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

##### 2 Answers

Jun 7, 2015

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:

y of vertex, or Max point : f(-2) =-4 + 8 - 3 = 1 graph{-x^2 - 4x - 3 [-10, 10, -5, 5]}

Jun 7, 2015