Consider the following rational functions. \[ f(x)=\frac{3 x}{x^{2}-9} \] \( h(x)=\frac{3}{x^{2}-x-2} \) Choose the graph of each function from the choices

We start by analyzing each function separately. --- **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.