What is the domain and range of y= 1/(x-7) -3?

Jan 28, 2018

$x \in \mathbb{R} , x \ne 7$
$y \in \mathbb{R} , y \ne - 3$

Explanation:

The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the value that x cannot be.

$\text{solve "x-7=0rArrx=7larrcolor(red)"excluded value}$

$\Rightarrow \text{domain is } x \in \mathbb{R} , x \ne 7$

$\left(- \infty , - 7\right) \cup \left(- 7 , + \infty\right) \leftarrow \textcolor{b l u e}{\text{in interval notation}}$

$\text{divide numerator/denominator of "1/(x-7)" by x}$

$y = \frac{\frac{1}{x}}{\frac{x}{x} - \frac{7}{x}} - 3 = \frac{\frac{1}{x}}{1 - \frac{7}{x}} - 3$

$\text{as } x \to \pm \infty , y \to \frac{0}{1 - 0} - 3$

$\Rightarrow y = - 3 \leftarrow \textcolor{red}{\text{excluded value}}$

$\text{range is } y \in \mathbb{R} , y \ne - 3$

$\left(- \infty , - 3\right) \cup \left(- 3 , + \infty\right) \leftarrow \textcolor{b l u e}{\text{in interval notation}}$
graph{1/(x-7)-3 [-10, 10, -5, 5]}