Let's clean up your linear function example. The transformation parameters are
y = -3\,f(2(x + 4)) + 6, right? When finding the transform equation, technically you should not reuse f(x) but instead write:
\begin{aligned}<br />
-3\,f(2(x + 4)) + 6 &= -3(2(x +4)) + 6 \\<br />
&= -3(2x +8) + 6 \\<br />
&= -6x - 24 + 6 \\<br />
&=-6x-18<br />
\end{aligned}
Regarding the next example, your basic function is f(x) = \frac{1}{x}. Your transformation parameters are
y = -\frac{3}{4}f(2(x + 3)) -7. Let's look at just the f part. Whatever is inside the f is what you plug into the function. So
f(2(x + 3)) = \frac{1}{2(x + 3)}.
There is a -3/4 outside the f, so you multiply this function by -3/4:
-\frac{3}{4}f(2(x + 3)) = -\frac{3}{4} \cdot \frac{1}{2(x + 3)}.
Finally, there is a -7 after the expression with f, so you subtract -7 from the function:
-\frac{3}{4}f(2(x + 3)) - 7 = -\frac{3}{4} \cdot \frac{1}{2(x + 3)} - 7.
This is where you stopped (and is there a typo?), but if your linear function example is any indication, I wouldn't consider this simplified. I would go ahead and multiply the two fractions:
-\frac{3}{4} \cdot \frac{1}{2(x + 3)} - 7
= -\frac{3}{8(x + 3)} - 7
And then subtract the fractions:
= -\frac{3}{8(x + 3)} - \frac{7(8(x + 3))}{8(x + 3)}
= -\frac{3}{8(x + 3)} - \frac{56x + 168}{8x + 24}
= -\frac{56x + 171}{8(x + 3)}
So based on that. What do you think the equation of this quadratic function will be?
\begin{align*}<br />
\text {Base function}\\<br />
f(x) &= x^2\\<br />
\text {Transformation parameters}\\<br />
y&= 2f[\frac{1}{4}(x + 3)] +2\\<br />
\text {Transformed equation}\\<br />
f(x)&=2(4x + 3)^2 +2<br />
\end{align*}
Where did the 1/4 go? Looks like another typo?
Again, I wouldn't stop here. I would multiply everything out and write my answer in standard form y = ax^2 + bx + c. Distribute the 1/4, square the resulting binomial, distribute the 2, and then add the 2.