What is the vertex of #y = 5(1/5x+2/25)^2 + 1#?

1 Answer
Roella W.
Dec 23, 2015

#(-7, -9.59)#

Explanation:

First expand the equation to simplify it, then use completing the square to get it into vertex form.
Expanding -
#y = 5(x^2/25 +14x/25 +4/625) +1#
#y = x^2/5 + 14x/5 + 4/125 +1#
#5y = x^2 +14x +129/25#
Using b/2 we complete the square and include another number to get the right value for c
#5y = (x+7)^2 - 49 +129/125#
#5y = (x+7)^2 - 47*(121/125)#
At the vertex #x= -7# and #y = (-47*121/125)/5 = -9.59#

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