How do you find the max and min of $y = - x^{2} + 6$ $y=-x^2+6$ by completing the square?

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How do you find the max and min of $y = - x^{2} + 6$ $y=-x^2+6$ by completing the square?

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Answer 1 · 2015-05-09T20:34:17

"Completing the squares" is a bit unusual for the given equation:
$y = - x^{2} + 6$ $y=-x^2+6$
since in "completed square form" it would be:
$y = (- 1) (x^{2} - 0 x + 0) + 6$ $y = (-1)(x^2-0x+0)+6$
or
$y = (- 1) {(x - 0)}^{2} + 6$ $y=(-1)(x-0)^2+6$

This is also the vertex form;
so the vertex is at $(0, 6)$ $(0,6)$

Because of the $(- 1)$ $(-1)$ factor we know that the parabola opens downward $\to$ $rarr$ the vertex is at a maximum.
As the magnitude of $x$ $x$ increases the value of $y$ $y$ decreases without bound.

So the maximum value is $6$ $6$ and the minimum is $- \infty$ $-oo$

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How do you find the max and min of $y = - x^{2} + 6$ $y=-x^2+6$ by completing the square?

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