For f(x) = (x^2+x+1)/x, the oblique (slant) asymptote as x approaches infinity is:

For f(x) = (x^2+x+1)/x, the oblique (slant) asymptote as x approaches infinity is:

• A. y = x
• B. y = 0
• C. y = x - 1
• D. y = x + 1

Answer: (D)

Oblique (slant) asymptotes show the line a function follows for very large x, found by dividing the numerator by the denominator and taking the linear part of the quotient. For f(x) = (x^2 + x + 1)/x, divide to get x + 1 with a remainder of 1, so f(x) = x + 1 + 1/x. As x grows, 1/x → 0, and the function behaves like the line y = x + 1. Therefore the oblique asymptote is y = x + 1. The other options don’t fit because they don’t reflect the +1 intercept produced by the division; y = x would miss the intercept, and a horizontal line or y = x − 1 doesn’t match the linear part revealed by the quotient.

Oblique (slant) asymptotes show the line a function follows for very large x, found by dividing the numerator by the denominator and taking the linear part of the quotient. For f(x) = (x^2 + x + 1)/x, divide to get x + 1 with a remainder of 1, so f(x) = x + 1 + 1/x. As x grows, 1/x → 0, and the function behaves like the line y = x + 1. Therefore the oblique asymptote is y = x + 1. The other options don’t fit because they don’t reflect the +1 intercept produced by the division; y = x would miss the intercept, and a horizontal line or y = x − 1 doesn’t match the linear part revealed by the quotient.