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For those of you who spend half an hour answering complicated integral questions, this will a little bit of breather.

Consider the graph of $f(x)=e^x$, how would the transformation of this graph, defined as $2f(3x+2)+1$ look? My logic tells me to execute the translations in $y$ first, then the scaling in $y$, then the translations in $x$ and then the scaling in $x$ (sort of like a reversed order of operations).

That being said, consider the point on the graph where $y=8$, when the translation in $y$ is applied, the new value is $y=9$, then after the scaling in $y$ is applied, $y=18$, now, the $x$ value at this point is $x \approx2.9$, and when the translation in $x$ is applied, the new value is $x\approx1.9$, and when the scaling in $x$ is applied, the new value is $x\approx0.6$ and hence the transformation is done.

The question is, is this a correct approach? And are there other approaches which might speed this process up slightly?

joshuaheckroodt
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1 Answers1

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First, rewrite it so that all the transformations can be easily identified, like when you complete the square of a quadratic:

$$y \;\; = \;\; 2 \cdot f\left(\, 3\left(x + \frac{2}{3} \right) \,\right) \; + \; 1 $$

or

$$(y \; - \; 1) \;\; = \;\; 2 \cdot f\left(\, 3\left(x + \frac{2}{3} \right) \,\right) $$

In general, for problems like this there are $4$ types of algebraic operations that can be performed, and their corresponding $4$ types of geometric transformations: add/subtract to input variable or output variable (horizontal or vertical shift), multiply/divide to input variable or output variable (horizontal or vertical stretch).

One possible ordered sequence of transformations to $y = f(x)$ that results in this is the following. (For #2 and #3, recall how trig. graphs behave.)

1. Horizontal shift left by $\frac{2}{3}$

$$ y \; = \; f(x) \;\;\;\; \text{becomes} \;\;\;\; y \; = \; f\left(x + \frac{2}{3}\right) $$

2. Horizontal compression by a factor of $3$

$$ y \; = \; f\left(x + \frac{2}{3}\right) \;\;\;\; \text{becomes} \;\;\;\; y \; = \; f\left(3\left(x + \frac{2}{3}\right)\right) $$

3. Vertical expansion by a factor of $2$

$$ y \; = \; f\left(3\left(x + \frac{2}{3}\right)\right) \;\;\;\; \text{becomes} \;\;\;\; y \; = \; 2\cdot f\left(3\left(x + \frac{2}{3}\right)\right) $$

4. Vertical shift up by $1$

$$ y \; = \; 2\cdot f\left(3\left(x + \frac{2}{3}\right)\right) \;\;\;\; \text{becomes} \;\;\;\; y \; = \; 2\cdot f\left(3\left(x + \frac{2}{3}\right)\right)\; + \; 1 $$

Dave L. Renfro
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  • And would you always apply them in this order once you get your transformation to a form where each transformation can be easily identified? – joshuaheckroodt Nov 15 '17 at 19:32
  • You can change the order, but doing so might require some adjustments. To be safe, it's always a good idea to keep track of the algebraic transformations that correspond to the geometric transformations. For example, if you switch #3 and #4, then the vertical shift needs to be $\frac{1}{2},$ since the next step multiplies everything on the right hand side by $2$ (so you need to start out with adding $\frac{1}{2}$ so that when you're done you'll be adding $1).$ The order I used is what I've always considered to be the safest for things like this. Sometimes one or more shifts will be absent. – Dave L. Renfro Nov 15 '17 at 19:42