Determine the values of x for which y is undefined, those would be when the denominator is 0.

`x^2-4=0`

Factor the expression to find the roots.

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

`x=-2` and

`x=2`

The asymptotes are x=-2 and x=2, thus the domain of the function is x<-2, -2<x<2 and x>2

Determine the...

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Determine the values of x for which y is undefined, those would be when the denominator is 0.

`x^2-4=0`

Factor the expression to find the roots.

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

`x=-2` and

`x=2`

The asymptotes are x=-2 and x=2, thus the domain of the function is x<-2, -2<x<2 and x>2

Determine the range: y`!=` 0, but y approaches 0 when x`->oo` or

x`->-oo` , observe that the graph is symmetric with respect to the x=0 axis because the asymptotes are symmetric.

Determine the value of y when x=0.

`y=1/(0-4)=-0.25`

Determine the value of y when x is close to 2, x=1.5 and x=-1.5

`y=1/(1.5^2-4)`

`y=1/(2.25-4)=-1/1.75=-0.57`

Thus y=-0.57 when x=1.5 and x=-1.5. Since the value of y at x=0 is higher (-0.25>-0-0.57), the value of y will become more negative as it approaches the asymptotes.

Determine the value of y when x is close to 2 on the other side of the asymptote. Compute y for x=2.999

`y=1/(2.1^2-4)=1/(4.41-4)=1/0.41=2.439`

The value of y is positive and cannot be 0, then the curve has asymptotes x=2 and y=0, same for x=-2 and y=0 on the other side.

Verify using graphing: