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