What is the vertex form of $6y=-x^2 + 9x$ ?

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Answer 1 · 2016-08-29T13:02:50

$y = -1/6(x-9/2)^2+27/8$

Divide both sides by $6$ to get:

$y = -1/6(x^2-9x)$

$=-1/6((x-9/2)^2-9^2/2^2)$

$=-1/6(x-9/2)^2+1/6*81/4$

$=-1/6(x-9/2)^2+27/8$

Taking the two ends together, we have:

$y = -1/6(x-9/2)^2+27/8$

which is in vertex form:

$y = a(x-h)^2+k$

with multiplier $a = -1/6$ and vertex $(h, k) = (9/2, 27/8)$

graph{(6y+x^2-9x)((x-9/2)^2+(y-27/8)^2-0.02) = 0 [-5.63, 14.37, -3.76, 6.24]}

What is the vertex form of 6y=-x^2 + 9x6y=-x^2 + 9x ?