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How do you write in vertex form for $y = x^{2} + .6 x - 11$ $y=x^2 + .6x - 11$ by completing the square?

1 Answer

Jun 2, 2015

The general squared binomial is ${(x + a)}^{2} = x^{2} + 2 a x + a^{2}$ $(x+a)^2 = x^2+2ax+a^2$

Given the equation $y = x^{2} + .6 x - 11$ $y= x^2+.6x-11$

If the coefficient of $x$ $x$ is $(0.6)$ $(0.6)$
then
in terms of the general form $a = 0.3$ $a=0.3$ and $a^{2} = 0.09$ $a^2 = 0.09$

Writing as a completed square:
$XXXXX$ $color(white)("XXXXX")$ $y = {(x + 0.3)}^{2} - 11 - 0.09$ $y = (x+0.3)^2 -11 -0.09$

$XXXXX$ $color(white)("XXXXX")$ $y = {(x - (- 0.3))}^{2} + (- 11.09)$ $y = (x - (-0.3))^2 + (-11.09)$

which is the vertex form
$XXXXX$ $color(white)("XXXXX")$ with the vertex at $(- 0.3, - 11.09)$ $(-0.3,-11.09)$

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