9514 1404 393
Answer:
(-7, -11)
Step-by-step explanation:
[tex]\dfrac{x^2+3x-28}{x+7}=\dfrac{(x+7)(x-4)}{x+7}=x-4\qquad x\ne-7[/tex]
The hole is at x=-7. The left- and right-limits at that point are -7-4 = -11. So, the coordinates of the hole are (-7, -11).
