To solve for the slope of the slant (oblique) asymptote of the rational function $f(x) = \frac{x^3 - x^2 - 4x}{x^2 - 4x}$, follow these steps:

Step 1: Identify the degrees of the numerator and the denominator

• The degree of the numerator, $x^3 - x^2 - 4x$, is 3.
• The degree of the denominator, $x^2 - 4x$, is 2.

Since the degree of the numerator (3) is greater than the degree of the denominator (2) by exactly 1, there will be a slant (oblique) asymptote.

Step 2: Perform polynomial long division

We divide $x^3 - x^2 - 4x$ by $x^2 - 4x$.

1. First term: $\frac{x^3}{x^2} = x$. Multiply $x$ by $x^2 - 4x$, which gives $x^3 - 4x^2$.

   • Subtract: $(x^3 - x^2 - 4x) - (x^3 - 4x^2) = 3x^2 - 4x$.

2. Next term: $\frac{3x^2}{x^2} = 3$. Multiply $3$ by $x^2 - 4x$, which gives $3x^2 - 12x$.

   • Subtract: $(3x^2 - 4x) - (3x^2 - 12x) = 8x$.

So, the quotient from the division is $x + 3$, and the remainder is $8x$.

Step 3: Identify the slant asymptote

The equation of the slant asymptote is given by the quotient obtained in the division, which is $y = x + 3$. Therefore, the slope of the slant asymptote is the coefficient of $x$, which is 1.

Answer:

The slope of the slant asymptote of the graph of $f(x)$ is 1.

Would you like further details or clarifications on any of the steps?

Here are five follow-up questions for deeper understanding:

1. How does the degree of the numerator and denominator affect the presence of a slant asymptote?
2. Why is polynomial long division necessary to find the slant asymptote?
3. How would you find the vertical asymptotes of the same function?
4. Can a rational function have both slant and horizontal asymptotes? Why or why not?
5. How would the result change if the degrees of the numerator and denominator were equal?

Tip: Slant asymptotes occur when the degree of the numerator is exactly one greater than the degree of the denominator in a rational function.