How do you find the max and min of $y = - x^{2} - 8 x + 2$ $y=-x^2-8x+2$ by completing the square?

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Answer 1 · 2015-05-31T11:28:19

$- x^{2} - 8 x + 2 = - (x^{2} + 8 x - 2)$ $-x^2-8x+2 = -(x^2+8x-2)$

$= - ({(x + 4)}^{2} - 16 - 2)$ $=-((x+4)^2-16-2)$

$= - ({(x + 4)}^{2} - 18)$ $=-((x+4)^2-18)$

$= 18 - {(x + 4)}^{2}$ $=18-(x+4)^2$

Being the square of a real number ${(x + 4)}^{2} \geq 0$ $(x+4)^2 >= 0$

So the maximum value $18$ $18$ occurs when ${(x + 4)}^{2} = 0$ $(x+4)^2 = 0$ (that is when $x = - 4$ $x=-4$ )

The value is unbounded in the other direction.

$y = - x^{2} - 8 x + 2 \to - \infty$ $y = -x^2-8x+2 -> -oo$ as $x \to \infty$ $x ->oo$ or $x \to - \infty$ $x->-oo$

How do you find the max and min of y=−x2−8x+2y=-x^2-8x+2 by completing the square?