# How do you find the domain and range of 5(x - 2) (x + 1)?

May 23, 2018

Domain: $\left(- \infty , + \infty\right)$ Range: $\left[- 11.25 , + \infty\right)$

#### Explanation:

Let $f \left(x\right) = 5 \left(x - 2\right) \left(x + 1\right)$

$= 5 {x}^{2} - 5 x - 10$

$f \left(x\right)$ is defined $\forall x \in \mathbb{R} \to$ Domain: $\left(- \infty , + \infty\right)$

$f \left(x\right)$ is a quadratic function of the form $a {x}^{2} + b x + c$

Since $a > 0$, $f \left(x\right)$ will have a minimum value at $x = \frac{- b}{2 a}$

$\frac{- b}{2 a} = \frac{- \left(- 5\right)}{2 \times 5} = \frac{1}{2}$

$\therefore f {\left(x\right)}_{\min} = f \left(\frac{1}{2}\right)$

$f {\left(x\right)}_{\min} = \frac{5}{4} - \frac{5}{2} - 10 = - \frac{5}{4} - 10 = - \frac{45}{4}$

$= 11.25$

$f \left(x\right)$ has no finite upper bound.

Hence, Range: $\left[- 11.25 , + \infty\right)$

We can visualise the domain and range of $f \left(x\right)$ from its graph below.

graph{5(x-2)(x+1) [-24.9, 26.43, -14.9, 10.75]}