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 find the function that has a vertex at (2, -9), we need to recall the vertex form of a quadratic function, which is:
f(x) = a(x – h)^2 + k
Where (h, k) represents the coordinates of the vertex
To find the function that has a vertex at (2, -9), we need to recall the vertex form of a quadratic function, which is:
f(x) = a(x – h)^2 + k
Where (h, k) represents the coordinates of the vertex. In this case, h = 2 and k = -9.
Now let’s examine each given function to see which one matches the vertex form:
1. f(x) = -(x – 3)^2
Comparing this with the vertex form, we notice that h = 3, not 2. Therefore, this function does not have a vertex at (2, -9).
2. f(x) = (x + 8)^2
Again, h here is -8, not 2. Hence, this function does not have a vertex at (2, -9).
3. f(x) = (x – 5)(x + 1)
This is not a vertex form of the quadratic function. It’s actually a factored form (or expanded form) of a quadratic equation. It represents a parabola but does not have a vertex at (2, -9).
4. f(x) = -(x – 1)(x – 5)
Comparing this with the vertex form, we can see that h = (1+5)/2 = 6/2 = 3, which does not match the given x-coordinate of the vertex, 2. Therefore, this function does not have a vertex at (2, -9) either.
Hence, none of the given functions match the vertex (2, -9). It’s possible that there was a typographical error or some missing information.
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