# How do you find the domain and range of f(x)=sqrt(5x+1)?

Apr 9, 2017

Domain is pretty easy. It just takes some logic: we cannot take a square root of a negative number, so $x$ cannot equal any value that would make the expression negative.

With that in mind, we just need to set the expression inside the square root equal to $0$ and solve for $x$. That will give us the limits on it's domain

$0 = 5 x + 1$
$- 1 = 5 x$
$- \frac{1}{5} = x$

So, If $x$ equals any number smaller than $- \frac{1}{5}$, the expression becomes negative and an imaginary number. So, the domain is $\left[- \frac{1}{5} , \infty\right)$.

We include the $\frac{1}{5}$ because we can take the square root of $0$. Therefore, $- \frac{1}{5}$ is within our domain.

Now on to the range. This one is also pretty easy. We just need to see if there are any additions or subtractions outside of the square root. Those would be vertical shifts, and are responsible for changes in the range.

In our case, there are no vertical shifts, so the range is the same as the parent graph $y = \sqrt{x}$, which is $\left[0 , \infty\right)$.

Let's graph our equation and see if we're right
graph{y=sqrt(5x-1)}
We are. The graph begins at $\left(- \frac{1}{5} , 0\right)$ and continues up and right forever, just like our domains stated.