# What is the domain and range of f(x)=sqrt(4x+2)?

Apr 10, 2016

$x \in \left[- \frac{1}{2} , + \infty\right)$

#### Explanation:

The function is a Square Root Function

To easily determine the domain and range, we should first convert the equation to General Form:

$y = a \cdot \sqrt{x - b} + c$
Where the point $\left(b , c\right)$ is the endpoint of the function (essentially the place at which the graph begins).

Let's now convert the given function to General Form:
$y = \sqrt{4 \left(x + \frac{1}{2}\right)}$
We can now simplify this by taking the square root of 4 outside:
$y = 2 \cdot \sqrt{x + \frac{1}{2}}$

Therefore, from general form, we can now see that the endpoint of the graph is present at the point $\left(- \frac{1}{2} , 0\right)$ due to the fact that $b = - \frac{1}{2}$ and $c = 0$.

Additionally from General Form we can see that neither $a$ is negative, nor is $x$ negative, therefore no reflections about the $x$ or $y$ axis are present. This implies that the function originates from the point $\left(- \frac{1}{2} , 0\right)$ and continues to positive infinity.

For reference, the graph of the function $\left(y = \sqrt{4 x + 2}\right)$ is below:
graph{sqrt(4x+2) [-10, 10, -5, 5]}

Therefore, the domain of the function can be expressed as:
1. Domain: $x \in \left[- \frac{1}{2} , + \infty\right)$
2. Domain: $x \ge - \frac{1}{2}$
3. Domain: $- \frac{1}{2} \le x < + \infty$