To find the vertical and horizontal asymptotes of the function

$f(x) = \frac{(2x - 1)(2x + 1)}{(x - 2)(x + 5)}$

1. Vertical Asymptotes:

Vertical asymptotes occur where the denominator of the function is zero (since division by zero is undefined), provided that the numerator is not also zero at those points.

Set the denominator equal to zero: $(x - 2)(x + 5) = 0$

This gives the solutions: $x - 2 = 0 \quad \Rightarrow \quad x = 2$ $x + 5 = 0 \quad \Rightarrow \quad x = -5$

Thus, the vertical asymptotes are at $x = 2$ and $x = -5$.

2. Horizontal Asymptotes:

Horizontal asymptotes depend on the degree of the numerator and the denominator.

• The degree of the numerator $(2x - 1)(2x + 1)$ is $2$ because it simplifies to a quadratic $4x^2 - 1$.
• The degree of the denominator $(x - 2)(x + 5)$ is $2$ because it simplifies to a quadratic $x^2 + 3x - 10$.

Since the degrees of both the numerator and denominator are the same, the horizontal asymptote is found by taking the ratio of the leading coefficients of the numerator and denominator.

• The leading coefficient of the numerator is $4$.
• The leading coefficient of the denominator is $1$.

Therefore, the horizontal asymptote is: $y = \frac{4}{1} = 4$

Conclusion:

• Vertical asymptotes: $x = 2$ and $x = -5$
• Horizontal asymptote: $y = 4$

Would you like more details or have any questions?

Here are 5 related questions:

1. How do slant asymptotes differ from horizontal asymptotes?
2. What happens to $f(x)$ near the vertical asymptotes?
3. How can we find the x-intercepts of $f(x)$?
4. What is the behavior of the function as $x \to \infty$ and $x \to -\infty$?
5. Can a function have more than one horizontal asymptote?

Tip: To quickly find horizontal asymptotes, compare the degrees of the numerator and denominator in a rational function.