Answer:
The point which is in the solution set of [tex]y<x^2-2x-8[/tex] is (-2,-1)
Step-by-step explanation:
Given inequality is [tex]y<x^2-2x-8[/tex]
To find the point that lies in the solution set of [tex]y<x^2-2x-8[/tex]:
Given points (4,0), (-2,-1), (0,-2)
Now verify that point (4,0)
Whether lies in the solution set or not
Let (x,y) be the point (4,0)
ie, put x=4 and y=0 in the solution set
[tex]y<x^2-2x-8[/tex]
[tex]0<(4)^2-2(4)-8[/tex]
[tex]0<16-8-8[/tex]
[tex]0<0[/tex] which is not applicable
now we verify with the point (-2,-1) put x=-2 and y=-1 values in the solution set
[tex]y<x^2-2x-8[/tex]
[tex](-1)<(-2)^2-2(-2)-8[/tex]
[tex]-1<4+4-8[/tex]
[tex]-1<0[/tex] it is applicable line [tex]0>-1[/tex]
Now verify with (0,-2)
[tex]y<x^2-2x-8[/tex]
[tex]-2<0^2-2(0)-8[/tex]
[tex]-2<-8[/tex] which is not applicable
so the point (-2,-1) only satisfies the inequality.
Therefore the point(-2,-1) lies in the solution set [tex]y<x^2-2x-8[/tex]
