Question

Which function has a horizontal asymptote at y = 3?

Answer 1

Answer: A)
`f(x)=(6x^2-x+4)/(2x^2-1)`

Explanation:

Line
`y=L` is a horizontal asymptote of the function
`y=f(x)`, if either
`\lim_(x \to \infty) f(x)=L` or
`\lim_(x \to \infty) f(x)=L`, and
`L` is finite.

calculate the limits:

`\lim_(x \to \infty) ((6x^2-x+4)/(2x^2-1))=3`

`\lim_(x \to -\infty) ((6x^2-x+4)/(2x^2-1))=3`

Thus, the horizontal asymptote is
`y=3`