Question 352169
Complete the square,
{{{f(x)=-3x^2+2x+5}}}
{{{f(x)=-3(x^2-(2/3)x)+5}}}
{{{f(x)=-3(x^2-(2/3)x+1/9)+5+3/9}}}
{{{f(x)=-3(x-1/3)^2+15/3+1/3}}}
{{{f(x)=-3(x-1/3)^2+16/3}}}
Now the equation is in vertex form,
({{{h}}},{{{k}}})=({{{1/3}}},{{{16/3}}})