The point outside the solution set are the false values of the inequality
The point outside the solution set of (y+4)^2/9-(x+1)^2/9>1 is (-1,-1)
How to determine the point outside the solution set
The inequality of the graph is given as:
[tex]\frac{(y+4)^2}{9}-\frac{(x+1)^2}{9} > 1[/tex]
Next, we test the options
This gives
[tex]\frac{(4+4)^2}{9}-\frac{(-2+1)^2}{9} > 1[/tex]
[tex]\frac{64}{9}-\frac{1}{9} > 1[/tex]
Evaluate the difference
[tex]7 > 1[/tex] -- this is true
This gives
[tex]\frac{(-12+4)^2}{9}-\frac{(2+1)^2}{9} > 1[/tex]
[tex]6.1 > 1[/tex] -- this is true
This gives
[tex]\frac{(-1+4)^2}{9}-\frac{(-1+1)^2}{9} > 1[/tex]
[tex]1 > 1[/tex] -- this is false
Hence, the point outside the solution set of (y+4)^2/9-(x+1)^2/9>1 is (-1,-1)
Read more about solution sets at:
https://brainly.com/question/13729904
