# How do you find the domain of h(x) = (2x^2 + 5 )/ (sqrt(x-2))?

Jun 2, 2018

Domain: $\left\{x | x > 2\right\}$ or in interval notation $\left(2 , \infty\right)$

#### Explanation:

$h \left(x\right) = \frac{2 {x}^{2} + 5}{\sqrt{x - 2}}$

The domain is restricted by the radical; what is within it cannot be negative and the denominator; it cannot equal zero:

$x - 2 \ge 0$

$x \ge 2$

and

$\sqrt{x - 2} \ne 0$

${\left(\sqrt{x - 2}\right)}^{2} \ne {0}^{2}$

$x - 2 \ne 0$

$x \ne 2$

put it together:

$x > 2$

Domain: $\left\{x | x > 2\right\}$ or in interval notation $\left(2 , \infty\right)$