# How do you find the domain and range of y = sqrt(9 + x)?

Jul 27, 2016

Domain: $\left[- 9 , \infty\right)$
Range: $\left[0 , \infty\right)$
Assuming we are working strictly within the reals, the square root function requires nonnegative values. Thus the domain of the function $f \left(x\right) = \sqrt{9 + x}$ is the set of values for $x$ such that $9 + x \ge 0$, that is, where $x \ge - 9$.
As $9 + x$ varies from $0$ to $\infty$, $\sqrt{x + 9}$ varies from $0$ to $\infty$.
Thus, the domain of $\sqrt{x + 9}$ is $\left[- 9 , \infty\right)$ and its range is $\left[0 , \infty\right)$