Answer:
[tex]\boxed{\bold{f(x)=\frac12(x-3)^2+15}}[/tex]
Step-by-step explanation:
[tex]f(x)=a(x-h)^2+k\qquad\qquad h=\dfrac{-b}{2a}\,,\ \ k=f(h)\\\\\\f(x)=\frac12x^2+3x+\frac32\quad\implies\quad a=\frac12\,,\ b=3\\\\h=\dfrac{-3}{2\cdor(-\frac12)}=\dfrac{-3}{-1}=3\\\\h=f(3)=\frac12(3)^2+3\cdot3+\frac32=4\frac12+9+\frac32=15\\\\\\ \underline{f(x)=\frac12(x-3)^2+15}[/tex]
