**Problem Statement:**Find an equation of a rational function $ f $ that satisfies the given conditions.- **Vertical Asymptotes:** $ x = -5, x = 3 $- **Horizontal Asymptote:** $ y = 0 $- **x-intercept:** $ -1 $; $ f(0) = -2 $- **Hole at:** $ x = 2 $**Explanation:**To construct a rational function that satisfies these conditions, consider the following components:1. **Vertical Asymptotes**: For vertical asymptotes at $ x = -5 $ and $ x = 3 $, the function's denominator must include factors that become zero at these points, such as $ (x + 5) $ and $ (x - 3) $.2. **Horizontal Asymptote**: For a horizontal asymptote at $ y = 0 $, the degree of the polynomial in the numerator must be less than the degree of the polynomial in the denominator.3. **x-intercept at $ x = -1 $**: The function must cross the x-axis at $ x = -1 $, which implies the numerator must be zero when $ x = -1 $, such as including the factor $ (x + 1) $.4. **The condition $ f(0) = -2 $**: After incorporating the other conditions, we can include a constant multiplier to ensure the function evaluates to -2 when $ x = 0 $.5. **Hole at $ x = 2 $**: The function must have a factor in both the numerator and the denominator that cancels out at $ x = 2 $, such as $ (x - 2) $.Using these components, the function $ f(x) $ can be assembled as follows:\[f(x) = k \frac{(x + 1)(x - 2)}{(x + 5)(x - 3)(x - 2)}\]Here, $ k $ is a constant that can be determined by using the condition $ f(0) = -2 $. Substituting $ x = 0 $ into the function, solve for $ k $ to ensure $ f(0) = -2 $.

Answered: Find an equation of a rational function…

Find an equation of a rational function f that satisfies the given conditions. vertical asymptotes: x = -5, x = 3 horizontal asymptote: y = 0 x-intercept: -1; f(0) = -2 hole at x = 2

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

**Problem Statement:**

Find an equation of a rational function $ f $ that satisfies the given conditions.

- **Vertical Asymptotes:** $ x = -5, x = 3 $
- **Horizontal Asymptote:** $ y = 0 $
- **x-intercept:** $ -1 $; $ f(0) = -2 $
- **Hole at:** $ x = 2 $

**Explanation:**

To construct a rational function that satisfies these conditions, consider the following components:

1. **Vertical Asymptotes**: For vertical asymptotes at $ x = -5 $ and $ x = 3 $, the function's denominator must include factors that become zero at these points, such as $ (x + 5) $ and $ (x - 3) $.

2. **Horizontal Asymptote**: For a horizontal asymptote at $ y = 0 $, the degree of the polynomial in the numerator must be less than the degree of the polynomial in the denominator.

3. **x-intercept at $ x = -1 $**: The function must cross the x-axis at $ x = -1 $, which implies the numerator must be zero when $ x = -1 $, such as including the factor $ (x + 1) $.

4. **The condition $ f(0) = -2 $**: After incorporating the other conditions, we can include a constant multiplier to ensure the function evaluates to -2 when $ x = 0 $.

5. **Hole at $ x = 2 $**: The function must have a factor in both the numerator and the denominator that cancels out at $ x = 2 $, such as $ (x - 2) $.

Using these components, the function $ f(x) $ can be assembled as follows:

\[
f(x) = k \frac{(x + 1)(x - 2)}{(x + 5)(x - 3)(x - 2)}
\]

Here, $ k $ is a constant that can be determined by using the condition $ f(0) = -2 $. Substituting $ x = 0 $ into the function, solve for $ k $ to ensure $ f(0) = -2 $.

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