Why does this rational function have a false slant/oblique asymptote?

Question

Let's examine the following rational function: $f(x) = \frac{3x^3+2}{x^2-x-7}$.

Considering that the degree of the polynomial in the numerator is 1 greater than that of the denominator, it can be assumed that the function possesses no horizontal asymptote, but possess a slant, or oblique, asymptote.

As a result of long division, the slant asymptote appears to be $y = 3x + 3$. However, on a graph, the function $y = 3x + 3$ intersects with the original function, $f(x) = \frac{3x^3+2}{x^2-x-7}$, at the coordinate $(-.9583333..., .125)$.

Why is this the case? What condition prevents $y = 3x + 3$ from being a true asymptote of the function $f(x) = \frac{3x^3+2}{x^2-x-7}$ if long division produces a result declaring otherwise?

Answer 1

Notice that intersection of the function with the asymptote does not prevent it from being an asymptote. Broadly speaking, "asymptote" for $x$ to infinity, for example, means that the farest we go with $x$, the closest the function goes to the line, but it may intersect it an infinite number of times.

For example, consider $\frac{x^2+\sin(x)}{x}$ and look what happens.

Answer 2

As you can see from the plot above, there are actually two vertical asymptotes at the roots of the denominator (approximately $x = 3.19, -2.19$) and the portion of the graph between those crosses every non-vertical line. Outside the roots of the denominator, the graph really does approach the asymptote as expected.