Answer:
Since the quadratic function has a minimum value at some point between x = 2 and x = 6, its graph is a downward-facing parabola.
Let's assume that the function is of form f(x) = ax^2 + bx + c, where a, b, and c are constants.
Since the minimum value of the function is −2, we know that the vertex of the parabola lies on the line y = -2. Also, we know that the x-coordinate of the vertex is the average of 2 and 6, which is 4.
Therefore, the equation of the parabola can be written as f(x) = a(x-4)^2 - 2.
Since the y-intercept is the value of y when x = 0, we can find it by plugging in x = 0 into the equation of the parabola:
f(0) = a(0-4)^2 - 2
f(0) = 16a - 2
We know that the function yields negative values between x = 2 and x = 6, so the parabola must intersect the y-axis below the x-axis. This means that the y-intercept is negative.
To find the y-intercept, we need to solve the equation 16a - 2 = 0, which gives us a = 1/8.
Therefore, the equation of the parabola is f(x) = (1/8)(x-4)^2 - 2.
Finally, we can find the y-intercept by plugging in x = 0:
f(0) = (1/8)(0-4)^2 - 2
f(0) = 8 - 2
f(0) = 6
So the coordinates of the y-intercept are (0, 6).