Answer:
Explanation:
We are given the quadratic function f(x) = -1x^2 - 2x - 1, which factors into:
f(x) = -1(x + 1)(x + 1) = -(x + 1)^2. Comparing this to the vertex equation of a quadratic, we get:
f(x) = -(x + 1)^2 + 0
f(x) = a(x - h)^2 + k
where (h, k) represents the vertex. We see readily that h = -1, k = 0 and a = -1. The vertex is at (-1, 0), and the graph of this quadratic/parabola opens down (because a is negative). The vertex lies on the x-axis 1 unit to the left of the y-axis.
Quickly sketch this parabola. It begins in Quadrant III and ends in Quadrant IV. From this sketch we can see that the function is increasing on the interval (-infinity, -1), at its maximum at (-1, 0) and decreasisng on the interval (-1, infinity).
Quadratic functions are defined for all x, so the domain of this one is
(-infinity, infinity).
As we have seen, the maximum value this function can have is zero (0). Therefore, the range (which represents all possible y values) is
(-infinity, 0).