What is the maximum value of #f(x) = -(x+3)^2+4#?

2 Answers
Jan 26, 2018

The maximum value of #f(x)# is 4.

Explanation:

To find the maximum value of an upside-down parabola, you must find the y-coordinate of its vertex.

Since our equation is already in vertex form, we can grab the vertex pretty easily:

Vertex form: #a(x-h)^2+k#
where #(h, k)# is the parabola's vertex

#f(x)=-(x+3)^2+4#

#=-(x-(-3))^2+4#

#=> h=-3" and " k=4#

#=>"vertex" = (-3,4)#

Our maximum value, in this case, is #k#, or 4.

Jan 26, 2018

The maximum value #=4#

Explanation:

Given -

#y=-(x+3)^2+4#

#dy/dx=[[-2(x+3)]].(1)#

#dy/dx=-2x-6#

#(d^2x)/(dy^2)=-2#

#dy/dx=0=>-2x-6=0#

#x=(6)/(-2)=-3#

At #x=-3; dy/dx=0# and #(d^2y)/(dx^2)<1#

Hence the function has a maximum at #x=-3#

Maximum Value of the function.

#y=f(-3)=-(-3+3)^2+4=-(0)^2+4=4#

The maximum value #=4#

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