# What is the range of the function f(x)=x^2-x-2?

Aug 31, 2017

$\left[- \frac{9}{4} , + \infty\right)$

#### Explanation:

$\text{complete the square on f(x)}$

$f \left(x\right) = {x}^{2} - 2 \left(\frac{1}{2}\right) x + \frac{1}{4} - \frac{1}{4} - 2$

$\textcolor{w h i t e}{f \left(x\right)} = {\left(x - \frac{1}{2}\right)}^{2} - \frac{9}{4}$

$\text{for all x} , {\left(x - \frac{1}{2}\right)}^{2} \ge 0$

$\text{the minimum value of f(x) is therefore } - \frac{9}{4}$

$\Rightarrow \text{range is } \left[- \frac{9}{4} , + \infty\right)$
graph{x^2-x-2 [-10, 10, -5, 5]}