Transformations of functions in 1/2 Methods

[redset8]

Hi,
Our teacher gave us some practice questions, and I'm fine with most, but I'm getting stuck on this one...
"Use mapping notation and state in words the sequence of transformations that takes the graph of \(y=\frac{2}{(x+3)}-6\) to \(y=\frac{1}{x}\)"

The teacher's provided answer is: dilation of \(\frac{1}{2}\) from the x-axis, followed by translation of 3 in the positive direction of the x-axis, and a translation of 2 in the negative direction of the y-axis.

Is their answer even correct? Either way, could someone please be kind enough to step through this question with me. (it should be really simple, but it's been frustrating me for a while)

[S_R_K]

The answer is incorrect. If you dilate from the x-axis by a factor of \(\frac{1}{2}\) then the horizontal asymptote will be \(y=-3\), hence a translation of 3 units up (or in the positive direction of the y-axis) is required to get the graph of \(y=\frac{1}{x+3}\). (Obviously the final step is to translate 3 units right).

The other option is to translate up/down first, then dilate from the x-axis. This would be: translate 6 units up, then dilate by a factor of \(\frac{1}{2}\) from the x-axis.

Using mapping notation, the method is:

First write \(\frac{y+6}{2}=\frac{1}{x+3}\) and \(y'=\frac{1}{x'}\). Then equate LHSs and RHSs giving \(\frac{y+6}{2}=y'\) and \(x+3=x'\), and solve for \(y'\) and \(x'\) to find the required transformations.

[redset8]

Thanks for your help. Just a quick question.

The answer is incorrect. If you dilate from the x-axis by a factor of \(\frac{1}{2}\) then the horizontal asymptote will be \(y=-3\), hence a translation of 3 units up (or in the positive direction of the y-axis) is required to get the graph of \(y=\frac{1}{x+3}\). (Obviously the final step is to translate 3 units right).

So the final answer would be dilation of \(\frac{1}{2}\) from the x-axis, then translation of 3 units up (y-axis), but then wouldn't it be a translation of 3 units left (x-axis), as it is \(y=\frac{1}{x+3}\)?

And just when writing mapping notation, is writing \(\frac{y}{2}+3=y'\) okay instead of \(\frac{y+6}{2}=y'\)? Which is preferred/more clear?

[S_R_K]

Thanks for your help. Just a quick question.
So the final answer would be dilation of \(\frac{1}{2}\) from the x-axis, then translation of 3 units up (y-axis), but then wouldn't it be a translation of 3 units left (x-axis), as it is \(y=\frac{1}{x+3}\)?

No. The graph of \(y=\frac{1}{x+3}\) has its vertical asymptote at \(x=-3\) - which can be seen by finding when the denominator is zero - so a translation 3 units right is required.

And just when writing mapping notation, is writing \(\frac{y}{2}+3=y'\) okay instead of \(\frac{y+6}{2}=y'\)? Which is preferred/more clear?

Both are correct. Some people prefer the first expression, because they prefer always writing transformations in the "DRT" order, but it's better to be flexible, so that you don't get caught out when things appear a bit different than what you're used to.

[redset8]

No. The graph of \(y=\frac{1}{x+3}\) has its vertical asymptote at \(x=-3\) - which can be seen by finding when the denominator is zero - so a translation 3 units right is required

Oh I'm such an idiot, I'm thinking of transforming \(y=\frac{1}{x}\) to \(y=\frac{1}{x+3}\), rather than the other way around. All good.

Thanks again for your help!