How do you write $y = {(x - 2)}^{2} + 6$ $y= (x-2)^(2)+6$ in standard form?

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Answer 1 · 2017-08-09T10:25:43

See a solution process below:

First, expand the squared term in parenthesis using this rule:

${(x - y)}^{2} = x^{2} - 2 x y + y^{2}$ $(color(red)(x) - color(blue)(y))^2 = color(red)(x)^2 - 2color(red)(x)color(blue)(y) + color(blue)(y)^2$

$y = {(x - 2)}^{2} + 6$ $y = (color(red)(x) - color(blue)(2))^2 + 6$

$y = x^{2} - (2 \cdot x \cdot 2) + 2^{2} + 6$ $y = color(red)(x)^2 - (2 * color(red)(x) * color(blue)(2)) + color(blue)(2)^2 + 6$

$y = x^{2} - 4 x + 4 + 6$ $y = x^2 - 4x + 4 + 6$

Now, combine like terms:

$y = x^{2} - 4 x + (4 + 6)$ $y = x^2 - 4x + (4 + 6)$

$y = x^{2} - 4 x + 10$ $y = x^2 - 4x + 10$

How do you write y= (x-2)^(2)+6y=(x−2)2+6y= (x-2)^(2)+6 in standard form?