What is the vertex of y=2x^2-12x+16 ?

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Answer 1 · 2018-04-25T01:54:48

#y = 2x^2 -12 x + 16 ##= 2(x^2 - 6x) + 16 ##= 2(x^2 - 6x + 9) - 2(9) + 16 ##= 2(x-3)^2 -2# and we read off the vertex #(3,-2)#.

Answer 2 · 2018-04-25T02:01:16

The vertex is #(3,-2)#

Given an equation of a parabola of the form:

#y = ax^2+bx+c#

The x-coordinate, #h#, of the vertex, is:

#h =-b/(2a)#

The y-coordinate, #k#, of the vertex, is:

#k = ah^2-bh+c#

From the given equation, #y=2x^2-12x+16#, we observe that, #a=2, b = -12 and c = 16#

Using the above formulas:

#h = -(-12)/(2(2))#

#h = 3#

#k = 2(3)^2-12(3)+16#

#k = -2#

The vertex is #(3,-2)#