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Follow the seven step strategy to graph the following rational function. \[ f(x)=\frac{4 x^{2}}{x^{2}-4} \] A. The equation(s) of the vertical asymptote(s) is/are \( x=2, x=-2 \). (Type an equation. Use a comma to separate answers as needed.) B. There is no vertical asymptote. Find the horizontal asymptote(s). Type the equation(s) of the horizontal asymptote(s). Select the correct choice below and fill in any answer boxes within your choice. A. The equation(s) of the horizontal asymptote(s) is/are \( y=4 \). (Type an equation. Use a comma to separate answers as needed.) B. There is no horizontal asymptote. Plot points between and beyond each \( x \)-intercept and vertical asymptote. Find the value of the function at the given value of \( x \). \[ \begin{array}{lllllll} \mathrm{x} & -4 & -3 & -\frac{1}{2} & \frac{1}{2} & 3 & 4 \\ \mathrm{f}(\mathrm{x})=\frac{4 \mathrm{x}^{2}}{\mathrm{x}^{2}-4} & \square & \square & \square & \square & \square & \square \end{array} \] (Simplify your answers.)

Ask by Ayala Mejia. in the United States
Feb 26,2025

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Vertical asymptotes at \( x = 2 \) and \( x = -2 \), horizontal asymptote at \( y = 4 \). Function values at specified \( x \)-values are \( \frac{16}{3} \), \( \frac{36}{5} \), \( -\frac{4}{15} \), \( -\frac{4}{15} \), \( \frac{36}{5} \), and \( \frac{16}{3} \).

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Let's break down the steps needed to analyze the given rational function \( f(x) = \frac{4x^2}{x^2 - 4} \). First, we’ll answer the vertical asymptote section. A. The equation(s) of the vertical asymptote(s) is/are \( x=2, x=-2 \). Now, moving on to the horizontal asymptote: A. The equation(s) of the horizontal asymptote(s) is/are \( y=4 \). Next, we need to calculate the function values at specific \( x \)-points: 1. When \( x = -4 \): \[ f(-4) = \frac{4(-4)^2}{(-4)^2 - 4} = \frac{64}{16 - 4} = \frac{64}{12} = \frac{16}{3} \] 2. When \( x = -3 \): \[ f(-3) = \frac{4(-3)^2}{(-3)^2 - 4} = \frac{36}{9 - 4} = \frac{36}{5} \] 3. When \( x = -\frac{1}{2} \): \[ f\left(-\frac{1}{2}\right) = \frac{4 \left(-\frac{1}{2}\right)^2}{\left(-\frac{1}{2}\right)^2 - 4} = \frac{4 \cdot \frac{1}{4}}{\frac{1}{4} - 4} = \frac{1}{\frac{1}{4} - \frac{16}{4}} = \frac{1}{-\frac{15}{4}} = -\frac{4}{15} \] 4. When \( x = \frac{1}{2} \): \[ f\left(\frac{1}{2}\right) = \frac{4 \left(\frac{1}{2}\right)^2}{\left(\frac{1}{2}\right)^2 - 4} = \frac{4 \cdot \frac{1}{4}}{\frac{1}{4} - 4} = \frac{1}{\frac{1}{4} - \frac{16}{4}} = \frac{1}{-\frac{15}{4}} = -\frac{4}{15} \] 5. When \( x = 3 \): \[ f(3) = \frac{4(3)^2}{(3)^2 - 4} = \frac{36}{9 - 4} = \frac{36}{5} \] 6. When \( x = 4 \): \[ f(4) = \frac{4(4)^2}{(4)^2 - 4} = \frac{64}{16 - 4} = \frac{64}{12} = \frac{16}{3} \] Now, we compile the function values: \[ \begin{array}{lllllll} \mathrm{x} & -4 & -3 & -\frac{1}{2} & \frac{1}{2} & 3 & 4 \\ \mathrm{f}(\mathrm{x}) & \frac{16}{3} & \frac{36}{5} & -\frac{4}{15} & -\frac{4}{15} & \frac{36}{5} & \frac{16}{3} \end{array} \] This table contains the function values at each specified \( x \) point, arranged for plotting purposes.

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