MemeLord

Thank you very much! That verified my thought process!

Yes I did get D. Thank you very much.

Thanks.

@CPhill But what about the requirement of there being a \(\pm\) outside of every square root?

I'm not sure how to explain how to graph it, but the correct graph is the bottom left one. My best suggestion for graphing would be to test it out on www.desmos.com/calculator.

Then I guess I can't remember. I just created a new one.

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 \((-\infty,\infty)\).

\(\sqrt[3]{250x^6}\)

So, in this case, a cube root is the same as that value to the power of 1/3.

\((250x^6)^{\frac{1}{3}}\)

Then, you just multiply the exponents by each other. 

\(6*\frac{1}{3}=\frac{6}{3}=2\)

So that makes it:

\(250x^2\)