Determining the Function with a Vertex at (2, -9) | Math Content and Analysis

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 general form of a quadratic function given by f(x) = a(x – h)^2 + k, where (h, k) represents the vertex of the parabola

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

1) f(x) = -(x – 3)^2:
To find the vertex, we compare this equation to the general equation, f(x) = a(x – h)^2 + k, and identify that h = 3 and k = 0. Thus, the vertex is (3, 0), not (2, -9).

2) f(x) = (x + 8)^2:
Here, h = -8 and k = 0. Therefore, the vertex is (-8, 0), so it does not match the given vertex of (2, -9).

3) f(x) = (x – 5)(x + 1):
This equation represents a quadratic function in factored form. To determine the vertex, we need to convert the equation to vertex form. Expanding the equation, we get f(x) = x^2 – 4x – 5.

To find the vertex, we use the formula: h = -b/2a, where a = 1 and b = -4. Substituting the values, we find h = -(-4)/(2*1) = 2. Then, to find k, we substitute h into the equation: f(2) = (2)^2 – 4(2) – 5 = 4 – 8 – 5 = -9.

Thus, the vertex for this function is (2, -9), which matches the given vertex.

4) f(x) = -(x – 1)(x – 5):
In this case, we have the quadratic function already written in factored form. To determine the vertex, we can use the same method as before. Expanding the equation, we get f(x) = -x^2 + 6x – 5.

Using the formula h = -b/2a, where a = -1 and b = 6, we find h = -6/(2*(-1)) = -3. Now, substituting h into the equation, we find k: f(-3) = -(-3)^2 + 6(-3) – 5 = -9 – 18 – 5 = -32.

Therefore, the vertex for this function is (-3, -32), which does not match the given vertex.

In conclusion, among the given functions, f(x) = (x – 5)(x + 1) has a vertex at (2, -9).

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