Given the equation,
[tex]y=3x^2+12x_{}+11[/tex]
We are to solve for the vertex first, in order to solve for the vertex.
[tex]3x^2+12x+11=y[/tex]
factor all through by 3
[tex]\begin{gathered} \frac{3x^2}{3}+\frac{12x}{3}+\frac{11}{3}=y \\ 3(x^2+4x+\frac{11}{3})=y\ldots\ldots.1 \end{gathered}[/tex][tex]x^2+4x=-\frac{11}{3}\text{ complete the square for the inner expression}[/tex][tex]\begin{gathered} x^2+4x+(\frac{4}{2})^2=-\frac{11}{3}+(\frac{4}{2})^2 \\ (x+2)^2=-\frac{11}{3}+4=\frac{1}{3} \\ =(x+2)^2-\frac{1}{3} \end{gathered}[/tex]
Put (x+2)²-1/3 into equation 1
[tex]3((x+2)^2-\frac{1}{3})=y\ldots\ldots2[/tex]
The vertex is at (-2,-1)
Note:
[tex]\begin{gathered} \text{vertex}=(h,k) \\ \text{focus}=(h,k+\frac{1}{4a}) \end{gathered}[/tex]
P is the distance between the focus and the vertex.
[tex]\begin{gathered} (h-h,k+\frac{1}{4a}-k)=(0,\frac{1}{4a}) \\ \end{gathered}[/tex]
where,
[tex]a=3\text{ from equation 2}[/tex]
Therefore,
[tex]\begin{gathered} p=(0,\frac{1}{4\times3})=(0,\frac{1}{12}) \\ p=(0,\frac{1}{12}) \end{gathered}[/tex]
Hence,
[tex]p=\frac{1}{12}[/tex]
The correct answer is 1/12 [option D].
