#1**+2 **

\(\text{It should be pretty clear that the singularity occurs when }x=-7\\ \text{That the problem identifies it as a hole means that }x^2+3x-28 \\ \text{Should be divisible by }x+7\\\)

\(x^2+3x-28 = (x+7)(x-4) \text{ and thus}\\ \dfrac{x^2+3x-28}{x+7} = x-4,~x\neq -7\\ \text{thus the hole appears at}\\ (-7,~-7-4)=(-7,-11)\)

.Rom Jan 26, 2019