Identify Vertical Asymptotes

t^+2/2t^2+4t-3

Since I can't factor the bottom half, what alternative do I take?
As posted (and assuming that the missing exponent is a two), the expression is this:

. . . . .\(\displaystyle t^2\, +\, \dfrac{2}{2t^2}\, +\, 4t\, -\, 3\)

I suspect, though, that you meant this:

. . . . .\(\displaystyle f(t)\, =\, \dfrac{t^2\, +\, 2}{2t^2\, +\, 4t\, -\, 3}\)

If so, then you're right to think about factoring, as this would be a simple way to find zeroes. But all you really need is to find the zeroes, easily or otherwise. So set the denominator equal to zero, and solve with the Quadratic Formula. ;)
 
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