To find and graph all asymptotes of the rational function

$f(x) = \frac{6x^2 + 7x - 9}{2x + 1},$

we need to determine both vertical and horizontal (or slant) asymptotes.

1. Vertical Asymptote

Vertical asymptotes occur where the denominator is zero, as long as these points do not cancel with terms in the numerator.

The denominator is $2x + 1$. Setting it equal to zero: $2x + 1 = 0 \Rightarrow x = -\frac{1}{2}.$ Therefore, there is a vertical asymptote at $x = -\frac{1}{2}$.

2. Slant Asymptote (Oblique Asymptote)

Since the degree of the numerator (2) is exactly one degree higher than the degree of the denominator (1), there will be a slant asymptote. We find it by performing polynomial long division on $6x^2 + 7x - 9$ divided by $2x + 1$.

Step-by-step division:

1. Divide the leading term $6x^2$ by $2x$, which gives $3x$.
2. Multiply $3x$ by $2x + 1$, resulting in $6x^2 + 3x$.
3. Subtract $6x^2 + 3x$ from $6x^2 + 7x - 9$ to get $4x - 9$.
4. Divide $4x$ by $2x$ to get $+2$.
5. Multiply $2$ by $2x + 1$, resulting in $4x + 2$.
6. Subtract $4x + 2$ from $4x - 9$ to get $-11$.

Thus, the division gives: $f(x) = 3x + 2 + \frac{-11}{2x + 1}.$

As $x \to \infty$ or $x \to -\infty$, the term $\frac{-11}{2x + 1} \to 0$, so the slant asymptote is: $y = 3x + 2.$

Summary

• Vertical Asymptote: $x = -\frac{1}{2}$.
• Slant Asymptote: $y = 3x + 2$.

Would you like a detailed graph of these asymptotes, or have any questions?

Related Questions

1. How do you find the vertical asymptotes for rational functions with more complex denominators?
2. What is the difference between horizontal and slant asymptotes?
3. Can a rational function have both a horizontal and a slant asymptote?
4. What are the steps to perform polynomial long division?
5. How do asymptotes affect the behavior of the graph of a rational function?

Tip

When the degree of the numerator is exactly one more than the degree of the denominator, always look for a slant asymptote by performing polynomial division.