How do find the vertex and axis of symmetry for a quadratic equation #y=4x^2-2x+2#?

1 Answer
Jun 7, 2015

The general form of a parabolic equation ins vertex form is:
#color(white)("XXXX")##y = m(x-a)^2+b# where the vertex is at #(a,b)#

#y=4x^2-2x+2#
can be converted into that form (most easily by a completion of the squares process)
#color(white)("XXXX")##y = 4(x^2- 1/2x) +2#

#color(white)("XXXX")##y= 4(x^2-1/2x+(1/4)^2) + 2 -4(1/4)^2#

#color(white)("XXXX")##y = 4(x-1/4)^2 + 7/4#

The vertex is at #(1/4, 7/4)#