\(\text{It should be pretty clear that the singularity occurs when }x=-7\\ \text{That the problem identifies it as a hole means that }x^2+3x-28 \\ \text{Should be divisible by }x+7\\\)
\(x^2+3x-28 = (x+7)(x-4) \text{ and thus}\\ \dfrac{x^2+3x-28}{x+7} = x-4,~x\neq -7\\ \text{thus the hole appears at}\\ (-7,~-7-4)=(-7,-11)\)
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