Question

How do you determine the min and max of `-x^2 - 4x - 3`?

Answer 1

f(x) = -(x^2 + 4x + 3)
There is max when the derivative f'(x) = 2x + 4 = 0 -> x = -2
Or by another way:
Since a = -1 is negative, the parabola opens downward, there is a max at vertex.
x of vertex, or max point:`x = (-b/2a) = -4/2 = -2.`
y of vertex, or Max point : f(-2) =-4 + 8 - 3 = 1 graph{-x^2 - 4x - 3 [-10, 10, -5, 5]}

Answer 2

The derivative of the given function is y'= -2x-4. For a maxima or minima y' =0. Hence at x=-2, there would be a critical point. Hence at x=-2, y would be =1. To determine which one it is apply second derivative test. y''= -2, which is a negative number. Hence at x= -2 it is a maxima y=-1. The minima of the function is undefined (-`oo`).