Determining the Function with a Vertex at (2, -9)

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)

To determine which function has a vertex at (2, -9), we need to examine the vertex form of a quadratic function, which is given by f(x) = a(x – h)2 + k, where (h, k) represents the coordinates of the vertex

To determine which function has a vertex at (2, -9), we need to examine the vertex form of a quadratic function, which is given by f(x) = a(x – h)2 + k, where (h, k) represents the coordinates of the vertex.

Comparing the given functions with the vertex form, we can determine the values of (h, k) for each function:

1. f(x) = -(x – 3)2
Here, we have h = 3 and k = 0, since the – is distributed across the binomial. Therefore, the vertex is located at (h, k) = (3, 0), not at (2, -9).

2. f(x) = (x + 8)2
In this case, h = -8 and k = 0. Thus, the vertex is at (-8, 0), not (2, -9).

3. f(x) = (x – 5)(x + 1)
Since this function is in factored form, we cannot directly identify the vertex coordinates. We need to convert it into vertex form. Expanding the equation, we get f(x) = x² – 4x – 5. Completing the square, we obtain f(x) = (x – 2)² – 9, which matches the vertex form.
This means that the function f(x) = (x – 5)(x + 1) has a vertex at (2, -9).

4. f(x) = -(x – 1)(x – 5)
Like the previous function, we need to expand it and complete the square to find the vertex form. Upon expanding, we get f(x) = -x² + 6x – 5. By completing the square, we have f(x) = -(x – 3)² – 4, which is the vertex form representation.
From this, we can conclude that the function f(x) = -(x – 1)(x – 5) has a vertex at (3, -4).

Therefore, the function with a vertex at (2, -9) is f(x) = (x – 5)(x + 1).

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