Question

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

Answer 1

Interior of this parabola, having size a = 1/4, vertex V(13/2, 25/4) and axis `x=13/2 darr`.. See graph.

Explanation:

The ( not included ) boundary `uarr` curve, for the map of the

solution `{(x, y)}`, is the parabola

`y=-x^2+13x-36=-(x-13/2)^2+169/4-25`,

giving the standard form

`(x-13/2)^2=-(y-25/4)`

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