The given function is a quadratic function and it is in the form of y = ax^2 + bx + c, where a = -1, b = -6, and c = -9.
The vertex form of a quadratic function is given as y = a(x-h)^2 + k, where (h, k) represents the coordinates of the vertex.
To find the vertex of the given function, we can use the equation h = -b/2a and substitute the values of a and b.
h = -(-6) / 2(-1)
h = 6 / 2
h = 3
Substituting h = 3 into the function, we can find k.
y = -(-3+3)^2
y = -0^2
y = 0
Therefore, the vertex of the function is (3, 0).
The function is a downward-opening parabola because the coefficient of x^2 is negative, which means a < 0. The vertex is the maximum point of the parabola.
The x-intercept can be found by setting y = 0 and solving for x.
0 = -(x+3)^2
0 = x+3
x = -3
Therefore, the x-intercept is -3.
The y-intercept can be found by setting x = 0 and solving for y.
y = -(0+3)^2
y = -3^2
y = -9
Therefore, the y-intercept is -9.
The graph of the function is a downward-opening parabola with the vertex at (3, 0), x-intercept at -3, and y-intercept at -9.
Consider the function y=-\left(x+3\right)^2
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