Question 1048459
{{{y=3x^2 + 15x + 9}}}

{{{y=3(x^2 + 5x) + 9}}}

{{{y=3(x^2 + 5x+b^2)-3b^2 + 9}}}

{{{y=3(x^2 + 5x+b^2)-3b^2 + 9}}}-> compare {{{(x^2 + 5x+b^2)}}} to {{{(a^2 + 2ab+b^2)}}}, you see that {{{a=1}}},{{{2ab=5}}}; so, find {{{b}}}

{{{2*1*b=5}}}->{{{b=5/2}}}

{{{y=3(x^2 + 5x+(5/2)^2)-3(5/2)^2 + 9}}}

{{{y=3(x+ 5/2)^2-3(25/4) + 9}}}

{{{y=3(x+ 5/2)^2-75/4 + 36/4}}}

{{{y=3(x+ 5/2)^2-39/4}}}

compare to {{{y=a(x- h)^2+k}}} where {{{h}}} and {{{k}}} are {{{x}}} and {{{y}}} coordinate of the vertex, and {{{a>0}}} which means function has minimum 

in your case {{{h=-5/2}}} and {{{k=-39/4}}} 

so, the vertex is at ({{{-5/2}}},{{{-39/4}}}) or ({{{-2.5}}},{{{-9.75}}}) and it is a minimum 

{{{ graph( 600, 600, -10, 10, -12, 10, 3(x+ 5/2)^2-39/4) }}}