What are the coordinates of the minimum point for the square root function that relates to the quadratic function f(x)=12(x−11)2+4 ?(1 point) (,)
To find the minimum point of the quadratic function, we need to convert it to vertex form by completing the square:
f(x) = 12(x-11)^2 + 4
First, distribute the 12:
f(x) = 12(x^2 - 22x + 121) + 4
Next, factor out the 12:
f(x) = 12(x^2 - 22x + 121) + 4
Now, take half of the coefficient of x (-22) and square it (-11):
f(x) = 12(x^2 - 22x + 121) + 4
= 12(x - 11)^2 + 4
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 vertex is at (11, 4).
Therefore, the coordinates of the minimum point for the square root function are (11, 4).