How do you express $y = {(x - 2)}^{2} + 4$ $y=(x-2)^2+4$ 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:

${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $(color(red)(a) - color(blue)(b))^2 = color(red)(a)^2 - 2color(red)(a)color(blue)(b) + color(blue)(b)^2$

$y = {(x - 2)}^{2} + 4$ $y = (color(red)(x) - color(blue)(2))^2 + 4$

$y = (x^{2} - (2 \cdot x \cdot 2) + 2^{2}) + 4$ $y = (color(red)(x)^2 - (2 * color(red)(x) * color(blue)(2)) + color(blue)(2)^2) + 4$

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

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

$y = x^{2} - 4 x + 8$ $y = x^2 - 4x + 8$

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