Question 999469
Equation or function?  As the equation mentioned and shown, it has ZEROS.


(3x --- 1)(x---  1)=0 seems not factorable.


{{{x=(1+- sqrt(1^2-4*3*(-1)))/(2*3)}}}


{{{highlight(x=(1+- sqrt(13))/6)}}}----------these are the zeros.



The FUNCTION  {{{f(x)=3x^2-x-1}}}, in general form, can be converted into standard form.  You can read and study to learn how to do that.  <a href="http://www.algebra.com/my/Completing-the-Square-to-Solve-General-Quadratic-Equation.lesson?content_action=show_dev">Lesson, how to complete the square to solve a quadratic equation</a>;  http://www.algebra.com/my/Completing-the-Square-to-Solve-General-Quadratic-Equation.lesson?content_action=show_dev


That will also give information about vertex and axis of symmetry.
Say if you have further trouble with this.



The standard form function through completing the square and arranging is {{{f(x)=3(x-1/6)^2-13/12}}}; and vertex is  ( 1/6, -13/12).  Symmetry axis is {{{x=1/6}}}.


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Not sure what happened to your posted equation, but it is missing from what you posted.  The original equation was  {{{3x^2-x-1=0}}}.  I am trying to adjust what you posted so the equation shows up there again, at least.