Question 922855
Convert to vertex form by completing the square.
{{{f(x)=-x^2+3x-5}}}
{{{f(x)=-(x^2-3x)-5}}}
{{{f(x)=-(x^2-3x+9/4)-5+9/4}}}
{{{f(x)=-(x-3/2)^2-20/4+9/4}}}
{{{f(x)=-(x-3/2)^2-11/4}}}
The vertex is (3/2,-11/4).
The parabola opens downwards so the maximum value of the function is {{{-11/4}}}.
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{{{graph(300,300,-2,8,-8,2,-x^2+3x-5)}}}