Math Problem Statement

Graph the following function. Determine the​ asymptotes, if any. List the domain and the​ x-intercepts and the​ y-intercepts. Check your work using a graphing calculator. f(x)= x^2-16 / x+1

Solution

To analyze the function f(x)=x216x+1f(x) = \frac{x^2 - 16}{x + 1}, we will determine its asymptotes, domain, x-intercepts, and y-intercepts.

1. Asymptotes

Vertical Asymptote:

  • Vertical asymptotes occur where the denominator is zero, and the numerator is not zero.
  • Setting the denominator equal to zero: x+1=0    x=1x + 1 = 0 \implies x = -1
  • Thus, there is a vertical asymptote at x=1x = -1.

Horizontal Asymptote:

  • For rational functions, the horizontal asymptote can be found by comparing the degrees of the numerator and the denominator.
  • The degree of the numerator x216x^2 - 16 is 2, and the degree of the denominator x+1x + 1 is 1.
  • Since the degree of the numerator is greater than that of the denominator, there is no horizontal asymptote. However, there is an oblique asymptote, which can be found through polynomial long division.

Polynomial Long Division: x216x+1=x1with a remainder of15.\frac{x^2 - 16}{x + 1} = x - 1 \quad \text{with a remainder of} \quad -15. This implies: f(x)x1as x or .f(x) \approx x - 1 \quad \text{as } x \to \infty \text{ or } -\infty. Thus, the oblique asymptote is y=x1y = x - 1.

2. Domain

  • The domain of f(x)f(x) includes all real numbers except where the denominator is zero: x1.x \neq -1. Thus, the domain is (,1)(1,)(-\infty, -1) \cup (-1, \infty).

3. Intercepts

X-Intercepts:

  • To find the x-intercepts, set f(x)=0f(x) = 0: x216x+1=0    x216=0.\frac{x^2 - 16}{x + 1} = 0 \implies x^2 - 16 = 0. Factoring gives: (x4)(x+4)=0    x=4 and x=4.(x - 4)(x + 4) = 0 \implies x = 4 \text{ and } x = -4. So, the x-intercepts are (4,0)(4, 0) and (4,0)(-4, 0).

Y-Intercept:

  • To find the y-intercept, evaluate f(0)f(0): f(0)=02160+1=161=16.f(0) = \frac{0^2 - 16}{0 + 1} = \frac{-16}{1} = -16. So, the y-intercept is (0,16)(0, -16).

Summary

  • Vertical Asymptote: x=1x = -1
  • Oblique Asymptote: y=x1y = x - 1
  • Domain: (,1)(1,)(-\infty, -1) \cup (-1, \infty)
  • X-Intercepts: (4,0)(4, 0) and (4,0)(-4, 0)
  • Y-Intercept: (0,16)(0, -16)

Graphing Check

You can verify these findings using a graphing calculator by plotting the function f(x)f(x).

If you have any questions or would like further details on any of the steps, let me know!

Further Exploration

  1. How do changes in the coefficients of the function affect the asymptotes?
  2. What would happen to the function if the denominator were quadratic instead of linear?
  3. Can you find the limits of the function as xx approaches the asymptotes?
  4. How do the intercepts relate to the behavior of the graph?
  5. What are the characteristics of rational functions that can help predict their graph?

Tip: When analyzing rational functions, always check for asymptotes first to understand the overall shape of the graph.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Rational Functions
Asymptotes
Intercepts
Domain

Formulas

Vertical Asymptote: x + 1 = 0
Oblique Asymptote: y = x - 1
X-Intercepts: f(x) = 0
Y-Intercept: f(0)

Theorems

Properties of Asymptotes for Rational Functions
Intercepts of Functions

Suitable Grade Level

Grades 9-11