the axis of symmetry of y=3x^2-6x+4

Question

the_fizicx99 · Answer

You can start by finding the vertex, the "x" coordinate in (x, y) of the vertex is the axis of symmetry. It's where the parabola can be cut in half.

Vertex: $\ \sf x = \dfrac{-b}{2a}$

Whatever you get using the vertex formula, substitute that into f(x) = 3x^2 - 6x + 4, simplify that. That'll be the "y" coordinate in (x, y).

anonymous · Answer

what is the axis symmetry of y=3x^2-6x+4

the_fizicx99 · Answer

I just answered it, explaining what to do to find the axis of symmetry. Is there something you're not understanding?

mathmale · Answer

Your function has the form of a quadratic equation:

y=3x^2-6x+4   (your function)

y=ax^2+bx+c.  (standard form of a quadratic equation)

First, look at the x^2 term.  What's your coefficient?  3.   This matches up with the coefficient 'a' of the x^2 term in the standard form.

Next, determine b.  Match up the coefficients of the x terms.

Last, determine c.  

To find the x-coordinate of the vertex, compute$$x=-b/(2a) =\frac{ -b }{ 2a }$$