The function has no vertical, horizontal or slant asymptotes.
You need to determine the extrema of the function, hence, you need to determine the zeroes of the first derivative:
`f'(x) = 4x^3-9x^2+6x-1 => f'(x) = 0 => (4x-1)(x-1)^2 = 0 => 4x -1 = 0 => x = 1/4`
`(x-1)^2 = 0 => x_1 = x_2 = 1`
Hence, the function has 2 extreme points, at `x = 1/4` and `x = 1.` Notice that `f'(x)<0` for `x in` `(-oo,1/4)` and `f'(x)>0` for `x in` `(1/4,oo)` , hence, the function decreases till `x = 1/4` and then it increases. Hence, the function has a minimum point at `x =1/4` .
You need to determine the inflection points, hence, you need to find the zeroes of the second derivative.
`f''(x) = 12x^2 - 18x + 6 => f''(x) = 0 => 12x^2 - 18x + 6 = 0`
You need to divide by 6:
`2x^2 - 3x + 1 = 0 => x_(1,2) = (3+-sqrt(9-8))/4 => x_(1,2) = (3+-1)/4`
`x_1 = 1, x_2 = 1/2`
Hence, since the second derivative is negative on `(1/2,1) ` and positive on `(-oo,1/2) U (1,oo)` , then the function is concave down on `(1/2,1)` and concave up on `(-oo,1/2) U (1,oo).`
The graph of the function is represented below: