What is the vertex of $y = x^{2} - 12 x + 16$ $y=x^2-12x+16$ ?

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What is the vertex of $y = x^{2} - 12 x + 16$ $y=x^2-12x+16$ ?

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Answer 1 · 2018-07-02T10:04:14

$vertex = (6, - 20)$ $"vertex "=(6,-20)$

Explanation:

$given a quadratic in standard form$ $"given a quadratic in "color(blue)"standard form"$

$∙ x y = a x^{2} + b x + c x; a \neq 0$ $•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0$

$then the x-coordinate of the vertex is$ $"then the x-coordinate of the vertex is"$

$∙ x x_{vertex} = - \frac{b}{2 a}$ $•color(white)(x)x_(color(red)"vertex")=-b/(2a)$

$y = x^{2} - 12 x + 16 is in standard form$ $y=x^2-12x+16" is in standard form"$

$with a = 1, b = - 12 and c = 16$ $"with "a=1,b=-12" and "c=16$

$x_{vertex} = - \frac{- 12}{2} = 6$ $x_("vertex")=-(-12)/2=6$

$substitute x = 6 into the equation for y-coordinate$ $"substitute "x=6" into the equation for y-coordinate"$

$y_{vertex} = 36 - 72 + 16 = - 20$ $y_("vertex")=36-72+16=-20$

$vertex = (6, - 20)$ $color(magenta)"vertex "=(6,-20)$

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What is the vertex of $y = x^{2} - 12 x + 16$ $y=x^2-12x+16$ ?

1 Answer

Explanation:

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What is the vertex of y=x^2-12x+16y=x2−12x+16y=x^2-12x+16?

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Explanation:

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What is the vertex of $y = x^{2} - 12 x + 16$ $y=x^2-12x+16$ ?