已知函数f(x)=ax2+(b-8)x-a-ab,当x∈(-3,2)时,f(x)>0,当x∈(-∞,-3)∪(2,+∞)时,f(x)
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![已知函数f(x)=ax2+(b-8)x-a-ab,当x∈(-3,2)时,f(x)>0,当x∈(-∞,-3)∪(2,+∞)时,f(x)](/uploads/image/z/8710321-49-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dax2%2B%28b-8%29x-a-ab%2C%E5%BD%93x%E2%88%88%28-3%2C2%29%E6%97%B6%2Cf%28x%29%3E0%2C%E5%BD%93x%E2%88%88%28-%E2%88%9E%2C-3%29%E2%88%AA%282%2C%2B%E2%88%9E%29%E6%97%B6%2Cf%28x%29)
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