The given function is:

$y = \frac{2x - 1}{x + 2}$

We need to determine the vertical and horizontal asymptotes to help graph the function.

1. Vertical Asymptote:

Vertical asymptotes occur where the denominator equals zero (because division by zero is undefined). For the function $y = \frac{2x - 1}{x + 2}$, the denominator is $x + 2$. Setting this equal to zero:

$x + 2 = 0 \quad \Rightarrow \quad x = -2$

So, there is a vertical asymptote at $x = -2$.

2. Horizontal Asymptote:

Horizontal asymptotes depend on the degrees of the numerator and denominator. Here, both the numerator $2x - 1$ and the denominator $x + 2$ are linear, meaning they have the same degree (degree 1).

For rational functions where the degrees are the same, the horizontal asymptote is found by dividing the leading coefficients of the numerator and denominator. In this case:

$\frac{\text{Leading coefficient of } 2x}{\text{Leading coefficient of } x} = \frac{2}{1} = 2$

Thus, there is a horizontal asymptote at $y = 2$.

Conclusion:

The function has a vertical asymptote at $x = -2$ and a horizontal asymptote at $y = 2$. Therefore, the correct answer is the third option:

• "The function has a vertical asymptote at $x = -2$ and a horizontal asymptote at $y = 2$."

Would you like more details on how to interpret asymptotes, or do you have any other questions?

Here are 5 related questions you might find useful:

1. How do you find slant (oblique) asymptotes for rational functions?
2. Can a function have more than one vertical or horizontal asymptote?
3. How does the behavior of a function change near an asymptote?
4. What is the relationship between asymptotes and limits?
5. How can you graph rational functions with multiple asymptotes?

Tip: Always check both the numerator and denominator of rational functions to find potential zeros and asymptotes!