How do you find the axis of symmetry and vertex point of the function: #y=3x^2+3 #?

1 Answer
Alan P.
Oct 8, 2015

Axis of symmetry: #x=0#
Vertex: #(0,3)#

Explanation:

Note that any parabolic equation of the general form:
#color(white)("XXX")y=ax^2+bx+c#
has a vertical axis of symmetry.

#y=3x^2+3color(white)("XXX")iffcolor(white)("XXX")y=3x^2+0x+3#
is in this general form.

Further, it can be re-written as
#color(white)("XXX")y=3(x-color(red)(0))^2+color(blue)(3)#
which is the vertex form with a vertex at #(color(red)(0),color(blue)(3))#

Its (vertical) axis of symmetry (i.e. #x=c# for some constant #c#) must pass through the vertex
therefore, the axis of symetry is
#color(white)("XXX")x=color(red)(0)#

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