How do you find the axis of symmetry, and the maximum or minimum value of the function # y=-2x^2 +4x +2 #?

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Answer 1 · 2017-03-20T13:31:56

The axis of symmetry is #x=1#
The maximum value is #V=(1,4)#

We need

#a^2-2ab+b^2=(a-b)^2#

We complete the square and factorise

#y=-2x^2+4x+2#

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

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

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

The axis of symmetry is #x=1#

The vertex is #V=(1,4)#

The y-intercept is when #x=0#

#y=-2*1+4=2#

Therefore,

The maximum value is #V=(1,4)#

graph{(y+2x^2-4x-2)(y-1000(x-1))=0 [-11.25, 11.25, -5.625, 5.625]}