course Mth 158 This is awesome! I actually understand this!
......!!!!!!!!...................................
14:46:22 4.3.20. Find the domain of G(x)= (x-3) / (x^4+1).
......!!!!!!!!...................................
RESPONSE --> I think the domain of this one is all real numbers
.................................................
......!!!!!!!!...................................
14:46:39 The denominator is x^4 + 1, which could be zero only if x^4 = -1. However x^4 being an even power of x, it cannot be negative. So the denominator cannot be zero and there are no vertical asymptotes.
......!!!!!!!!...................................
RESPONSE --> okay
.................................................
......!!!!!!!!...................................
14:48:10 4.3.43. Find the vertical and horizontal asymptotes, if any, of H(x)= (x^4+2x^2+1) / (x^2-x+1).
......!!!!!!!!...................................
RESPONSE --> The denominator= (x-1)^2, so there is a vertical asymptote at x=1. Because the degree of the numerator is higher than that of the denominator, there are no horizontal asymptotes.
.................................................
......!!!!!!!!...................................
14:48:19 The function (x^4+2x^2+1) / (x^2-x+1) factors into (x^2 + 1)^2 / (x-1)^2. The denominator is zero if x = 1. The numerator is not zero when x = 1 so there is a vertical asymptote at x = 1. The degree of the numerator is greater than that of the denominator so the function has no horizontal asymptotes.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
14:50:27 4.3.50. Find the vertical and horizontal asymptotes, if any, of R(x)= (6x^2+x+12) / (3x^2-5x-2).
......!!!!!!!!...................................
RESPONSE --> The denominator can be factored into (3x+1)(x-2), so there are vertical asymptotes at x=-1/3 and x=2. The degree of the numerator= degree of denominator here so there is a horizontal asymptote at y=2
.................................................
......!!!!!!!!...................................
14:50:42 The expression R(x)= (6x^2+x+12) / (3x^2-5x-2) factors as (6·x^2 + x + 12)/((x - 2)·(3·x + 1)). The denominator is zero when x = 2 and when x = -1/3. The numerator is not zero at either of these x values so there are vertical asymptotes at x = 2 and x = -1/3. The degree of the numerator is the same as that of the denominator so the leading terms yield horizontal asymptote y = 6 x^2 / (3 x^2) = 2.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
"