To find the vertical asymptotes of a rational function, I need to factor the denominator and set it equal to zero. I notice that the denominator is a cubic polynomial, which can be hard to factor. However, I also notice that the constant term is -1, which means that 1 is a possible root of the polynomial by the rational root theorem. So I will try to divide the denominator by x-1 using synthetic division and see if it works. I write the coefficients of the polynomial in a row: 1 -3 3 -1. Then I write 1 below the first coefficient and bring it down. I multiply 1 by 1 and write it below the second coefficient. I add -3 and 1 and write -2 below. I multiply 1 by -2 and write it below the third coefficient. I add 3 and -2 and write 1 below. I multiply 1 by 1 and write it below the last coefficient. I add -1 and 1 and write 0 below. Since the remainder is 0, x-1 is a factor of the denominator. The quotient is 1 -2 1, which means that the other factor is x^2-2x+1. I recognize that this is a perfect square, and it factors as (x-1)^2. So the denominator can be written as (x-1)^3. To find the vertical asymptotes, I set this equal to zero and solve for x. The only solution is x = 1, since it has multiplicity 3. Therefore, there is only one vertical asymptote, and it is x = 1. 
How many vertical asymptotes does the graph of $y = \frac{1}{x^3-3x^2+3x-1}$ have?