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

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Answer 1 · 2017-01-27T11:01:43

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

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

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

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

giving the standard form

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

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

How do you graph the inequality y<-x^2+13x-36y<−x2+13x−36y<-x^2+13x-36?