What is the domain and range of `f(x) = -2 * sqrt(x-3) + 1`?

Let's look at the parent function: `sqrt(x)`

The domain of `sqrt(x)` is from `0` to `oo`. It starts at zero because we cannot take a square root of a negative number and be able to graph it. `sqrt(-x)` gives us `isqrtx`, which is an imaginary number.

The range of `sqrt(x)` is from `0` to `oo`

This is the graph of `sqrt(x)`

graph{y=sqrt(x)}

So, what is the difference between `sqrtx` and `-2 * sqrt(x-3) + 1`?

Well, let's start with `sqrt(x-3)`. The `-3` is a horizontal shift, but it is to the right, not the left. So now our domain, instead of from `[0, oo)`, is `[3, oo)`.

graph{y=sqrt(x-3)}

Let's look at the rest of the equation. What does the `+1` do? Well, it shifts our equation up one unit. That doesn't change our domain, which is in the horizontal direction, but it does change our range. Instead of `[0, oo)`, our range is now `[1, oo)`

graph{y=sqrt(x-3)+1}

Now let's see about that `-2`. This is actually two components, `-1` and `2`. Let's deal with the `2` first. Whenever there is a positive value in front of the equation, it is a vertical stretching factor.

That means, instead of having the point `(4, 2)`, where `sqrt(4)`
equals `2`, now we have `sqrt(2*4)` equals `2`. So, it changes how our graph looks, but not the domain or the range.

graph{y=2 * sqrt(x-3)+1}

Now we've got that `-1` to deal with. A negative in the front of the equation means a refection across the `x`-axis. That won't change our domain, but our range goes from `[1, oo)` to `(-oo, 1]`

graph{y=-2sqrt(x-3)+1}

So, our final domain is `[3,oo)` and our range is `(-oo, 1]`