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

Answer 1

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

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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Answer 2

Genesis Allen

Axis of symmetry formula for a quadratic function: x = -b / (2a) Vertex point formula for a quadratic function: (x, y) = (-b / (2a), f(-b / (2*a)))

Given the function: y = 3x^2 + 3 a = 3, b = 0, c = 3

Using the formulas: Axis of symmetry: x = -0 / (2*3) = 0 Vertex point: (0, f(0)) = (0, 3(0)^2 + 3) = (0, 3)

Therefore, the axis of symmetry is x = 0, and the vertex point is (0, 3).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.