How do you write $y = 2(x + 4)^2 - 21$ in standard form?

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Answer 1 · 2015-05-21T01:35:41

Standard form relates to writing a function such as yours as $ax^n+bx^(n-1)+...+a'y^n+b'y^(n-1)+...$ constant

Thus, let's just distribute your product and then move everything to the same side:

$y=2(x^2+8x+16)-21$
$y=2x^2+16x+32-21$
$y=2x^2+16x+11$

$2x^2+16x-y+11=0$

How do you write y = 2(x + 4)^2 - 21y = 2(x + 4)^2 - 21 in standard form?