Vertical Asymptote Took a test and got an answer of x = 5/3 and it was marked wrong

Determine the vertical asymptote(s) of f(x) = [(3x + 5) / (9x^(2) - 25) ]. Give the exact answer (not a decimal). For each asymptote, use the answer format: x = #

So what I did was factor the denominator into (3x + 5) * (3x - 5)
f(x) = [ (3x + 5)] / [(3x + 5) * (3x - 5)]

Cross out the (3x + 5) terms.

f(x) = [ (1) ] / [(3x - 5)]

Set the denominator equal to zero and solve for x. 3x - 5 = 0

3x = 5

x = 5/3

Therefore the answer for the vertical asymptote is at x = 5/3

Then why was did my teacher mark it wrong .... Supposedly the answer is x = - 5/3 ??? If so why???

The function you show does have an asymptote at x=5/3. Perhaps the function on the test had 3x-5 in the numerator? (Or the instructor intended for it to have 3x-5 in the numerator?) That function would have the asymtote at x=-5/3.

I believe that is correct. You should go ask what you did wrong because i see nothing wrong with it.