Consider the following polynomial function. \[ f(x)=(x+2)(x-3)^{2} \] Step 3 of 3 : Find the zero(s) at which $f$ "flattens out". Express the zero(s) as ordered pair(s). Answer Select the number of zero(s) at which $f$ "flattens out". Selecting an option will display any text boxes needed to complete your answer. O none 1 1 2 3 4

Step 1 ：Consider the polynomial function \(f(x)=(x+2)(x-3)^{2}\).

Step 2 ：The zeros at which the function 'flattens out' are the zeros of the derivative of the function. The derivative of a function gives us the slope of the function at any given point. When the function 'flattens out', the slope is zero. Therefore, we need to find the derivative of the function and set it equal to zero to find the x-values at which the function 'flattens out'.

Step 3 ：The zeros of the derivative are -1/3 and 3. These are the x-values at which the function 'flattens out'. However, the question asks for the zeros as ordered pairs, which means we need to find the corresponding y-values for these x-values. We can do this by substituting the x-values back into the original function.

Step 4 ：The y-values corresponding to the x-values -1/3 and 3 are 500/27 and 0 respectively.

Step 5 ：Therefore, the zeros at which the function 'flattens out' are \(\boxed{(-\frac{1}{3}, \frac{500}{27})}\) and \(\boxed{(3, 0)}\).