Applying Transformations

Section 9.2 Applying Transformations

In Section 9.1, we learned how to determine what transformations were applied to a function. In this section, we will go backwards: we tell you what tranformations need to be applied, and you have to apply them to the function. It’s important to note that Table 9.1.4 will still be very helpful in organizing the information! Specifically, you need to still think about three key things:

Where does the transformation go/what direction is it going in? Vertical transformations are applied to the outputs, and horizontal transformations are applied to the inputs.

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What kind of transformation/operation is being applied? Shifts involve adding or subtracting, scales involve multiplying or dividing by a positive number, and reflections involve multiplying by \(-1\text{.}\)

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Horizontal transformations act backwards from what we expect.

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Example 9.2.1.

Suppose the graph of \(g(x)\) is the same as \(f(x)\text{,}\) but shifted up by 2 and stretched horizontally by 7. Write the formula for \(g(x)\) in terms of \(f(x)\text{.}\)

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When the question asks for the equation "in terms of" \(f(x)\text{,}\) that means that you are just applying it to the function notation. There should still be the letter \(f\) in your answer, and there is no explicit formula for \(f(x)\) that you need to worry about.

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So, the key to this problem is to figure out how each transformation changes the function notation:

Shifting up by 2:
- Up is vertical, so this is applied to the outside.
  
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- Shifts are adding/subtracting.
  
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- We will add 2 to the outside of the function.
  
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Stretching horizontally by 7:
- Horizontally is applied to the inside.
  
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- Stretch is a scale, so we are multiplying or dividing.
  
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- Remember: horizontal transformations are backwards! So, we will divide the inputs by 7.
  
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We put this all together to get the final answer: \(g(x)=f\left(\frac{x}{7}\right)+2\text{,}\) which can also be written as \(g(x)=f\left(\frac{1}{7}x\right)+2\text{.}\)

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Checkpoint 9.2.2.

Suppose the graph of \(g(x)\) is the same as \(f(x)\text{,}\) but shifted left by 5 and flipped vertically over the \(x\)-axis. Write the formula for \(g(x)\) in terms of \(f(x)\text{.}\)

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Answer.

\(g(x)=-f(x+5)\)

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Solution.

The key to this problem is to figure out how each transformation changes the function notation:

Shifting left by 5:
- Left is horizontal, so this is applied to the inside.
  
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- Shifts are adding/subtracting.
  
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- Remember: horizontal transformations are backwards! So, we will add 5 to the inputs, since left is in the negative \(x\)-direction.
  
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Vertical flip:
- Vertical is applied to the outside.
  
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- Flipping involves multiplying by \(-1\text{.}\)
  
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- So, we will multiply the outside by \(-1\text{.}\)
  
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We put this all together to get the final answer: \(g(x)=-f(x+5)\text{.}\)

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Just as we saw with identifying transformations, we could either have questions where we are just working with function notation, or where we have explicit formulas for the functions. Let’s see an example of applying transformations to a function when you are given explicit formulas.

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Example 9.2.3.

Suppose \(f(x)=4x^3-7x^2+5x-3\text{,}\) and the graph of \(g(x)\) is the same as \(f(x)\text{,}\) but shifted to the left by 2 and up by 9. Write the formula for \(g(x)\text{,}\) but do not simplify your answer.

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Just like before, we need to figure out how each transformation changes the formula:

Shifting left by 2:
- Left is horizontal, so this is applied to the inside.
  
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- Shifts are adding/subtracting.
  
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- Remember: horizontal transformations are backwards! So, we will add 2 to the inputs, since left is in the negative \(x\)-direction.
  
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Shifting up by 9:
- Vertical is applied to the outside.
  
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- Shifts are adding/subtracting.
  
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- So, we will add \(9\) to the outside.
  
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At this point, we should write the function notation like we did when we didn’t have an explicit formula. This makes it a lot easier to know how to change the formula later. In this case, our function notation is

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\begin{equation*} g(x)=f(x+2)+9 \end{equation*}

Now that we know how the transformations affect the formula, we are ready to apply it to the formula we were given for \(f(x)\text{.}\) Often, the hardest part of this question is keeping all of the parentheses in the right place for the inputs. So, we are going to use the trick from Section 2.1, where we replace the original inputs with empty parentheses before we plug in the new inputs:

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\begin{align*} g(x)\amp =f(x+2)+9\\ \amp=\left(4({\color{red}{\underline{\hspace{0.5in}}}})^3-7({\color{red}{\underline{\hspace{0.5in}}}})^2+5({\color{red}{\underline{\hspace{0.5in}}}})-3\right)+9\\ \amp= \left(4({\color{red}{x+2}})^3-7({\color{red}{x+2}})^2+5({\color{red}{x+2}})-3\right)+9 \end{align*}

At this point, we are done with the question. Since we were told not to simplify our answer, we are not going to distribute any of the parentheses. We could combine the \(-3\) and \(9\text{,}\) because that is a very small step. However, it is still possible to make an arithmetic error when doing that. So, unless we are told to simplify, we should not do any simplification and just leave our answer exactly as it is.

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Checkpoint 9.2.4.

Suppose \(f(x)=2x^4-5x^2+8\text{,}\) and the graph of \(g(x)\) is the same as \(f(x)\) but vertically stretched by 2, horizontally stretched by 6, and vertically flipped over the \(x\)-axis. Write the formula for \(g(x)\text{,}\) but do not simplify your answer.

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Answer.

\(g(x)=-2\left(2\left(\frac{x}{6}\right)^4-5\left(\frac{x}{6}\right)^2+8\right)\)

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Solution.

We need to figure out how each transformation changes the formula:

Vertical stretch by 2:
- Vertical is applied to the outside.
  
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- Stretch is a scale, so we are multiplying/dividing.
  
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- We will multiply by 2 on the outside.
  
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Horizontal stretch by 6:
- Horizontal is applied to the inside.
  
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- Stretch is a scale, so we are multiplying/dividing.
  
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- Remember: horizontal transformations are backwards! So, to make the graph bigger horizontally, we will divide by 6 on the inside.
  
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Vertical flip:
- Vertical is applied to the outside.
  
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- Flipping involves multiplying by \(-1\text{.}\)
  
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- So, we will multiply the outside by \(-1\text{.}\)
  
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\begin{equation*} g(x)=-2f\left(\frac{x}{6}\right) \end{equation*}

Now that we know how the transformations affect the formula, we are ready to apply it to the formula we were given for \(f(x)\text{,}\) using empty parentheses to help us with our substitution:

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\begin{align*} g(x)\amp =-2f\left(\frac{x}{6}\right)\\ \amp=-2\left(2({\color{red}{\underline{\hspace{0.5in}}}})^4-5({\color{red}{\underline{\hspace{0.5in}}}})^2+8\right)\\ \amp= -2\left(2\left({\color{red}{\frac{x}{6}}}\right)^4-5\left({\color{red}{\frac{x}{6}}}\right)^2-3\right) \end{align*}

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