I'm litteraly confounded as to how to turn this equation

y=x^2+18x-4

and I have no idea how to annihilate this inequality

-x^2+x+6 is greater than or equal to 0

- Jun 12th 2008, 09:05 PMmygodishendrixConverting Quadratic equations to Vertex form and solving quadratic inequalities
I'm litteraly confounded as to how to turn this equation

y=x^2+18x-4

and I have no idea how to annihilate this inequality

-x^2+x+6 is greater than or equal to 0 - Jun 12th 2008, 09:30 PMearboth
1. Complete the square:

$\displaystyle y = x^2+18x-4 ~\iff~ y = x^2+18x + {\color{blue} 81 - 81 } - 4 ~\iff~ \boxed{y = (x+9)^2-85}$

Therefore the vertex is at V(-9, -85)

2. $\displaystyle -x^2+x+6 \geq 0 ~\iff~ x^2-x \leq 6$

Complete the square now;

$\displaystyle x^2-x + {\color{blue} \frac14 }\leq 6+ {\color{blue} \frac14 } ~\iff~ \left(x-\frac12 \right)^2 \leq \left(\frac52 \right)^2$

Calculate the square root on both sides:

$\displaystyle \left|x-\frac12 \right| \leq \frac52 ~\implies~ -\frac52 \leq x-\frac12 \leq \frac52$

Finally you get:

$\displaystyle \boxed{-2\leq x \leq 3}$