Which function has a vertex at (2, -9)?f(x) = -(x – 3)2f(x) = (x + 8)2f(x) = (x – 5)(x + 1)f(x) = -(x – 1)(x – 5)
C f(x) = (x – 5)(x + 1)
The given quadratic functions have the following forms:
f(x) = -(x – 3)^2 => Vertex form: f(x) = -a(x – h)^2 + k, where (h,k) is the vertex
Therefore, a = -1, h = 3, and k = 0.
Since k = 0 and the negative sign in front of the function ensures that the vertex is a maximum point, therefore, the vertex is (3,0).
Thus, f(x) = -(x-3)^2 does not have a vertex at (2,-9).
f(x) = (x + 8)^2 => Vertex form: f(x) = a(x – h)^2 + k, where (h,k) is the vertex
Therefore, a = 1, h = -8, and k = 0.
Since k = 0 and the positive sign in front of the function ensures that the vertex is a minimum point, therefore, the vertex is (-8,0).
Thus, f(x) = (x+8)^2 does not have a vertex at (2,-9).
f(x) = (x – 5)(x + 1) => Standard form: f(x) = ax^2 + bx + c, where the vertex is at (-b/2a, f(-b/2a))
Therefore, a = 1, b = -4, and c = -5.
The x-coordinate of the vertex is -b/2a = -(-4)/(2*1) = 2.
To find the y-coordinate, we substitute x = 2 into the function: f(2) = (2 – 5)(2 + 1) = -9.
Thus, f(x) = (x-5)(x+1) has a vertex at (2,-9).
f(x) = -(x – 1)(x – 5) => Standard form: f(x) = ax^2 + bx + c, where the vertex is at (-b/2a, f(-b/2a))
Therefore, a = -1, b = 6, and c = -5.
The x-coordinate of the vertex is -b/2a = -6/(-2) = 3.
To find the y-coordinate, we substitute x = 3 into the function: f(3) = -(3-1)(3-5) = -4.
Thus, f(x) = -(x-1)(x-5) has a vertex at (3,-4).
Therefore, the function f(x) = -(x-1)(x-5) has a vertex at (2,-9).
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