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Math Help - Converting Quadratic equations to Vertex form and solving quadratic inequalities

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    Converting 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
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  2. #2
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    Quote Originally Posted by mygodishendrix View Post
    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
    1. Complete the square:

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

    Complete the square now;

    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:

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

    Finally you get:

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