In a quadratic model, the relationship between the feature and the target is no longer constant and differentiation is required to find the point where the target variable is maximized?

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Non-linear terms introduce curvature in the relationship between the feature and the target. In a quadratic model, the squared feature makes the slope vary with x, so you can’t rely on a constant rate of change. To find where the target is maximized, you differentiate with respect to the feature and set the derivative to zero. For y = a x^2 + b x + c, the derivative is dy/dx = 2 a x + b. Solving 2 a x + b = 0 gives the candidate x* = -b/(2 a). This point corresponds to the vertex of the parabola, and it yields a maximum when a is negative (a < 0). The concept here is tied to the presence of non-linear terms, which is why differentiation becomes the tool to locate the optimum. The other options don’t specifically describe the reason the maximum is found or the curvature introduced by the quadratic term.