What is the axis of symmetry and vertex for the graph #y=4x^2-2x+2#?

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Answer 1 · 2017-11-26T05:15:21

Axis of symmetry: #x=1/4#

Vertex is at #(1/4, 1 3/4)#

The equation of a parabola is #y = ax^2 +bx+c#

#y =4x^2 - 2x+2# is a the equation of a parabola

To find the axis of symmetry use: #x = (-b)/(2a)#

#x = (-(-2))/(2(4)) = 2/8 = 1/4#

Therefore, the #x#-co-ordinate of the vertex is #1/4#.

Substitute #1/4# into the equation to find the #y#-value.

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

#y = 4xx1/16 -2/4+2#

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

#y = 1 3/4#

Vertex is #(1/4, 1 3/4)#