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

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Answer 1 · 2018-03-30T04:04:39

Vertex is at #(-0.5,1.25) #

#y=-(x+2)^2+3x+5 or y=-(x^2+4x+4)+3x+5#

or # y=-x^2-4x-4+3x+5 or y=-x^2-x+1# or

#y=-(x^2+x)+1 or y=-(x^2+x+0.5^2)+0.5^2+1# or

#y=-(x+0.5)^2+1.25# . Comparing with vertex form of

equation #f(x) = a(x-h)^2+k ; (h,k)# being vertex we find

here #h=-0.5 , k=1.25 :.# Vertex is at #(-0.5,1.25) #

graph{-(x+2)^2+3x+5 [-10, 10, -5, 5]}