The equation was given in its standard form, as follows:
[tex]y=2x^2+6x+11[/tex]To rewrite it in the vertex form, we need to complete the square to write it in the following form:
[tex]y=a(x-h)^2+k[/tex]If we rewrite the standard form using the, we get:
[tex]\begin{gathered} y=2(x^2+3x+\frac{11}{2})=2(x^2+2\times x\times\frac{3}{2}+(\frac{3}{2}_{})^2-(\frac{3}{2})^2+\frac{11}{2}) \\ y=2((x^2+2\times x\times\frac{3}{2}+\frac{9}{3})-\frac{9}{4}+\frac{11}{2})=2((x+\frac{3}{2})^2-\frac{9}{4}+\frac{22}{4})_{} \\ y=2((x+\frac{3}{2})^2+\frac{13}{4})=2(x+\frac{3}{2})^2+\frac{13}{2} \\ \\ y=2(x+\frac{3}{2})^2+\frac{13}{2} \end{gathered}[/tex]