Question 164879
For any quadratic {{{y=ax^2+bx+c}}}, the x-coordinate of the vertex can be found by the formula



{{{x=(-b)/(2a)}}}



From {{{f(x)=x^2+6x-2}}} we can see that {{{a=1}}}, {{{b=6}}} and {{{c=-2}}}



{{{x=(-6)/(2(1))}}} Plug in {{{a=1}}} and {{{b=6}}}



{{{x=(-6)/(2)}}} Multiply



{{{x=-3}}} Reduce



So the x-coordinate of the vertex is {{{x=-3}}}