From the image, the question asks to find the asymptotes of the given function:

$f(x) = \frac{5x^2 - 4x + 3}{4 - x}.$

Let’s analyze and calculate the vertical, horizontal, and oblique (if applicable) asymptotes for this rational function.

1. Vertical Asymptote

A vertical asymptote occurs where the denominator equals zero (and the numerator is nonzero). Here, the denominator is $4 - x$.

$4 - x = 0 \implies x = 4.$

Thus, there is a vertical asymptote at $x = 4$.

2. Horizontal Asymptote

Horizontal asymptotes depend on the degrees of the numerator and denominator:

• Degree of numerator: 2 ($5x^2$).
• Degree of denominator: 1 ($4 - x$).

Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

3. Oblique (Slant) Asymptote

When the degree of the numerator is exactly one higher than the denominator, there is an oblique asymptote. To find it, divide the numerator by the denominator:

$\text{Divide } 5x^2 - 4x + 3 \text{ by } 4 - x.$

Perform polynomial division:

$\frac{5x^2 - 4x + 3}{4 - x} = -5x - 9 + \frac{-33}{4 - x}.$

The oblique asymptote is the quotient (ignoring the remainder):

$y = -5x - 9.$

Summary

• Vertical Asymptote: $x = 4$.
• Horizontal Asymptote: None.
• Oblique Asymptote: $y = -5x - 9$.

Would you like step-by-step details on the division or any clarification?

Here are 5 related questions for practice:

1. How do you determine if a function has a horizontal asymptote?
2. What are the conditions for an oblique asymptote to exist?
3. How do you find vertical asymptotes of a rational function?
4. Can a function have both a horizontal and an oblique asymptote? Why or why not?
5. How would the asymptotes change if the numerator’s degree was lower than the denominator’s?

Tip: Always check the degrees of the numerator and denominator first to decide