Question 115442
The x-coordinate of the vertex (h, k)  of a parabola expressed in the form {{{y=a^2+bx+c}}} is given by {{{h=(-b)/2a}}}.


So, for {{{f(x)=x^2+6x-2}}}, {{{h=(-6)/2=-3}}}


The whole number values immediately to the left and right of this point are -4 and -2, so you need to evaluate f(-4), f(-3), and f(-2).


{{{f(-4)=(-4)^2+6(-4)-2=-10}}}
{{{f(-3)=(-3)^2+6(-3)-2=-11}}}
{{{f(-2)=(-2)^2+6(-2)-2=-10}}}


You can make your own table.


{{{graph(600,600,-5,1,-12,1,x^2+6x-2)}}}