# What is the domain and range of h(x)=-sqrt(x^2-16) -3?

Dec 29, 2017

Domain for $h \left(x\right)$ is $x \le - 4$ and $x \ge 4$. Range for $h \left(x\right)$ is $\left(- \infty , - 3\right)$.

#### Explanation:

It is apparent that ${x}^{2} - 16 > 0$, hence we must $x \le - 4$ or $x \ge 4$ and that is the domain for $h \left(x\right)$.

Further the least value for $\sqrt{{x}^{2} - 16}$ is $0$ and it can up to $\infty$.

Hence range for $h \left(x\right) = - \sqrt{{x}^{2} - 16} - 3$ is from a minimum of $- \infty$ to maximum of $- 3$ i.e. $\left(- \infty , - 3\right)$.