# How do you find the minimum and maximum value of y=-(x-1)(x+4)?

Jul 28, 2018

$y = - \left(x - 1\right) \left(x + 4\right) = - \left({x}^{2} + 3 x - 4\right)$

$= - \left({\left(x + \frac{3}{2}\right)}^{2} - \frac{9}{4} - 4\right)$

$= - \left({\left(x + \frac{3}{2}\right)}^{2} - \frac{25}{4}\right)$

$\implies y = - {\left(x + \frac{3}{2}\right)}^{2} + \frac{25}{4}$

Now:

${\left(x + \frac{3}{2}\right)}^{2} \ge 0 q \quad \forall x$

$\therefore - {\left(x + \frac{3}{2}\right)}^{2} \le 0 q \quad \forall x$

$\therefore y \le \frac{25}{4} q \quad \forall x$

$\implies \left\{\left({y}_{\text{max") = 25/4),(y_("max}} = - \infty\right)\right.$

graph{-(x-1)(x+4) [-5, 5, -9.01, 9.01]}