How do you write the quadratic in vertex form given $y=-4x^2-5x+3$ ?

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Answer 1 · 2015-05-05T00:28:15

$y= (-4)(x+5/8)^2 + 73/16$

The general vertex form of a quadratic is
$y = m(x-a)^2+b$
where $(a,b)$ is the vertex

$y= -4x^2-5x +3$

$y= (-4)(x^2+5/4x) +3 " extract the "m" factor"$

$y=(-4)(x^2+5/4x+(5/8)^2) -(-4)(5/8)^2 + 3$

$y= (-4)(x+5/8)^2 + 73/16$

...and (assuming I haven't made any mistakes) the vertex is at $(-5/8,73/16)=(-5/8,4 9/16)$

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How do you write the quadratic in vertex form given y=-4x^2-5x+3y=-4x^2-5x+3?