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]}