What is the standard form of $y = {(x - 6)}^{2} + 11$ $y= (x-6)^2+11$ ?

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Answer 1 · 2018-06-13T21:05:50

$y = x^{2} - 12 x + 47$ $y=x^{2}-12x+47$

Standard form of a quadratic is when the equation is given in the form:

$y = a x^{2} + b x + c$ $y=ax^{2}+bx+c$ where a, b and c are constants

To achieve, simply simplify the above equation

$y = (x - 6) (x - 6) + 11$ $y=(x-6)(x-6)+11$
$y = x^{2} - 12 x + 36 + 11$ $y=x^{2}-12x+36+11$
$y = x^{2} - 12 x + 47$ $y=x^{2}-12x+47$

What is the standard form of y= (x-6)^2+11y=(x−6)2+11y= (x-6)^2+11?