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How do you write $y = 4 {(x - 2)}^{2} - 1$ $y = 4(x - 2)^2 - 1$ in standard form?

1 Answer

Feb 28, 2017

$y = 4 x^{2} - 16 x + 15$ $y=4x^2-16x+15$

Explanation:

Standard form is as follows:
$y = a x^{2} + b x + c$ $y=ax^2+bx+c$

$y = 4 {(x - 2)}^{2} - 1$ $y=4(x-2)^2-1$
$y = 4 (x - 2) (x - 2) - 1$ $y=4(x-2)(x-2)-1$
$y = 4 (x^{2} - 4 x + 4) - 1$ $y=4(x^2-4x+4)-1$
$y = 4 x^{2} - 16 x + 16 - 1$ $y=4x^2-16x+16-1$
$y = 4 x^{2} - 16 x + 15$ $y=4x^2-16x+15$

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