# How do you solve y^3-2y^2-9y+18>=0 using a sign chart?

Aug 4, 2018

Solution: $- 3 \le y \le 2 \mathmr{and} y \ge 3 \mathmr{and} \left[- 3 , 2\right] \cup \left[3 , \infty\right)$

#### Explanation:

$f \left(y\right) = {y}^{3} - 2 {y}^{2} - 9 y + 18 \ge 0$

${y}^{3} - 2 {y}^{2} - 9 y + 18$

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

$= \left(y - 2\right) \left({y}^{2} - 9\right) = \left(y - 2\right) \left(y + 3\right) \left(y - 3\right) \therefore$ zeros are

$y = - 3 , y = 2 , y = 3 \therefore f \left(- 3\right) = 0 , f \left(2\right) = 0 , f \left(3\right) = 0$

$f \left(y\right) = \left(y - 2\right) \left(y + 3\right) \left(y - 3\right)$

Sign chart:

When, $y < - 3$, sign of $f \left(y\right)$ is  (-) * (-)* (-) = (-) ; < 0

When, $- 3 < y < 2$, sign of $f \left(y\right)$ is (-) * (+) *(-) = (+) ; > 0

When, $2 < y < 3$, sign of $f \left(y\right)$ is  (+) * (+)* (-) = (-) ; < 0

When $y > 3$, sign of $f \left(y\right)$ is  (+) * (+)* (+) = (+) ; > 0

Solution: $- 3 \le y \le 2 \mathmr{and} y \ge 3 \mathmr{and} \left[- 3 , 2\right] \cup \left[3 , \infty\right)$

graph{x^3-2 x^2-9 x+18 [-80, 80, -40, 40]} [Ans]