Plz help!!!!

Part 1:

When shifting an equation, if the value shifting is inside the equation, you go the opposite direction. So, for example, if it was:

\(f(x)=|x+7|\), then you would shift 7 to the left.

So, for your problem of \(f(x)= |x-3|+2\), when you replace x with x+2, the graph will shift 2 units to the left.

Part 2:

\(f(x)=\frac{1}{2}\sqrt{x+5}\)

So, square roots have specific domain restrictions, such that the value inside can't be negative because no number, when squared, can be negative, so when doing the opposite it can't be negative. 

So, with that in mind, x can be -5 but no lower.

So the domain would be \([-5,\infty)\)

The range is all possible y values. Since the square root can put out either positive or negative numbers, the range is \((-\infty,\infty)\).

Question 2

The domain  is  [-5, inf )

However.....from a positive square root we can only get 0  or a positive 

So.....the range is  [0, inf)

here's a graph to show this : https://www.desmos.com/calculator/quxajtdn9b