Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 21

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: