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

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Answer 1 · 2015-12-20T05:44:16

Vertex : #(-13/4 , -49/8)#

Vertex form: #y = 2(x+13/4)^2 -49/8#

Step 1: Expand/multiply the function so it can be int eh standard form of #y = ax^2 + bc+c#

Given
#y = 4(x+2)^2 -2x -3x -1#

#= 4(x+2)(x+2)-2x^2 -3x-1#

#= 4(x^2 +2x+2x+4) -2x^2 -3x-1#

#=4(x^2 +4x+4) = 2x^2 -3x -1#

#=4x^2 +16 x +16 -2x^2 -3x -1#

#=2x^2 + 13x+15#

#a = 2, " " " b= 13, " " " c= 15#

The formula for vertex is #(-b/(2a), f(-b/(2a)))#

#x_(vertex) = -b/(2a) = h #

#x_(vertex) = (-13)/(2*2) = -13/4#

#y_(vertex) = f(-b/(2a)) = k#

#f(-13/4) = 2(-13/4)^2 +13(-13/4)+15#

#= 2(169/16)-169/4 +15 = -49/8#

Vertex form: #y = a(x-h)^2 + k#

#y = 2(x+13/4)^2 -49/8#