How do you write the quadratic in vertex form given $y= -3x^2 + 21x$ ?

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Answer 1 · 2015-05-10T13:50:40

$y = -3(x - 7/2)^2+147/4 = -3(x - 3.5)^2+36.75$

$y = -3x^2+21x = -3(x^2-7x)$
$= -3((x-(7/2))^2 - (7/2)^2)$
$= -3(x-(7/2))^2+3(7/2)^2$
$= -3(x - 7/2)^2+147/4$
$= -3(x - 3.5)^2+36.75$

The vertex of the parabola is $(3.5, 36.75)$

How do you write the quadratic in vertex form given y= -3x^2 + 21xy= -3x^2 + 21x?