Find all the values for x for which the function is not continuous: f(x)=x^2-4/x^2-3x+2.

What type of discontinuity is each and why? Appreciate the help.Thanks in advance :).

Assuming this is the eq: f(x) = x^2 - (4/x^2) - 3x+2

x^2 is continuous for all x.

- (4/x^2) is continuous for all x EXCEPT 0. This is because you can't, technically, divide by 0
This discontinuity is a vertical asymptote.
As x goes to 0, y goes to negative infinity.

-3x+2 is continuous for all x.

Therefore the only discontinuity in that equation is the vertical asymptote at x=0
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Assuming this is the eq: f(x) = (x^2 -4)/(x^2- 3x+2)

x^2-4 is continuous for all x.

We must factor x^2-3x+2 to determine when it equals 0.

x^2-3x+2=(x-1)*(x-2)
(Can use the quadratic equation if you don't know another way)

Therefore there are discontinuities at x=1 and x=2. Either will cause there to be division by 0.

At x=1 there is a vertical asymptote. At x=2 there is a horizontal one.

Is it (x^2-4)/(x^2-3x+2).? or x^2-(4/x^2)-3x+2.? or x^2-(4/x^2-3x+2).? The answers may be different in these cases.