Question 138228
Algebra solution:


{{{f(x)=ax^2+bx+c}}} is a parabola that opens upward when {{{a>0}}} or downward when {{{a<0}}} (and is no longer a parabola if {{{a=0}}}).  Your value for a is -1, so the parabola opens downward and the vertex is then a maximum.


The x-coordinate of the vertex is given by {{{(-b)/2a}}}, so for your function:
{{{(-(-6))/2(-1)=3}}}.  The value of the function at that point is {{{f(3)=-(3)^2-6(3)-3=-9-18-3=-30}}}.


Calculus solution:
A continuous function has a local extrema wherever the first derivitive is equal to zero.  It is a maximum if the 2nd derivitive is negative at that point and a minimum if the 2nd derivitive is positive at that point.


{{{f(x)=y=-x^2-6x-3}}} => {{{dy/dx=-2x-6}}}


Let {{{-2x-6=0}}} => local extrema at {{{x=3}}}, and the value of the function at that point is found the same way as in the algebra solution: {{{f(3)=-(3)^2-6(3)-3=-9-18-3=-30}}}.


{{{d^2y/dx^2=-2}}} => the 2nd derivitive is everywhere negative, so the local extreme point is a maximum.