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Summarize all pertinent information obtained by applying the graphing strategy and sketch the graph of \( y=f(x) \). \[ f(x)=\frac{x^{2}+10 x+24}{x^{2}+8 x+16} \] Find any oblique asymptotes of \( f(x) \). Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. The function has one oblique asymptote, \( \square \) (Type an equation.) B. The function has two different oblique asymptotes. The oblique asymptote as \( x \) approaches \( -\infty \) is \( \square \) and the oblique asymptote as x approaches \( \infty \) is \( \square \) (Type equations.) C. There are no oblique asymptotes. ind the intervals where \( f(x) \) is increasing or decreasing. Select the correct choice below and fill in the answer box(es) to complete your choice. A. The function is increasing on \( \square \) It is never decreasing. (Type an exact answer. Type your answer in interval notation. Use a comma to separate answers as needed.) B. The function is increasing on \( \square \) It is decreasing on \( \square \) . (Type exact answers. Type your answers in interval notation. Use commas to separate answers as needed.) C. The function is decreasing on \( \square \) It is never increasing. (Type an exact answer. Type your answer in interval notation. Use a comma to separate answers as needed.)

Ask by O'Quinn Wheeler. in the United States
Mar 17,2025

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There are no oblique asymptotes. The function is strictly decreasing on \((-\infty,-4)\) and \((-4,\infty)\).

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Beyond the Answer

To analyze the function \( f(x)=\frac{x^{2}+10 x+24}{x^{2}+8 x+16} \), let's start with factoring the numerator and the denominator: The numerator factors to \( (x + 4)(x + 6) \) and the denominator factors to \( (x + 4)(x + 4) \), or \( (x + 4)^2 \). Therefore, we can simplify: \[ f(x) = \frac{x + 6}{x + 4} \quad \text{where } x \neq -4 \] ### Asymptotes: To find oblique asymptotes, perform polynomial long division, or note that since the degrees of the numerator and denominator are equal in the simplified form, the horizontal asymptote can be directly determined. This leads to: 1. The function has a horizontal asymptote as \( x \to \infty \) or \( x \to -\infty \) given by \( y = 1 \). So, the correct answer is: **C. There are no oblique asymptotes.** ### Increasing and Decreasing Intervals: Next, we will determine where the function is increasing or decreasing by finding the derivative: Let \( g(x) = x + 6 \) and \( h(x) = x + 4 \). By applying the quotient rule: \[ f'(x) = \frac{(g'(x)h(x) - g(x)h'(x))}{h(x)^2} = \frac{(1)(x + 4) - (x + 6)(1)}{(x + 4)^2} \] Simplifying yields: \[ f'(x) = \frac{x + 4 - x - 6}{(x + 4)^2} = \frac{-2}{(x + 4)^2} \] Since \( (x + 4)^2 > 0 \) for all \( x \neq -4 \), \( f'(x) < 0 \) everywhere except at the vertical asymptote. Thus, the function is always decreasing over the intervals \( (-\infty, -4) \) and \( (-4, \infty) \). So, the correct answer is: **B. The function is increasing on \( \emptyset \) It is decreasing on \( (-\infty, -4), (-4, \infty) \).**

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