# What is the domain and range of y=-sqrt(x^2-3x-10)?

Aug 31, 2015

Domain: $\left(- \infty , - 2\right] \cup \left[5 , + \infty\right)$
Range: $\left(- \infty , 0\right]$

#### Explanation:

Right from the start, you know that the expression under the square root must be positive, because for real numbers you can only take the square root of positive numbers.

Make the expression inside the square root equal to zero to determine its roots

${x}^{2} - 3 x - 10 = 0$

${x}_{1 , 2} = \frac{- \left(- 3\right) + - \sqrt{{\left(- 3\right)}^{2} - 4 \cdot 1 \cdot \left(- 10\right)}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{3 \pm \sqrt{49}}{2}$

${x}_{1 , 2} = \frac{3 \pm 7}{2} = \left\{\begin{matrix}{x}_{1} = \frac{3 + 7}{2} = 5 \\ {x}_{2} = \frac{3 - 7}{2} = - 2\end{matrix}\right.$

The quadratic can be factored using its two roots

$\left(x - 5\right) \left(x + 2\right) = 0$

In order for this to be greater than or equal to zero, you need either both $\left(x - 5\right)$ and $\left(x + 2\right)$ to be positive or both to be negative. Do not forget to include the values that makes each term equal to zero.

For any value of $x \ge 5$ you have

$\left\{\begin{matrix}x - 5 \ge 0 \\ x + 2 > 0\end{matrix}\right. \implies \left(x - 5\right) \left(x + 2\right) \ge 0$

Likewise, for any value of $x \le - 2$ you have

$\left\{\begin{matrix}x - 5 < 0 \\ x + 2 \le 0\end{matrix}\right. \implies \left(x - 5\right) \left(x + 2\right) \ge 0$

The domain of the original function will thus be $\left(- \infty , - 2\right] \cup \left[5 , + \infty\right)$.

To find the range of the function, keep in mind that the square root of any positive real number is always positive. Since you have

$f \left(x\right) = - \sqrt{{x}^{2} - 3 x - 10}$

you get that

$f \left(x\right) \le 0 \text{, } \left(\forall\right) x \in \left(- \infty , - 2\right] \cup \left[5 , + \infty\right)$

The range of the function will thus be $\left(- \infty , 0\right]$.

graph{-sqrt(x^2 -3x - 10) [-10, 10, -5, 5]}