Question 308022
You can either differentiate and find when the derivative is zero. 
f(x)={{{3x^2+12x+8 }}}
f'(x)={{{6x+12=0 }}}
{{{x=-2 }}}
{{{f(-2)=3(4)+12(-2)+8 }}}
{{{f(-2)=12-24+8=-4 }}}
Use the second derivative to find whether its min or max.
f''(x)={{{6>0 }}}
Since the second derivative is positive, the value is a minimum.
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Or algebraically complete the square to find the vertex form,
{{{f(x)=3x^2+12x+8 }}}
{{{f(x)=3(x^2+4x)+8 }}}
{{{f(x)=3(x^2+4x+4)+8-12 }}}
 {{{ f(x)=3(x+2)^2-4 }}}

Since the {{{x^2}}} coefficient is positive (3), the parabola opens upwards and the value at the vertex is a minimum and occurs at (-2,-4). 
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b) is the correct answer.