Find any asymptotes and holes in the graph of the rational function. Verify your answers by using a graphing utility. (Enter answers as a comma-separated list. If an answer does not exist, enter DNE.)

Question

f(x) = x²+2x+1 / 8x²-x-9

vertical asymptote(s): x=

horizontal asymptote(s): y=

hole(s): x=

Answer 1

Vertical asymptotes occur where then denominator equals zero so,solve

8x^2-x-9=0 Which factors to (8x-9)(x+1)=0 So x=-1 and x= 9/8 are where your vertical asymptotes occur unless there is a hole at one of these points. To determine this need to factor the numerator as well which gives us (x+1)(x+1)/ (8x-9)(x+1)

You have an (x+1) in both the numerator and denominator so there is a hole at x=-1 and a vertical asymptote at x=9/8

To determine the horizontally asymptote you need to compare the terms with the highest power in the numerator and denominator. In this case the highest power in both is equal( highest power is second power)

Take the ratio of them x^/8x^= 1/8 so horizontal asymptote is y= 1/8

Answer 2

f(x) = (x² + 2x + 1)/(8x² - x - 9) + (x+1)²/((8x-9)(x+1))

vertical asymptote: x = 9/8 (8x - 9 = 0); not x+1 = 0, because that term is in the numerator.

horizontal asymptote: 1/8 (the ratio of the leading coefficients)

hole: x = -1 (x + 1 = 0; because it is quadratic in the numerator and linear in the denominator, to remove the hole, we would have f(-1) = 0)