In question 1b of the midterm in the pdf attached below, I don't understand why the solution should be what it stated in the other pdf attached rather than what I did in the picture, which basically flips 1a along y-axis.
Flipping about the axis does not apply to the shift.
You have h(t), and you want to compute h(y(t)) where y(t) = -t-2.
h(-y(t)) [tex]\neq[/tex] h(y(-t))
The former is what you did, which is to compute y(t+2) and then flip, but that gives you y(-t+2), so then you need to shift by 4 to get y(-t-2).
