How do you express #y=(x-2)^2+4# in standard form?

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Answer 1 · 2017-07-23T22:31:57

See a solution process below:

First, square the term in parenthesis using this rule:

#(color(red)(a) - color(blue)(b))^2 = color(red)(a)^2 - 2color(red)(a)color(blue)(b) + color(blue)(b)^2#

#y = (color(red)(x) - color(blue)(2))^2 + 4#

#y = (color(red)(x)^2 - (2 * color(red)(x) * color(blue)(2)) + color(blue)(2)^2) + 4#

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

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

#y = x^2 - 4x + 8#