Question 747568
The equation of the parabola given in general form does not make graphing into any easy process.  Completing the Square and converting to standard form does make graphing much easier to do.


The trinomial {{{x^2+4x-2}}} contains an expression which is like the area of a rectangle and a term can be added to turn this expression into a factorable square expression.


{{{y=x^2+4x-2}}}
{{{y=(x^2+4x)-2}}}, and we want to add a square term to the grouped expression, and this term will be {{{(4/2)^2=4}}}.  We also must subtract the same term.


{{{y=(x^2+4x+4)-2-4}}}
{{{highlight(y=(x+2)^2-6)}}}--------STANDARD FORM.
Compare to {{{y=(x-h)^2+k}}} for which the vertex is (h,k) and is a minimum.  If it were as {{{y=-(x-h)^2+k}}}, then the vertex would be a maximum.


Vertex is (-2,-6), and this vertex is the parabola's minimum.  You can find the zeros using the general solution to quadratic formula, or you could choose to solve for x starting from your standard form equation just found.