I am in a debate with higher ups in my school district, and before I step go over peoples heads to find someone who agrees with me I wanted to run this by you guys.
I have added a picture of the answer key that they provided for our next exam. My opinion is that this is wrong because a vertical compression should make the graph wider.(the magnitude of the slope should be less than the function before the compresssion).
I also explained to them that the original function could be defined as, $f(x)=2|x+1|+4$
It is my understanding that to perform a vertical compression on this we should multiply the coefficient of the absolute value by the compression factor of 1/2 $(\frac{1}{2})(2|x+1|)+4$ which would give $|x+1|+4.$ I explained this and was told that we should multiply the 1/2 times the entire function like this $(\frac{1}{2})(2|x+1|+4)$ giving us $|x+1|+2$. To which I responded by explaining that this would cause the vertex to move. The person told me that this is what happens. The vertical compression will also move the vertex.
My question is simple. Am I losing my mind or am I just working with people who don't understand this? If I am right, any explination or resource you could share to help me convince them to fix this before the test would be great.
To be accurate, the graphed function in the image is $f(x)=-2|x+1|+4$.
But looking at the problem, $P$ need not belong to any function at all. You can apply transformations to points as you do functions. The answer key is correct -- one reason why you may be confused is because the transformed function is not graphed.
If you were to create an equation for a function after the transformations above, it would be $$y=\frac12f[-(x-3)].$$
Since $f(x)=-2|x+1|+4$, it can be simplified to the following: \begin{align} y&=\frac12f[-(x-3)]\\ y&=\frac12\big(-2|(-(x-3))+1|+4\big)\\ y&=-|-x+3+1|+2\\ y&=-|x-4|+2 \end{align}
If you graph $y=-|x-4|+2$, you should expect the point $P(5,1)$ to be on it.
Here is a visual via Desmos.