What is the vertex form of #y=-1/3(x-3)(x/3+7) #?

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Answer 1 · 2017-12-15T17:05:33

Vertex form of equation is #y=-1/9(x+9)^2+16# and vertex is #(-9,16)#

Vertex form of equation is #y=a(x- h)^2+k#, whose vertex is #(h,k)#. Hence

#y=-1/3(x-3)(x/3+7)#

= #-1/9(x-3)(x+21)#

= #-1/9(x^2-3x+21x-63)#

= #-1/9(x^2+18x+81-81-63)#

= #-1/9((x+9)^2-144)#

= #-1/9(x+9)^2+16#

and vertex is #(-9,16)#

graph{ -1/9(x+9)^2+16 [-19, 1, 8, 18]}