get started, here are links to Parent Function Transformations in other sections:

You may not be familiar with all the functions and characteristics in the tables; here are some topics to review:

# Basic Parent Functions

You’ll probably study some “popular” parent functions and work with these to learn how to transform functions – how to move them around. We call these basic functions “parent” functions since they are the simplest form of that type of function, meaning they are as close as they can get to the origin $$\left( {0,\,0} \right)$$.

The chart below provides some basic parent functions that you should be familiar with. I’ve also included the anchor points, or critical points, the points with which to graph the parent function.

 Parent Function Graph Parent Function Graph $$y=x$$ Linear, Odd Domain: $$\left( {-\infty ,\infty } \right)$$ Range: $$\left( {-\infty ,\infty } \right)$$ End Behavior**: $$\begin{array}{l}x\to -\infty \text{, }\,y\to -\infty \\x\to \infty \text{, }\,\,\,y\to \infty \end{array}$$Critical points: $$\displaystyle \left( {-1,\,-1} \right),\,\left( {0,\,0} \right),\,\left( {1,\,1} \right)$$ $$y=\left| x \right|$$ Absolute Value, Even Domain: $$\left( {-\infty ,\infty } \right)$$ Range: $$\left[ {0,\infty } \right)$$ End Behavior: $$\begin{array}{l}x\to -\infty \text{, }\,y\to \infty \\x\to \infty \text{, }\,\,\,y\to \infty \end{array}$$Critical points: $$\displaystyle \left( {-1,\,1} \right),\,\left( {0,\,0} \right),\,\left( {1,\,1} \right)$$ $$y={{x}^{2}}$$ Quadratic, Even Domain: $$\left( {-\infty ,\infty } \right)$$ Range: $$\left[ {0,\infty } \right)$$ End Behavior: $$\begin{array}{l}x\to -\infty \text{, }\,y\to \infty \\x\to \infty \text{, }\,\,\,y\to \infty \end{array}$$Critical points: $$\displaystyle \left( {-1,\,1} \right),\,\left( {0,\,0} \right),\,\left( {1,\,1} \right)$$ $$y=\sqrt{x}$$ Radical (Square Root), Neither Domain: $$\left[ {0,\infty } \right)$$ Range: $$\left[ {0,\infty } \right)$$ End Behavior: $$\displaystyle \begin{array}{l}x\to 0,\,\,\,\,y\to 0\\x\to \infty \text{, }\,\,\,y\to \infty \end{array}$$Critical points: $$\displaystyle \left( {-1,\,1} \right),\,\left( {0,\,0} \right),\,\left( {1,\,1} \right)$$ $$y={{x}^{3}}$$ Cubic, Odd Domain: $$\left( {-\infty ,\infty } \right)$$ Range: $$\left( {-\infty ,\infty } \right)$$ End Behavior: $$\begin{array}{l}x\to -\infty \text{, }\,y\to -\infty \\x\to \infty \text{, }\,\,\,y\to \infty \end{array}$$Critical points: $$\displaystyle \left( {-1,\,-1} \right),\,\left( {0,\,0} \right),\,\left( {1,\,1} \right)$$ $$y=\sqrt[3]{x}$$ Cube Root, Odd Domain: $$\left( {-\infty ,\infty } \right)$$ Range: $$\left( {-\infty ,\infty } \right)$$ End Behavior: $$\begin{array}{l}x\to -\infty \text{, }\,y\to -\infty \\x\to \infty \text{, }\,\,\,y\to \infty \end{array}$$Critical points: $$\displaystyle \left( {-1,\,-1} \right),\,\left( {0,\,0} \right),\,\left( {1,\,1} \right)$$ $$\begin{array}{c}y={{b}^{x}},\,\,\,b>1\,\\(y={{2}^{x}})\end{array}$$Exponential, Neither Domain: $$\left( {-\infty ,\infty } \right)$$ Range: $$\left( {0,\infty } \right)$$ End Behavior: $$\begin{array}{l}x\to -\infty \text{, }\,y\to 0\\x\to \infty \text{, }\,\,\,y\to \infty \end{array}$$Critical points: $$\displaystyle \left( {-1,\,\frac{1}{b}} \right),\,\left( {0,\,1} \right),\,\left( {1,\,b} \right)$$Asymptote:  $$y=0$$ $$\begin{array}{c}y={{\log }_{b}}\left( x \right),\,\,b>1\,\,\,\\(y={{\log }_{2}}x)\end{array}$$Log, Neither Domain: $$\left( {0,\infty } \right)$$ Range: $$\left( {-\infty ,\infty } \right)$$ End Behavior: $$\begin{array}{l}x\to {{0}^{+}}\text{, }\,y\to -\infty \\x\to \infty \text{, }\,y\to \infty \end{array}$$ Critical points: $$\displaystyle \left( {\frac{1}{b},\,-1} \right),\,\left( {1,\,0} \right),\,\left( {b,\,1} \right)$$Asymptote: $$x=0$$ $$\displaystyle y=\frac{1}{x}$$ Rational (Inverse), Odd Domain: $$\left( {-\infty ,0} \right)\cup \left( {0,\infty } \right)$$ Range: $$\left( {-\infty ,0} \right)\cup \left( {0,\infty } \right)$$ End Behavior: $$\begin{array}{l}x\to -\infty \text{, }\,y\to 0\\x\to \infty \text{, }\,\,\,y\to 0\end{array}$$Critical points: $$\displaystyle \left( {-1,\,-1} \right),\,\left( {1,\,1} \right)$$Asymptotes: $$y=0,\,\,x=0$$ $$\displaystyle y=\frac{1}{{{{x}^{2}}}}$$ Rational (Inverse Squared), Even Domain: $$\left( {-\infty ,0} \right)\cup \left( {0,\infty } \right)$$ Range: $$\left( {0,\infty } \right)$$ End Behavior: $$\begin{array}{l}x\to -\infty \text{, }\,y\to 0\\x\to \infty \text{, }\,\,\,y\to 0\end{array}$$Critical points: $$\displaystyle \left( {-1,\,1} \right),\,\left( {1,\,1} \right)$$Asymptotes: $$x=0,\,\,y=0$$ $$y=\text{int}\left( x \right)\text{=}\left[ \text{x} \right]$$ Greatest Integer, NeitherDomain:$$\left( {-\infty ,\infty } \right)$$ Range: $$\{y:y\in \mathbb{Z}\}\text{ (integers)}$$End Behavior: $$\begin{array}{l}x\to -\infty \text{, }\,y\to -\infty \\x\to \infty \text{, }\,\,\,y\to \infty \end{array}$$Critical points: $$\displaystyle \begin{array}{l}x:\left[ {-1,\,0} \right)\,\,\,y:-1\\x:\left[ {0,\,1} \right)\,\,\,y:0\\x:\left[ {1,\,2} \right)\,\,\,y:1\end{array}$$ $$y=C$$   ($$y=2$$) Constant, EvenDomain: $$\left( {-\infty ,\infty } \right)$$ Range: $$\{y:y=C\}$$End Behavior: $$\begin{array}{l}x\to -\infty \text{, }\,y\to C\\x\to \infty \text{, }\,\,\,y\to C\end{array}$$Critical points: $$\displaystyle \left( {-1,\,C} \right),\,\left( {0,\,C} \right),\,\left( {1,\,C} \right)$$

**Notes on End Behavior: To get the end behavior of a function, we just look at the smallest and largest values of $$x$$, and see which way the $$y$$ is going. Not all functions have end behavior defined; for example, those that go back and forth with the $$y$$ values and never really go way up or way down (called “periodic functions”) don’t have end behaviors.

Most of the time, our end behavior looks something like this:$$\displaystyle \begin{array}{l}x\to -\infty \text{, }\,y\to \,\,?\\x\to \infty \text{, }\,\,\,y\to \,\,?\end{array}$$ and we have to fill in the $$y$$ part. So the end behavior for a line with a positive slope is: $$\begin{array}{l}x\to -\infty \text{, }\,y\to -\infty \\x\to \infty \text{, }\,\,\,y\to \infty \end{array}$$.

There are a couple of exceptions; for example, sometimes the $$x$$ starts at 0 (such as in the radical function), we don’t have the negative portion of the $$x$$ end behavior. Also, when $$x$$ starts very close to zero (such as in in the log function), we indicate that $$x$$ is starting from the positive (right) side of 0 (and the $$y$$ is going down); we indicate this by $$\displaystyle x\to {{0}^{+}}\text{, }\,y\to -\infty$$.

# Generic Transformations of Functions

Again, the “parent functions” assume that we have the simplest form of the function; in other words, the function either goes through the origin $$\left( {0,0} \right)$$, or if it doesn’t go through the origin, it isn’t shifted in any way.

When a function is shifted, stretched (or compressed), or flipped in any way from its “parent function“, it is said to be transformed, and is a transformation of a function.

T-charts are extremely useful tools when dealing with transformations of functions. For example, if you know that the quadratic parent function $$y={{x}^{2}}$$ is being transformed 2 units to the right, and 1 unit down (only a shift, not a stretch or a flip yet), we can create the original t-chart, following by the transformation points on the outside of the original points.Then we can plot the “outside” (new) points to get the newly transformed function:

Transformation

T-chart

Graph

$$y={{x}^{2}}$$

Transform function 2 units to the right, and 1 unit down.

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

Domain:  $$\left( {-\infty ,\infty } \right)$$

Range:   $$\left[ {-1,\,\,\infty } \right)$$

When looking at the equation of the moved function, however, we have to be careful.

When functions are transformed on the outside of the $$f(x)$$ part, you move the function up and down and do the “regular” math, as we’ll see in the examples below. These are vertical transformations or translations, and affect the $$y$$ part of the function.

When transformations are made on the inside of the $$f(x)$$ part, you move the function back and forth (but do the “opposite” math – since if you were to isolate the x, you’d move everything to the other side). These are horizontal transformations or translations, and affect the $$x$$ part of the function.

There are several ways to perform transformations of parent functions; I like to use t-charts, since they work consistently with ever function.

## Vertical Transformations

Here are the rules and examples of when functions are transformed on the “outside” (notice that the $$y$$ values are affected). The t-charts include the points (ordered pairs) of the original parent functions, and also the transformed or shifted points.

Notice that the first two transformations are translations, the third is a dilation, and the last is a reflection.

Transformation

What It Does

Example

Graph

$$f\left( x \right)+b$$

Translation

Move graph up  $$b$$  units

Every point on the graph is shifted up  $$b$$  units.

The $$x$$’s stay the same; add  $$b$$  to the $$y$$ values.

Parent: $$y={{x}^{2}}$$

Transformed: $$y={{x}^{2}}+ \,2$$

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

Domain:  $$\left( {-\infty ,\infty } \right)$$     Range:  $$\left[ {2,\,\,\infty } \right)$$

$$f\left( x \right)-b$$

Translation

Move graph down  $$b$$  units

Every point on the graph is shifted down  $$b$$  units.

The $$x$$’s stay the same; subtract  $$b$$  from the $$y$$ values.

Parent: $$y=\sqrt{x}$$

Transformed: $$y=\sqrt{x}- \,3$$

 x y   y–3 0 0   –3 1 1   –2 4 2   –1 9 3     0

Domain:  $$\left[ {0,\,\,\infty } \right)$$     Range: $$\left[ {-3,\,\,\infty } \right)$$

$$a\,\cdot f\left( x \right)$$

Dilation

Stretch graph vertically by a scale factor of  $$a$$ (sometimes called a dilation).

Note that if $$a<1$$, the graph is compressed or shrinked.

Every point on the graph is stretched  $$a$$  units.

The $$x$$’s stay the same; multiply the $$y$$ values by $$a$$.

Parent: $$y={{x}^{3}}$$

Transformed: $$y={{4x}^{3}}$$

 x y   4y –1 –1  –4 0 0    0 1 1    4

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

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

Reflection

Flip graph around the $$x$$-axis.

Every point on the graph is flipped vertically.

The $$x$$’s stay the same; multiply the $$y$$ values by $$-1$$.

Parent: $$y=\left| x \right|$$

Transformed: $$y=-\left| x \right|$$

 x y   –y –2 2   –2 –1 1   –1 0 0     0 1 1   –1 2 2   –2

Domain: $$\left( {-\infty ,\infty } \right)$$     Range: $$\left( {-\infty ,\,\,0} \right]$$

## Horizontal Transformations

Here are the rules and examples of when functions are transformed on the “inside” (notice that the $$x$$ values are affected). Notice that when the $$x$$ values are affected, you do the math in the “opposite” way from what the function looks like: if you you’re adding on the inside, you subtract from the $$x$$; if you’re subtracting on the inside, you add to the $$x$$; if you’re multiplying on the inside, you divide from the $$x$$; if you’re dividing on the inside, you multiply to the $$x$$. If you have a negative value on the inside, you flip across the $$\boldsymbol{y}$$ axis (notice that you still multiply the $$x$$ by $$-1$$ just like you do for with the $$y$$ for vertical flips).

Notice that the first two transformations are translations, the third is a dilation, and the last is a reflection.

(You may find it interesting is that a vertical stretch behaves the same way as a horizontal compression, and vice versa, since when stretch something upwards, we are making it skinnier.)

Transformation

What It Does

Example

Graph

$$f\left( {x+b} \right)$$

Move graph left  $$b$$  units

(Do the “opposite” when change is inside the parentheses or underneath radical sign.)

Every point on the graph is shifted left  $$b$$  units.

The $$y$$’s stay the same; subtract  $$b$$  from the $$x$$ values.

Parent: $$y={{x}^{2}}$$

Transformed: $$y={{\left( {x+\,\,2} \right)}^{2}}$$

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

Domain: $$\left( {-\infty ,\infty } \right)$$     Range: $$\left[ {0,\,\,\infty } \right)$$

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

Move graph right  $$b$$  units

Every point on the graph is shifted right  $$b$$  units.

The $$y$$’s stay the same; add  $$b$$  to the $$x$$ values.

Parent: $$y=\sqrt{x}$$

Transformed: $$y=\sqrt{{x- \,3}}$$

 x + 3   x y 3      0 0 4      1 1 7      4 2 12     9 3

Domain: $$\left[ {-3,\infty } \right)$$      Range: $$\left[ {0,\,\,\infty } \right)$$

$$f\left( {a\cdot x} \right)$$

Compress graph horizontally by a scale factor of  $$a$$  units

(stretch or multiply by $$\displaystyle \frac{1}{a}$$)

Every point on the graph is compressed  $$a$$  units horizontally.

The $$y$$’s stay the same; multiply the $$x$$ values by $$\displaystyle \frac{1}{a}$$.

Parent:  $$y={{x}^{3}}$$

Transformed: $$y={{\left( {4x} \right)}^{3}}$$

 $$\frac{1}{4}x$$      x y $$-\frac{1}{4}$$       –1 –1 0         0 0 $$\frac{1}{4}$$         1 1

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

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

Flip graph around the $$y$$ axis

Every point on the graph is flipped around the $$y$$ axis.

The $$y$$’s stay the same; multiply the $$x$$ values by $$-1$$.

Parent: $$y=\sqrt{x}$$

Transformed: $$y=\sqrt{{-x}}$$

 –x       x y 0      0 0 –1      1 1 –4      4 2 –9      9 3

Domain:  $$\left( {-\infty ,0} \right]$$     Range: $$\left[ {0,\,\,\infty } \right)$$

# Mixed Transformations

Most of the problems you’ll get will involve mixed transformations, or multiple transformations, and we do need to worry about the order in which we perform the transformations.

It usually doesn’t matter if we make the x changes or the y changes first, but within the x’s and y’s, we need to perform the transformations in the following order. Note that this is sort of similar to the order with PEMDAS (parentheses, exponents, multiplication/division, and addition/subtraction).

When performing these rules, the coefficients of the inside x must be 1; for example, we would need to have $$y={{\left( {4\left( {x+2} \right)} \right)}^{2}}$$ instead of $$y={{\left( {4x+8} \right)}^{2}}$$ (by factoring). If you didn’t learn it this way, see IMPORTANT NOTE below.

Here is the order:

1. Perform Flipping across the axes first (negative signs).
2. Perform Stretching and Shrinking next (multiplying and dividing).
3. Perform Horizontal and Vertical shifts last (adding and subtracting).

But we can do steps 1 and 2 together (order doesn’t actually matter), since we can think of the first two steps as a “negative stretch/compression.”

I like to take the critical points and maybe a few more points of the parent functions, and perform all the transformations at the same time with a t-chart! We just do the multiplication/division first on the x or y points, followed by addition/subtraction. It makes it much easier!

Note again that since we don’t have an x “by itself” (coefficient of 1) on the inside, we have to get it that way by factoring! For example, we’d have to change $$y={{\left( {4x+8} \right)}^{2}}\text{ to }y={{\left( {4\left( {x+2} \right)} \right)}^{2}}$$.

Let’s try to graph this “complicated” equation and I’ll show you how easy it is to do with a t-chart:

$$\displaystyle f(x)=-3{{\left( {2x+8} \right)}^{2}}+10$$

(Note that for this example, we could move the $${{2}^{2}}$$ to the outside to get a vertical stretch of $$3\left( {{{2}^{2}}} \right)=12$$, but we can’t do that for many functions.)

We first need to get the $$x$$ by itself on the inside by factoring, so we can perform the horizontal translations. This is what we end up with:

$$\displaystyle f(x)=-3{{\left( {2\left( {x+4} \right)} \right)}^{2}}+10$$

Now, what we need to do is to look at what’s done on the “outside” (for the $$y$$’s) and make all the moves at once, by following the exact math. Then we can look on the “inside” (for the $$x$$’s) and make all the moves at once, but do the opposite math. We do this with a t-chart.

We’re starting with the parent function $$f(x)={{x}^{2}}$$. If we look at what we’re doing on the outside of what is being squared, which is the $$\displaystyle \left( {2\left( {x+4} \right)} \right)$$, we’re flipping it across the $$x$$-axis (the minus sign), stretching it by a factor of 3, and adding 10 (shifting up 10). These are the things that we are doing vertically, or to the $$y$$.

Now if we look at what we are doing on the inside of what we’re squaring, we’re multiplying it by 2, which means we have to divide by 2 (horizontal compression by a factor of $$\frac{1}{2}$$), and we’re adding 4, which means we have to subtract 4 (a left shift of 4). Remember that we do the opposite when we’re dealing with the $$x$$.

Also remember that we always have to do the multiplication or division first with our points, and then the adding and subtracting (sort of like PEMDAS).

Note: You might see mixed transformations in the form $$g\left( x \right)=a\cdot f\left( {\left( {\frac{1}{b}} \right)\left( {x-h} \right)} \right)+k$$, or with a coordinate rule $$\displaystyle \left( {x,\,y} \right)\to \left( {bx+h,\,\,ay+k} \right)$$, 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. For the transformation $$\displaystyle f(x)=-3{{\left( {2\left( {x+4} \right)} \right)}^{2}}+10$$, we have $$a=-3$$, $$b=\frac{1}{2}\,\,\text{or}\,\,.5$$, $$h=-4$$, and $$k=10$$. Our transformation $$g\left( x \right)=-3f\left( {2\left( {x+4} \right)} \right)+10=g\left( x \right)=-3f\left( {\left( {\frac{1}{{\frac{1}{2}}}} \right)\left( {x-\left( {-4} \right)} \right)} \right)+10$$ would be $$\boldsymbol{{\left( {x,\,y} \right)\to \left( {.5x-4,\,-3y+10} \right)}}$$.

Here is the t-chart with the original function, and then the transformations on the outsides. Now we can graph the outside points (points that aren’t crossed out) to get the graph of the transformation:

t-chart

Transformed Graph

Parent:  $$y={{x}^{2}}$$  (Quadratic)

Transformed:  $$\displaystyle f(x)=-3{{\left( {2\left( {x+4} \right)} \right)}^{2}}+10$$

y changes:      $$\displaystyle f(x)=\color{blue}{{-3}}{{\left( {2\left( {x+4} \right)} \right)}^{2}}\color{blue}{+10}$$

x changes:    $$\displaystyle f(x)=-3{{\left( {\color{blue}{2}\left( {x\text{ }\color{blue}{{+\text{ }4}}} \right)} \right)}^{2}}+10$$

Opposite for x, “regular” for y:

$$\left( {x,\,y} \right)\to \left( {.5x-4,\,-3y+10} \right)$$

 .5x – 4       x y   –3y + 10 –5        –2 4        –2 –4.5         1 1          7 –4          0 0        10 –3.5         1 1          7 –3          2 4        –2

Domain:    $$\left( {-\infty ,\infty } \right)$$   Range: $$\left( {-\infty ,10} \right]$$

Note: Since this is a parabola, we could have also graphed the transformation by noticing (by the vertex form) that the vertex is $$\left( {-4,\,10} \right)$$, like we did in the Introduction to Quadratics section here. Then, by moving the $${{2}^{2}}$$ to the outside to make a vertical stretch of 12, we could go over (and back) 1 and down 12 from the vertex to get other points. Cool!

IMPORTANT NOTE In some books, for $$\displaystyle f\left( x \right)=-3{{\left( {2x+8} \right)}^{2}}+10$$, they may NOT have you factor out the 2 on the inside, but just switch the order of the transformation on the $$\boldsymbol{y}$$.

In these cases, the order of transformations would be horizontal shifts, horizontal reflections/stretches, vertical reflections/stretches, and then vertical shifts. For example, for this problem, you would move to the left 8 first for the $$\boldsymbol{y}$$, and then compress with a factor of $$\displaystyle \frac {1}{2}$$ for the $$\boldsymbol{x}$$ (which is opposite of PEMDAS). Then you would perform the $$\boldsymbol{y}$$ (vertical) changes the regular way – reflect and stretch by 3 first, and then shift up 10. So you would have $$\displaystyle \boldsymbol{{\left( {x,\,y} \right)\to \left( {\frac{1}{2}\left( {x-8} \right),\,-3y+10} \right)}}$$. Try a t-chart; you’ll get the same t-chart as above!

# Transformations in Function Notation (based on Graph and/or Points).

You may also be asked to perform a transformation of a function using a graph and individual points; in this case, you’ll probably be given the transformation in function notation. Note that we may need to use several points from the graph and “transform” them, to make sure that the transformed function has the correct “shape”.

Let’s do an example.  Let’s use the transformation $$\displaystyle f\left( {-\frac{1}{2}\left( {x-1} \right)} \right)-3$$, given the following points and graph. Remember that the transformations inside the parentheses are done to the x (doing the opposite math), and outside are done to the y.

Note that this transformation takes the original function, flips it around the y axis, performs a horizontal stretch by 2, moves it right by 1, and then down by 3.

Graph and Points of Function

Transformation

Function:

Domain: $$\left[ {-4,5} \right]$$      Range: $$\left[ {-7,5} \right]$$

Key Points:

 x y –4 5 0 1 2 2 5 –7

Transformation:  $$\displaystyle f\left( {-\frac{1}{2}\left( {x-1} \right)} \right)-3$$

$$y$$ changes: $$\displaystyle f\left( {-\frac{1}{2}\left( {x-1} \right)} \right)\color{blue}{{-\text{ }3}}$$

$$x$$ changes:  $$\displaystyle f\left( {\color{blue}{{-\frac{1}{2}}}\left( {x\text{ }\color{blue}{{-\text{ }1}}} \right)} \right)-3$$

(we’ll do the “opposite” math with the “$$x$$”)

Key Points Transformed:

 –2x + 1      x y       y – 3 9         –4 5          2 1           0 1        –2 –3           2 2        –1 –9           5 –7       –10

Transformed Function:

Domain:  $$\left[ {-9,9} \right]$$     Range: $$\left[ {-10,2} \right]$$

Draw the points in the same order as the original to make it easier! And remember if you’re having trouble drawing the graph from the transformed ordered pairs, just take more points from the original graph to map to the new one!

Note: You may also be given a random point, say $$\left( {8,-2} \right)$$, and give the transformed coordinates for the point of the graph $$y=-6g\left( {-2x} \right)-2$$. To do this, to get the transformed $$y$$, multiply the $$y$$ part of the point by –6 and then subtract 2. To get the transformed $$x$$, multiply the $$x$$ part of the point by $$-\frac{1}{2}$$ (opposite math). The new point is $$\left( {-4,\,10} \right)$$.

## Functional Notation Transformation Using Algebra

Let’s say we want to use a “function notation” transformation to transform a parent or non-parent equation.  We can do this without using a t-chart, but by using substitution and algebra.

For example, if we want to transform $$f\left( x \right)={{x}^{2}}+4$$ using the transformation $$\displaystyle -2f\left( {x-1} \right)+3$$, we can just substitute “$$x-1$$” for  “$$x$$” in the original equation, multiply by –2, and then add 3.

For example: $$\displaystyle -2f\left( {x-1} \right)+3=-2\left[ {{{{\left( {x-1} \right)}}^{2}}+4} \right]+3=-2\left( {{{x}^{2}}-2x+1+4} \right)+3=-2{{x}^{2}}+4x-7$$.

We used this method to help transform a piecewise function here.

## More Examples of Mixed Transformations:

Here are a couple more examples (using t-charts), with different parent functions.

Don’t worry if you are totally lost with the exponential and log functions; they will be discussed in the Exponential Functions and Logarithmic Functions sections, and their transformations are discussed in more detail in the Transformations, Inverses, Compositions, and Inequalities of Exponents/Logs section.

Also, the last type of function is a rational function that will be discussed in the Rational Functions section.

Here’s a mixed transformation with the Greatest Integer Function (sometimes called the Floor Function). Note how we can use intervals as the $$x$$ values to make the transformed function easier to draw:

# Writing Transformed Equations from Graphs

You might be asked to write a transformed equation, give a graph. A lot of times, you can just tell by looking at it, but sometimes you have to use a point or two. And you do have to be careful and check your work, since the order of the transformations can matter.

Note that when figuring out the transformations from a graph, it’s difficult to know whether you have an “$$a$$” (vertical stretch) or a “$$b$$” (horizontal stretch) in the equation $$g\left( x \right)=a\cdot f\left( {\left( {\frac{1}{b}} \right)\left( {x-h} \right)} \right)+k$$.

Sometimes the problem will indicate what parameters ($$a$$, $$b$$, and so on) to look for. For others, like polynomials (such as quadratics and cubics), a vertical stretch mimics a horizontal compression, so it’s possible to factor out a coefficient to turn a horizontal stretch/compression to a vertical compression/stretch.

(For more complicated graphs, you may want to take several points and perform a regression in your calculator to get the function, if you’re allowed to do that).

We will find a transformed equation from an absolute value graph in the Absolute Value section below.

Here are some problems:

 Graph Getting Equation Write the general equation for the cubic equation in the form:  $$\displaystyle y={{\left( {\frac{1}{b}\left( {x-h} \right)} \right)}^{3}}+k$$ We see that that the center point, or critical point is at $$\left( {-4,\,-5} \right)$$, so the cubic is in the form: $$\displaystyle y={{\left( {\frac{1}{b}\left( {x+4} \right)} \right)}^{3}}-5$$. Notice that to get back and over to the next points, we go back/over $$3$$ and down/up $$1$$, so we see there’s a horizontal stretch of $$3$$, so $$b=3$$. (It turns out that this is also a vertical compression of $$\displaystyle \frac{1}{{{{3}^{3}}}}=\frac{1}{{27}}$$). So we have $$\displaystyle y={{\left( {\frac{1}{3}\left( {x+4} \right)} \right)}^{3}}-5$$. Try it – it works! Find the equation of this graph in any form: Here’s a generic method you can typically use: We see that this is a cubic polynomial graph (parent graph $$y={{x}^{3}}$$), but flipped around either the $$x$$ the $$y$$-axis, since it’s an odd function; let’s use the $$x$$-axis for simplicity’s sake. The equation will be in the form $$y=a{{\left( {x+b} \right)}^{3}}+c$$, where $$a$$ is negative, and it is shifted up $$2$$, and to the left $$1$$. Now we have $$y=a{{\left( {x+1} \right)}^{3}}+2$$. We need to find $$a$$; use the point $$\left( {1,\,-10} \right)$$:       \begin{align}-10&=a{{\left( {1+1} \right)}^{3}}+2\\-10&=8a+2\\8a&=-12;\,\,\,\,\,\,a=-\frac{{12}}{8}=-\frac{3}{2}\end{align}      The equation of the graph is: $$\displaystyle y=-\frac{3}{2}{{\left( {x+1} \right)}^{3}}+2$$.Be sure to check your answer by graphing or plugging in more points! √ Find the equation of this graph in any form: The graph looks like a quadratic with vertex is $$\left( {-1,\,-8} \right)$$, which is a shift of $$8$$ down and $$1$$ to the left. This will give us an equation of the form $$y=a{{\left( {x+1} \right)}^{2}}-8$$, which is (not so coincidentally!) the vertex form for quadratics.We need to find $$a$$; use the point $$\left( {1,\,0} \right)$$:    \begin{align}y&=a{{\left( {x+1} \right)}^{2}}-8\\\,\,\,\,0&=a{{\left( {1+1} \right)}^{2}}-8\\8&=4a;\,\,\,\,\,a=2\end{align}       The equation of the graph then is: $$y=2{{\left( {x+1} \right)}^{2}}-8$$. Note: we could have also noticed that the graph goes over $$1$$ and up $$2$$ from the vertex, instead of over $$1$$ and up $$1$$ normally with $$y={{x}^{2}}$$. This would mean that our vertical stretch is $$2$$. Find the equation of this graph in any form: We see that this graph is exponential because of its horizontal asymptote at $$y=-3$$.  From this, we know that it will have a vertical shift of $$-3$$, since normally the asymptote is $$y=0$$. (It’s difficult to see if there is a horizontal shift, but let’s use the vertical one to see if that works.)Our equation will then be in the form $$y=a{{b}^{x}}-3$$.Use a system of equations with the given points: $$\left( {0,\,1} \right)$$ and $$\left( {1,\,-1} \right)$$. When you have a problem like this, first use any point that has a “$$0$$” in it if you can; it will be easiest to solve the system. $$\begin{array}{l}\text{Solve for }a\text{ first, using }\left( {0,\,1} \right):\\1=a{{b}^{0}}-3;\,\,\,\,\,a\left( 1 \right)=1+3;\,\,\,\,a=4\\y=4{{b}^{x}}-3\end{array}$$                $$\begin{array}{l}\text{Plug in }\left( {1,-1} \right)\text{for }b:\\-1=4{{b}^{1}}-3;\,\,\,\,\,4b=2;\,\,\,\,\,b=.5\end{array}$$ The exponential function is $$y=4{{\left( {.5} \right)}^{x}}-3$$;  It works!   √

# Rotational Transformations

You may be asked to perform a rotation transformation on a function (you usually see these in Geometry class). A rotation of 90° counterclockwise involves replacing $$\left( {x,\,y} \right)$$ with $$\left( {-y,\,x} \right)$$, a rotation of 180° counterclockwise involves replacing $$\left( {x,\,y} \right)$$ with $$\left( {-x,\,-y} \right)$$, and a rotation of 270° counterclockwise involves replacing $$\left( {x,\,y} \right)$$ with $$\left( {y,\,-x} \right)$$.

Here is an example:

# Transformations of Inverse Functions

We learned about Inverse Functions here, and you might be asked to compare original functions and inverse functions, as far as their transformations are concerned.    Remember that an inverse function is one where the $$x$$ is switched by the $$y$$, so the all the transformations originally performed on the $$x$$ will be performed on the $$y$$:

Problem:

If a cubic function is vertically stretched by a factor of 3, reflected over the y axis, and shifted down 2 units, what transformations are done to its inverse function?

Solution:

We need to do transformations on the opposite variable.  So the inverse of this function will be horizontally stretched by a factor of 3, reflected over the $$x$$ axis, and shifted to the left 2 units.

Here is a graph of the two functions:

Note that examples of Finding Inverses with Restricted Domains can be found here.

# Absolute Value Transformations

Now let’s try some of the absolute value shifts, first using 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:

Transformation

T-chart

Graph

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

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

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

Transformation

T-chart

Graph

Original Function

(Points)

 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:

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

 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.

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  –9, replace the $$y$$ with a  6, since the $$y$$ value for positive  9  is  6.

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

Note: The boxed $$y$$ is the $$y$$ value associated with the absolute value of that $$x$$ value.

Let’s try:

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

Note: The boxed $$y$$ is the $$y$$ value associated with the absolute value of that $$x$$ value.

Here’s an example of a mixed absolute value transformation; you can see that this can get complicated. It looks like when we have absolute values on the inside (affecting the $$x$$’s), we do those first:

 Transformation T-chart Graph $$y=\sqrt{{\left| {2\left( {x+3} \right)} \right|}}+4$$ With this mixed transformation, we need to perform the inner absolute value first: For any 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:$$y=\sqrt{x}$$($$x$$ 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.

Here are more absolute value examples with parent functions:

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?

## 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!

# Applications of Parent Function Transformations

You may see a “word problem” that used Parent Function Transformations, and you may just have to use what you know about how to shift the functions (instead of coming up with the solution off the top of your head).

Here is an example:

Learn these rules, and practice, practice, practice!

For Practice: Use the Mathway widget below to try a Transformation problem. Click on Submit (the blue arrow to the right of the problem) and click on Describe the Transformation to see the answer.

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on Tap to view steps, or Click Here, you can register at Mathway for a free trial, and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

On to Piecewise Functions – you are ready!