How do you write #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:

#(color(red)(x) - color(blue)(y))^2 = color(red)(x)^2 - 2color(red)(x)color(blue)(y) + color(blue)(y)^2#

#y = (color(red)(x) - color(blue)(2))^2 + 6#

#y = color(red)(x)^2 - (2 * color(red)(x) * color(blue)(2)) + color(blue)(2)^2 + 6#

#y = x^2 - 4x + 4 + 6#

Now, combine like terms:

#y = x^2 - 4x + (4 + 6)#

#y = x^2 - 4x + 10#