What is the vertex of #y= 2x^2 -4x - 12#?

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Answer 1 · 2017-03-18T13:00:57

Vertex#" "->" "(x,y)=(1,-14)#

I am going to use part of the process of completing the square.

Write as:#" "y=2(x^2-4/2x)-12#

#x_("vertex")=(-1/2)xx(-4/2) = + 1#

So by substitution:

#y_("vertex")=2(1)^2-4(1)-12 = -14#

Vertex#" "->" "(x,y)=(1,-14)#

Tony B

Answer 2 · 2017-03-18T13:06:52

The vertex is #=(1,-14)#

We need

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

Let complete the squares and factorise

#y=2x^2-4x-12#

#y=2(x^2-2x)-12#

#y=2(x^2-2x+1)-12-2#

#y=2(x-1)^2-14#

Therefore,

the vertex is #=(1,-14)#

graph{(y-(2x^2-4x-12))((x-1)^2+(x+14)^2-0.01)=0 [-7.8, 6.25, -14.32, -7.295]}