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

Jul 2, 2017

domain is $\left[3 , \infty\right)$ and our range is $\left(- \infty , 1\right]$

#### Explanation:

Let's look at the parent function: $\sqrt{x}$

The domain of $\sqrt{x}$ is from $0$ to $\infty$. 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 $i \sqrt{x}$, which is an imaginary number.

The range of $\sqrt{x}$ is from $0$ to $\infty$

This is the graph of $\sqrt{x}$

graph{y=sqrt(x)}

So, what is the difference between $\sqrt{x}$ and $- 2 \cdot \sqrt{x - 3} + 1$?

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 $\left[0 , \infty\right)$, is $\left[3 , \infty\right)$.

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 $\left[0 , \infty\right)$, our range is now $\left[1 , \infty\right)$

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 $\left(4 , 2\right)$, where $\sqrt{4}$
equals $2$, now we have $\sqrt{2 \cdot 4}$ 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 $\left[1 , \infty\right)$ to $\left(- \infty , 1\right]$

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

So, our final domain is $\left[3 , \infty\right)$ and our range is $\left(- \infty , 1\right]$