To graph the vertical and horizontal asymptotes of the given rational function $f(x) = \frac{x^2 + 2}{-2x + 5}$, let's analyze the asymptotes step by step:

1. Vertical Asymptote(s)

Vertical asymptotes occur when the denominator is equal to zero, and the function is undefined at those values of $x$.

For the denominator: $-2x + 5 = 0$ Solving for $x$: $x = \frac{5}{2}$ So, there is a vertical asymptote at $x = \frac{5}{2}$.

2. Horizontal Asymptote(s)

To determine horizontal asymptotes, we compare the degrees of the numerator and the denominator:

• The degree of the numerator $(x^2 + 2)$ is 2.
• The degree of the denominator $(-2x + 5)$ is 1.

Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. Instead, this function will have an oblique (slant) asymptote.

3. Oblique (Slant) Asymptote

To find the slant asymptote, divide the numerator by the denominator using polynomial long division.

$\frac{x^2 + 2}{-2x + 5}$

When you perform the long division, the quotient is: $y = -\frac{1}{2}x - \frac{5}{4}$

So, the slant asymptote is given by the equation: $y = -\frac{1}{2}x - \frac{5}{4}$

Plot Points

Now, the important information for the graph:

• Vertical asymptote: $x = \frac{5}{2}$
• Slant asymptote: $y = -\frac{1}{2}x - \frac{5}{4}$

Would you like me to provide further details or clarify anything?

Follow-up Questions:

1. How does the degree of the polynomial affect the type of asymptote?
2. What happens to the graph near the vertical asymptote?
3. How do you find the intercepts of the rational function?
4. What is the behavior of the graph as $x \to \infty$?
5. How does the slant asymptote affect the end behavior of the function?

Tip:

For rational functions, always check the degree of the numerator and denominator to determine if you have horizontal, vertical, or slant asymptotes.