Question 327053
Complete the square to put the equation in vertex form, {{{y=a(x-h)^2+k}}}, the min or max value of the function occurs at the vertex.
{{{f(x)=-x^2-6x-8}}}
{{{f(x)=-(x^2+6x)-8}}}
{{{f(x)=-(x^2+6x+9)-8+9}}}
{{{f(x)=-(x+3)^2+1}}}
Since the coefficient of the {{{x^2}}} term is negative, the parabola opens downward and the value at the vertex is a maximum.
The vertex (h,k) is (-3,1).
{{{ymax=1}}}
.
.
.
{{{drawing(300,300,-8,2,-10,10,grid(1),circle(-3,1,.2),graph(300,300,-8,2,-10,10,-x^2-6x-8))}}}