# How do you find the domain and range for y = sqrt(x^2 -3x +2)?

Oct 22, 2017

Domain: $x \le 1 \mathmr{and} x \ge 2 \mathmr{and} x | \left(- \infty , 1\right] \cup \left[2 , \infty\right)$
Range: $y \ge 0 \mathmr{and} y | \left[0 , \infty\right)$

#### Explanation:

y= sqrt(x^2-3x+2) = sqrt((x-1)(x-2)); Domain: under

root should be $\ge 0 \therefore \left(x - 1\right) \left(x - 2\right) \ge 0$

When $1 < x < 2$ sign of $y$ is $\left(+\right) \left(-\right) = \left(-\right) \therefore < 0$

Therefore for 1 < x < 2 ; y  is undefined .

Domain: $x \le 1 \mathmr{and} x \ge 2 \mathmr{and} x | \left(- \infty , 1\right] \cup \left[2 , \infty\right)$

Range: $y \ge 0 \mathmr{and} y | \left[0 , \infty\right)$ since square root of positive quantity

is also positive.

graph{(x^2-3x+2)^0.5 [-10, 10, -5, 5]}