The graph of the equation is a parabola with vertex #(3,-4)#.
When a parabola is in vertex form, #y=a(x-h)^2 +k#, the vertex (maximum or minimum) is given by the point #(h,k)#. Because #a<0#, the parabola opens downward, so it must have a maximum. It does not have a minimum point because the parabola extends downward forever to #-∞#. In any parabola there is either a minimum or a maximum, so your question should be rephrased: "How do you find the minimum or maximum value of #y=-(x-3)^2 -4#." graph{y=-(x-3)^2 -4 [-262.6, 262.6, -131.3, 131.3]}
One way to find the maximum or vertex is graphically. As depicted in the graph above, the maximum value of #f# occurs when #x=3# and #y=-4#. The vertex or maximum is (#3,-4#)
Another way to find the vertex (#h,k#) is by putting the equation in vertex form: #y=a(x-h)^2 +k#. Your equation is already in this form, so just solve for #h# and #k#. Solve for #h# by setting #x-3=0# and solving for #x=h=3#. Then, find #k = -4#. The vertex or maximum is (#3,-4#)
If your quadratic function was in standard form (#ax^2+bx+c=0#), the vertex is given by the point (#-b/(2a), f(-b/(2a))#). For example, if you were asked to find the maximum or vertex of a quadratic function with equation #y=-x^2+6x-13#, the x-value would be #-b/(2a) = -6/(2*-1) = 3#). The y-value would be #f(3) = -(3)^2 +6(3)-13 = -4#). The vertex or maximum is (#3,-4#)
