which function had a vertex at (2,6)?
well, if it's a quadratic, then it will be
y = a(x-2)^2 + 6
or some variant on that.
To determine which function has a vertex at (2,6), we need to consider quadratic functions in vertex form. The vertex form of a quadratic function is given by:
f(x) = a(x - h)^2 + k
Where (h, k) represents the coordinates of the vertex. In this case, the given vertex is (2,6), so h = 2 and k = 6.
Plugging these values into the vertex form equation, we have:
f(x) = a(x - 2)^2 + 6
Therefore, any quadratic function in the form f(x) = a(x - 2)^2 + 6 will have a vertex at (2,6). The value of 'a' determines the shape and opening direction of the parabola.
To find the function with a vertex at (2,6), we need to identify the form of the quadratic function that represents a parabola with a vertex. The general form of a quadratic function is:
f(x) = ax^2 + bx + c
In this case, since the vertex is given as (2,6), we can plug these values into the equation to solve for the unknowns:
6 = a(2)^2 + b(2) + c
Simplifying further:
6 = 4a + 2b + c
Due to the lack of additional information, we cannot determine the exact coefficients of the function. However, we can provide an equation in the general form that has a vertex at (2,6):
f(x) = a(x - 2)^2 + 6
The value 'a' can take any real number, which determines the shape and direction of the parabola.