Pregunta
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In Silicon Valley, a number of computer firms were found to be contaminating underground water supplies with toxic chemicals stored in leaking underground containers. A water quality control agency ordered the companies to take immediate corrective action and contribute to a monetary pool for the testing and cleanup of the underground contamination. Suppose that the required monetary pool (in millions of dollars) is given by the following function, where \( x \) is the percentage (expressed as a decimal fraction) of the total contaminant removed. Complete parts (A) through (E). \( P(x)=\frac{11 x}{1-x} \quad 0 \leq x<1 \) A. The vertical asymptote(s) is (are) \( x=1 \). (C) Find any horizontal and vertical asymptote. Find the vertical asymptote(s), if any. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. B. There are no vertical asymptotes. Find the horizontal asymptote(s), if any. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The horizontal asymptote(s) is (are) \( y=\square \) (Simplify your answer. Use a comma to separate answers as needed.) B. There are no horizontal asymptotes.

Ask by Lewis Montgomery. in the United States
Mar 17,2025

Solución de inteligencia artificial de Upstudy

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The vertical asymptote is \( x = 1 \), and the horizontal asymptote is \( y = -11 \).

Solución

To find the vertical asymptote(s) of the function \( P(x) = \frac{11x}{1-x} \), we need to determine the values of \( x \) for which the function is undefined. The function is undefined when the denominator is equal to zero. Therefore, we set the denominator equal to zero and solve for \( x \): \[ 1 - x = 0 \] \[ x = 1 \] So, the vertical asymptote is \( x = 1 \). Now, let's find the horizontal asymptote(s) of the function. To find the horizontal asymptote, we need to determine the behavior of the function as \( x \) approaches positive or negative infinity. As \( x \) approaches positive infinity, the term \( \frac{11x}{1-x} \) approaches \( \frac{11x}{-x} = -11 \). Therefore, the horizontal asymptote is \( y = -11 \).

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Bonus Knowledge

To analyze the function \( P(x) = \frac{11x}{1-x} \), let’s dive into the vertical and horizontal asymptotes. The vertical asymptote occurs where the denominator equals zero. Setting \( 1 - x = 0 \) gives us \( x = 1 \), confirming that there is a vertical asymptote at this point. For horizontal asymptotes, we look at the behavior of \( P(x) \) as \( x \) approaches the limits. As \( x \) approaches 1 (from the left), \( P(x) \) trends towards infinity, indicating no horizontal asymptote. However, if we examine \( P(x) \) as \( x \) approaches 0, we see that \( P(0) = 0 \). Thus, while we don’t see a horizontal asymptote in the conventional sense, we can interpret that as \( x \) trends towards 0, \( P(x) \) approaches \( 0 \). So, the vertical asymptote is \( x = 1 \), and the horizontal asymptote does not exist in the traditional sense, though \( P(x) \) approaches \( 0 \) at \( x = 0 \). Enjoy diving deeper into the world of functions and asymptotic behavior—it's all about understanding how these mathematical concepts interact with one another!

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