# How do you find the domain and range of f(x)=(x-2)/(x^2-6x+9)?

Jun 14, 2018

The domain is $x \in \left(- \infty , 3\right) \cup \left(3 , + \infty\right)$. The range is $y \in \left[- \frac{1}{4} , + \infty\right)$.

#### Explanation:

The function is

$f \left(x\right) = \frac{x - 2}{{x}^{2} - 6 x + 9} = \frac{x - 2}{x - 3} ^ 2$

The denominator must be $\ne 0$

Therefore,

${\left(x - 3\right)}^{2} \ne 0$, $\implies$, $x \ne 3$

The domain is $x \in \left(- \infty , 3\right) \cup \left(3 , + \infty\right)$

To find the range, let

$y = \frac{x - 2}{{x}^{2} - 6 x + 9}$

Cross multiply,

$y \left({x}^{2} - 6 x + 9\right) = x - 2$

$y {x}^{2} - 6 y x - x + 9 y + 2 = 0$

$y {x}^{2} - \left(6 y + 1\right) x + 9 y + 2 = 0$

This is a quadratic equation in $x$, and in order to have solutions,

the discriminant $\Delta \ge 0$

$\Delta = {\left(6 y + 1\right)}^{2} - 4 \left(y\right) \left(9 y + 2\right) \ge 0$

$36 {y}^{2} + 12 y + 1 - 36 {y}^{2} - 8 y \ge 0$

$4 y + 1 \ge 0$

$y \ge - \frac{1}{4}$

The range is $y \in \left[- \frac{1}{4} , + \infty\right)$.

graph{(x-2)/(x^2-6x+9) [-5.24, 8.81, -2.67, 4.353]}