# How do you find the domain and range of (x-2)/(x^2+3x-10)?

Nov 25, 2017

The domain is $x \in \mathbb{R} - \left\{- 5\right\}$. The range is $y \in \left(- \infty , 0\right) \cup \left(0 , + \infty\right)$

#### Explanation:

The denominator is

${x}^{2} + 3 x - 10 = \left(x - 2\right) \left(x + 5\right)$

Therefore,

$\frac{x - 2}{{x}^{2} + 3 x - 10} = \frac{x - 2}{\left(x - 2\right) \left(x + 5\right)} = \frac{1}{x + 5}$

As the denominator $\ne 0$, so

$\left(x + 5\right) \ne 0$

So,

The domain is $x \in \mathbb{R} - \left\{- 5\right\}$

To calculate the range, proceed as follows

Let $y = \frac{1}{x + 5}$

$y \left(x + 5\right) = 1$

$y x + 5 y = 1$

$x = \frac{1 - 5 y}{y}$

So,

$y \ne 0$

Therefore,

The range is $y \in \left(- \infty , 0\right) \cup \left(0 , + \infty\right)$

graph{1/(x+5) [-10, 10, -5, 5]}

Nov 25, 2017

$x \in \mathbb{R} , x \ne - 5$
$y \in \mathbb{R} , y \ne 0$

#### Explanation:

$\text{let } y = \frac{x - 2}{{x}^{2} + 3 x - 10}$

$\text{factorise numerator/denominator and simplify}$

$y = \frac{\cancel{\left(x - 2\right)}}{\left(x + 5\right) \cancel{\left(x - 2\right)}} = \frac{1}{x + 5}$

$\text{the denominator cannot equal zero as this would make}$
$\text{y undefined. Equating the denominator to zero and}$
$\text{solving gives the value that x cannot be}$

$\text{solve "x+5=0rArrx=-5larrcolor(red)"excluded value}$

$\Rightarrow \text{domain is } x \in \mathbb{R} , x \ne - 5$

$\text{to find the range rearrange making x the subject}$

$y \left(x + 5\right) = 1 \leftarrow \textcolor{b l u e}{\text{cross-multiply}}$

$\Rightarrow x y + 5 y = 1$

$\Rightarrow x y = 1 - 5 y$

$\Rightarrow x = \frac{1 - 5 y}{y}$

$\text{the denominator cannot equal zero}$

$\Rightarrow \text{range is } y \in \mathbb{R} , y \ne 0$