Question

Complete the inequality so that it will be true for any value of x: -x^2+18x-81 ... 0

Answer 1

first off let's notice the leading term's coefficient is negative, so that means the parabola is opening downwards, let's find its vertex.

`\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{\textit{\small a}}{\downarrow }}{-1}x^2\stackrel{\stackrel{\textit{\small b}}{\downarrow }}{+18}x\stackrel{\stackrel{\textit{\small c}}{\downarrow }}{-81} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)`

`\left(-\cfrac{ 18}{2(-1)}~~~~ ,~~~~ -81-\cfrac{ (18)^2}{4(-1)}\right) \implies\left( - \cfrac{ 18 }{ -2 }~~,~~-81 - \cfrac{ 324 }{ -4 } \right) \\\\\\ \left( \cfrac{ 18 }{ 2 }~~,~~-81 + \cfrac{ 324 }{ 4 } \right)\implies \left( \cfrac{ 18 }{ 2 }~~,~~-81 + 81 \right)\implies (~9~~,~~ 0~)`

so it has a vertex at (9 , 0), so the highest point f(x) reaches is 0 really, from there on, is all going south, Check the picture below.

so we can say that f(x) or "y" is either 0 or less than 0

-x² + 18x - 81 ⩽ 0