As shown in the figure, the minimum area of △ AOB can be obtained when the image of a function of degree passes through the point P (2,3), the positive half axis and a of the intersection x-axis, and the positive half axis and B of the intersection y-axis

As shown in the figure, the minimum area of △ AOB can be obtained when the image of a function of degree passes through the point P (2,3), the positive half axis and a of the intersection x-axis, and the positive half axis and B of the intersection y-axis

Let y = KX + B, then 3 = 2K + B, then B = 3-2k, let y = 0, then x = - BK, then OA = - Bk. let x = 0, then y = B, then ob = B. s △ AOB = 12 × (- BK) × B = 12 × (3 − 2K) 2 − k = 12 × 4k2 − 12K + 9 − k = 12 [(2 − K-3 − K) 2 + 24] ≥ 12