Let's analyze the rational function $f(x) = \frac{x^2 + 2x + 1}{x^2 - x - 6}$.

1. Domain

The domain of a rational function is determined by the values of $x$ for which the denominator is not equal to zero.

Denominator:

$x^2 - x - 6$ We need to find the roots of the denominator by solving the quadratic equation: $x^2 - x - 6 = 0$

Factor the quadratic expression: $(x - 3)(x + 2) = 0$

The roots are: $x = 3 \quad \text{and} \quad x = -2$

So, the denominator is zero at $x = 3$ and $x = -2$. These values must be excluded from the domain.

Domain: $x \in \mathbb{R}$ except $x \neq 3$ and $x \neq -2$.

2. Range

The range of a rational function is determined by the possible values of $y = f(x)$. Since the function is continuous except at the points where the denominator is zero, the range typically covers all real numbers except for any horizontal asymptotes.

3. Intercepts

x-intercepts:

The x-intercepts occur where $f(x) = 0$. For a rational function, this occurs when the numerator is zero (as long as the denominator is not zero at the same point).

Numerator: $x^2 + 2x + 1 = 0$

Factor the quadratic expression: $(x + 1)^2 = 0$

So, $x = -1$ is the only x-intercept.

x-intercept: $(-1, 0)$

y-intercepts:

The y-intercept occurs where $x = 0$.

Evaluate $f(0)$: $f(0) = \frac{0^2 + 2(0) + 1}{0^2 - 0 - 6} = \frac{1}{-6} = -\frac{1}{6}$

y-intercept: $\left(0, -\frac{1}{6}\right)$

4. Asymptotes

Vertical Asymptotes:

Vertical asymptotes occur where the denominator is zero, provided the numerator is not zero at those points.

From the factorization of the denominator: $x = 3 \quad \text{and} \quad x = -2$

Thus, the vertical asymptotes are at: $x = 3 \quad \text{and} \quad x = -2$

Horizontal Asymptote:

To determine the horizontal asymptote, compare the degrees of the numerator and the denominator.

• Degree of the numerator: 2 (because $x^2 + 2x + 1$ is quadratic)
• Degree of the denominator: 2 (because $x^2 - x - 6$ is also quadratic)

When the degrees of the numerator and denominator are equal, the horizontal asymptote is given by the ratio of the leading coefficients.

The leading coefficient of both the numerator and denominator is 1, so the horizontal asymptote is: $y = \frac{1}{1} = 1$

Horizontal Asymptote: $y = 1$

Summary:

• Domain: $x \in \mathbb{R}$ except $x \neq 3$ and $x \neq -2$
• Range: $y \in \mathbb{R}$, excluding potential values based on horizontal asymptote analysis (typically requires further exploration or graphing).
• x-intercept: $(-1, 0)$
• y-intercept: $\left(0, -\frac{1}{6}\right)$
• Vertical Asymptotes: $x = 3$ and $x = -2$
• Horizontal Asymptote: $y = 1$

Would you like to delve into any of these details further? Here are some related questions:

1. How do we find slant asymptotes if they exist?
2. Can we find exact values of the function around the vertical asymptotes?
3. How does the behavior of the function change near the intercepts?
4. How do horizontal asymptotes affect the range of the function?
5. What is the significance of holes in rational functions?

Tip: Always check for removable discontinuities (holes) in rational functions where both the numerator and denominator are zero at the same point.