How do you find the vertex of #f(x)= 2x^2-4x+6#?

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Answer 1 · 2017-03-21T11:36:56

Vertex#->(x,y)=(1,4)#

Using part of the process of completing the square to determine the value of #x#. Then determining the value of #y# by substitution.

Given:#" "y=2x^2-4x+6#

Write as:#" "y=2(x^2color(red)(-4/2)x)+6#

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

By substitution:

#y_("vertex")=2(1)^2-4(1)+6 = 4#

Vertex#->(x,y)=(1,4)#

Compare to the standardised form of #y=ax^2+bx+c#

Notice that in the graph that #y_("intercept") = c=6#

Tony B