107學測數學選填第17題詳解

$y = x^2 + 2x - 3 = (x+1)^2 - 4$，故頂點 $C = (-1, -4)$。 令 $y = 0 \implies x^2 + 2x - 3 = 0 \implies (x+3)(x-1) = 0 \implies x = -3, 1$，故 $A = (-3, 0), B = (1, 0)$。 取向量 $\overset{\large\rightharpoonup}{CA} = (-3 - (-1), 0 - (-4)) = (-2, 4)$，$\overset{\large\rightharpoonup}{CB} = (1 - (-1), 0 - (-4)) = (2, 4)$。 $$\cos \angle ACB = \dfrac{\overset{\large\rightharpoonup}{CA} \cdot \overset{\large\rightharpoonup}{CB}}{|\overset{\large\rightharpoonup}{CA}| |\overset{\large\rightharpoonup}{CB}|} = \dfrac{(-2)(2) + (4)(4)}{\sqrt{20} \sqrt{20}} = \dfrac{12}{20} = \dfrac{3}{5}$$ 故填 $\dfrac{3}{5}$。