Question: User: 2. Rational Function Analysis
Given the rational function:
R(x) = \frac{x^2 + 2x - 8}{x^2 - 16}
(a)
i. Find the domain.
ii. Identify all vertical asymptotes and holes (if any).
(b) i. Find all horizontal or oblique asymptotes.
ii. Find all x-intercepts and y-intercepts.
(c)
i. Determine the behavior of the graph near each vertical asymptote.
ii. Sketch a rough graph showing all asymptotes, intercepts, and any removable discontinuities.
1. **State the problem:**
We analyze the rational function $$R(x) = \frac{x^2 + 2x - 8}{x^2 - 16}$$ to find its domain, vertical asymptotes, holes, horizontal/oblique asymptotes, intercepts, and graph behavior near vertical asymptotes.
2. **Find the domain:**
The domain excludes values where the denominator is zero.
$$x^2 - 16 = 0 \implies (x-4)(x+4) = 0 \implies x = 4, -4$$
Domain: $$\{x \in \mathbb{R} : x \neq 4, -4\}$$
3. **Factor numerator and denominator:**
Numerator:
$$x^2 + 2x - 8 = (x+4)(x-2)$$
Denominator:
$$x^2 - 16 = (x-4)(x+4)$$
4. **Simplify the function:**
$$R(x) = \frac{(x+4)(x-2)}{(x-4)(x+4)}$$
Cancel common factor $(x+4)$:
$$R(x) = \frac{\cancel{(x+4)}(x-2)}{(x-4)\cancel{(x+4)}} = \frac{x-2}{x-4}, \quad x \neq -4$$
5. **Identify vertical asymptotes and holes:**
- Hole at $$x = -4$$ because factor canceled.
- Vertical asymptote at $$x = 4$$ because denominator zero remains.
6. **Find horizontal or oblique asymptotes:**
Degree numerator = 1, degree denominator = 1.
Since degrees are equal, horizontal asymptote is ratio of leading coefficients:
$$y = \frac{1}{1} = 1$$
7. **Find x-intercepts and y-intercepts:**
- x-intercept: numerator zero (after simplification):
$$x - 2 = 0 \implies x = 2$$
- y-intercept: evaluate $$R(0)$$:
$$R(0) = \frac{0 - 2}{0 - 4} = \frac{-2}{-4} = \frac{1}{2}$$
8. **Behavior near vertical asymptote $$x=4$$:**
- As $$x \to 4^-$$, denominator $$x-4 \to 0^-$$, numerator $$x-2 \to 2$$, so $$R(x) \to \frac{2}{0^-} = -\infty$$
- As $$x \to 4^+$$, denominator $$x-4 \to 0^+$$, numerator $$x-2 \to 2$$, so $$R(x) \to \frac{2}{0^+} = +\infty$$
9. **Summary:**
- Domain: $$x \neq 4, -4$$
- Hole at $$x = -4$$ with value:
$$\lim_{x \to -4} R(x) = \frac{-4 - 2}{-4 - 4} = \frac{-6}{-8} = \frac{3}{4}$$
- Vertical asymptote at $$x=4$$
- Horizontal asymptote at $$y=1$$
- x-intercept at $$x=2$$
- y-intercept at $$y=\frac{1}{2}$$
10. **Graph sketch notes:**
- Plot hole at $(-4, \frac{3}{4})$
- Draw vertical asymptote line at $$x=4$$
- Draw horizontal asymptote line at $$y=1$$
- Mark intercepts at $(2,0)$ and $(0, \frac{1}{2})$
This completes the analysis of the rational function.