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

1 Answer
Alan P.
Feb 25, 2016

Vertex #(45,-4)#

Explanation:

There are a couple ways of doing this; perhaps the most obvious is to convert the given equation into standard vertex form:
#color(white)("XXX")y=m(x-a)^2+b# with its vertex at #(a,b)#

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

#rarr y=5((x-45)/3)^2-4#

#rarr 5/9(x-45)^2+(-4)#
#color(white)("XXX")#which is the vertex form with vertex at #(45,-4)#

Alternately think of replacing #hatx=x/3# and the given equation is in vertex form for #(hatx,y)=(15,-4)#
and since #x=3*hatx# the vertex using #x# is #(x,y)=(3xx15,-4)#
graph{5(x/3-15)^2-4 [35.37, 55.37, -6.36, 3.64]}

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