How do you graph the inequality y<-x^2+13x-36?

Jan 27, 2017

Interior of this parabola, having size a = 1/4, vertex V(13/2, 25/4) and axis $x = \frac{13}{2} \downarrow$.. See graph.

Explanation:

The ( not included ) boundary $\uparrow$ curve, for the map of the

solution $\left\{\left(x , y\right)\right\}$, is the parabola

$y = - {x}^{2} + 13 x - 36 = - {\left(x - \frac{13}{2}\right)}^{2} + \frac{169}{4} - 25$,

giving the standard form

${\left(x - \frac{13}{2}\right)}^{2} = - \left(y - \frac{25}{4}\right)$

graph{y+x^2-13x+36<0 [-20, 20, -10, 10]}