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What is the vertex of # y= 1/3(x/5+1)^2+4/15 #?

1 Answer

Jan 2, 2016

The vertex form is #y = a(x-h)^2 +k# where #(h,k)# is the vertex. For our problem the vertex is #(-5,4/15)#

Explanation:

#y=1/3(x/5+1)^2 + 4/15#

#y=1/3((x+5)/5)^2 + 4/15#

#y=1/75(x+5)^2 + 4/15#

Compare with
#y=a(x-h)^2 + k#

#h=-5# and #k=4/15#

The vertex #(h,k)# is #(-5,4/15)#

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