Question

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Answer 1

The point which is in the solution set of
`y<x^2-2x-8` is (-2,-1)

Step-by-step explanation:

Given inequality is
`y<x^2-2x-8`

To find the point that lies in the solution set of
`y<x^2-2x-8`:

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

`y<x^2-2x-8`

`0<(4)^2-2(4)-8`

`0<16-8-8`

`0<0` which is not applicable

now we verify with the point (-2,-1) put x=-2 and y=-1 values in the solution set

`y<x^2-2x-8`

`(-1)<(-2)^2-2(-2)-8`

`-1<4+4-8`

`-1<0` it is applicable line
`0>-1`

Now verify with (0,-2)

`y<x^2-2x-8`

`-2<0^2-2(0)-8`

`-2<-8` which is not applicable

so the point (-2,-1) only satisfies the inequality.

Therefore the point(-2,-1) lies in the solution set
`y<x^2-2x-8`