Let's calculate the first and second derivatives

#f(x)=x^3+6x^2+12x-1#

#f'(x)=3x^2+12x+12#

#=3(x^2+4x+4)#

#=3(x+2)(x+2)#

#=3(x+2)^2#

and

#f''(x)=6x+12#

The critical point is when #f'(x)=0#

That is,

#x+2=0#

#x=-2#

Let's build a chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-2##color(white)(aaaaaa)##+oo#

#color(white)(aaaa)##f'(x)##color(white)(aaaaaa)##+##color(white)(aaaaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##↗##color(white)(aaaaaa)##↗#

Let's look at #f''(x)#

#f''(x)=0# when #x+2=0#

That is when #x=-2#

There is a point of inflection at #(-2,-9)#

Let's do a chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaa)##-2##color(white)(aaaaaa)##+oo#

#color(white)(aaaa)##f''(x)##color(white)(aaaaaa)##-##color(white)(aaaaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##nn##color(white)(aaaaaa)##uu# graph{x^3+6x^2+12x-1 [-38.27, 26.68, -22.07, 10.37]}