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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

Ask by Collins Schmidt. in the United States
Mar 14,2025

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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. --- **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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The Deep Dive

To analyze the graphs of the functions \( f(x) \) and \( h(x) \), let's start with their characteristics. For \( f(x) = \frac{3x}{x^2 - 9} \), we notice that the denominator factors to \( (x-3)(x+3) \), indicating vertical asymptotes at \( x = 3 \) and \( x = -3 \). As \( x \) approaches these values, the function heads toward infinity or negative infinity, depending on the direction. Now, considering \( h(x) = \frac{3}{x^2 - x - 2} \), we can factor the denominator as \( (x-2)(x+1) \), leading to vertical asymptotes at \( x = 2 \) and \( x = -1 \). The function will also have horizontal asymptotes that we can determine by looking at the degrees of the polynomial in the numerator and denominator. So, when you choose the graphs, look for the key features: vertical asymptotes for both functions and their general behavior as \( x \) approaches those asymptotes! Happy graph hunting!

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