This is a curved line.When x=-3,y=x^2-4.There are seven points on the line and the seventh point is x=3.So,draw the tangent at x=-2 and find its gradient.Thanks

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Given equation of the curve

`y=x^2-4` (i)

differentiate (i) with respect to x

we have

`(dy)/(dx)=2x`

`(dy)/(dx)}_(x=-2)=2xx(-2)=-4`

Thus gradient (slope)=-4

The equation of tangent at `(x_1,y_1)` and slope m is .

`y-y_1=m(x-x_1)`

when x=-2 then y=0

Thus equation of tangent at (-2,0) and slope -4 is

`y-0=-4(x+2)`

`y=-4x-8`

`4x+y+8=0`

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