Answer
The graph of \( f(x) = \frac{3x}{x^2 - 9} \) has vertical asymptotes at \( x = -3 \) and \( x = 3 \), a horizontal asymptote at \( y = 0 \), passes through the origin, and is symmetric about the origin. The graph of \( h(x) = \frac{3}{x^2 - x - 2} \) has vertical asymptotes at \( x = -1 \) and \( x = 2 \), a horizontal asymptote at \( y = 0 \), a \( y \)-intercept at \( (0, -\frac{3}{2}) \), and no \( x \)-intercepts.
Solution
We start by analyzing each function separately.
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**1. Function \(\displaystyle f(x)=\frac{3x}{x^2-9}\)**
- **Factor the denominator:**
\[
x^2-9=(x-3)(x+3)
\]
The domain of \( f(x) \) excludes values that make the denominator zero. Thus, we exclude
\[
x-3=0\quad\Rightarrow\quad x=3,\qquad x+3=0\quad\Rightarrow\quad x=-3.
\]
So, \( x\neq 3 \) and \( x\neq -3 \).
- **Vertical asymptotes:**
Since there is no cancellation between factors in the numerator and denominator, \( f(x) \) has vertical asymptotes at
\[
x = 3 \quad \text{and} \quad x = -3.
\]
- **Horizontal asymptote:**
The degree of the numerator (1) is less than the degree of the denominator (2), so
\[
y=0
\]
is the horizontal asymptote.
- **Intercepts:**
- **\(x\)-intercept:** Set the numerator equal to zero:
\[
3x = 0\quad\Rightarrow\quad x=0.
\]
Since \( 0 \) is in the domain (denom \(=0^2-9=-9\neq0\)), the function passes through
\[
(0, 0).
\]
- **\(y\)-intercept:** Evaluated by setting \( x=0 \) (which we just did), giving
\[
f(0)=0.
\]
- **Symmetry:**
Check whether the function is even or odd by substituting \(-x\):
\[
f(-x)=\frac{3(-x)}{(-x)^2-9}=\frac{-3x}{x^2-9}=-f(x).
\]
Therefore, \( f(x) \) is an odd function (symmetric about the origin).
- **Graph summary for \( f(x) \):**
- Vertical asymptotes: \( x = 3 \) and \( x = -3 \)
- Horizontal asymptote: \( y = 0 \)
- \( x\)-intercept: \( (0, 0) \) (also the \(y\)-intercept)
- Odd symmetry
---
**2. Function \(\displaystyle h(x)=\frac{3}{x^2-x-2}\)**
- **Factor the denominator:**
Factor the quadratic:
\[
x^2-x-2=(x-2)(x+1)
\]
The values that make the denominator zero are
\[
x-2=0\quad\Rightarrow\quad x=2,\qquad x+1=0\quad\Rightarrow\quad x=-1.
\]
So, the domain excludes \( x=2 \) and \( x=-1 \).
- **Vertical asymptotes:**
There is no cancellation with the numerator, so the vertical asymptotes are at
\[
x = 2 \quad \text{and} \quad x = -1.
\]
- **Horizontal asymptote:**
Since the degree of the numerator (0, since \(3\) is constant) is less than the degree of the denominator (2), we have
\[
y = 0.
\]
- **Intercepts:**
- **\(x\)-intercept:** The numerator is \(3\) (which is never zero), so there is no \(x\)-intercept.
- **\(y\)-intercept:** Set \( x=0 \):
\[
h(0)=\frac{3}{0^2-0-2}=\frac{3}{-2}=-\frac{3}{2}.
\]
So the \(y\)-intercept is
\[
\left(0,-\frac{3}{2}\right).
\]
- **Symmetry:**
Testing for symmetry by replacing \( x \) with \(-x\) does not yield a relationship like \( h(-x)=h(x) \) or \( h(-x)=-h(x) \); hence, \( h(x) \) is not even or odd.
- **Graph summary for \( h(x) \):**
- Vertical asymptotes: \( x = 2 \) and \( x = -1 \)
- Horizontal asymptote: \( y = 0 \)
- \(y\)-intercept: \( \left(0, -\frac{3}{2}\right) \)
- No \(x\)-intercepts
---
**Conclusion:**
- The graph of \( f(x)=\frac{3x}{x^2-9} \) has vertical asymptotes at \( x=-3 \) and \( x=3 \), a horizontal asymptote at \( y=0 \), passes through the origin, and is symmetric about the origin (odd function).
- The graph of \( h(x)=\frac{3}{x^2-x-2} \) has vertical asymptotes at \( x=-1 \) and \( x=2 \), a horizontal asymptote at \( y=0 \), a \(y\)-intercept at \( \left(0,-\frac{3}{2}\right) \), and it does not cross the \(x\)-axis.
When choosing the graphs from the options provided, look for these characteristics in each graph.
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