Answer :
Let's examine the key features of the given functions step by step.
### Function \( f(x) = \frac{-2x + 4}{x - 6} \)
1. Horizontal Asymptote:
To find the horizontal asymptote, we take the limit of \( f(x) \) as \( x \) approaches infinity.
[tex]\[ \lim_{{x \to \infty}} \frac{-2x + 4}{x - 6} = -2 \][/tex]
Thus, the horizontal asymptote is \( y = -2 \).
2. Oblique Asymptote:
To identify the oblique asymptote, we perform polynomial long division of the numerator by the denominator:
[tex]\[ \frac{-2x + 4}{x - 6} = -2 + \frac{16}{x - 6} \][/tex]
As \( x \) approaches infinity, the term \(\frac{16}{x-6}\) approaches 0.
Therefore, the oblique asymptote is given by \( y = -2 \).
### Function \( g(x) = \frac{x^2 - 2x}{x^2 + x - 6} \)
1. Horizontal Asymptote:
Again, we find the limit as \( x \) approaches infinity:
[tex]\[ \lim_{{x \to \infty}} \frac{x^2 - 2x}{x^2 + x - 6} = 1 \][/tex]
Thus, the horizontal asymptote is \( y = 1 \).
2. Oblique Asymptote:
Here, we perform polynomial long division of \( x^2 - 2x \) by \( x^2 + x - 6 \):
[tex]\[ \frac{x^2 - 2x}{x^2 + x - 6} = 1 + \frac{-3x + 6}{x^2 + x - 6} \][/tex]
Thus, the oblique asymptote is represented by remainders once the polynomial division is performed, indicating unique regions for simplified form analysis.
### Function \( h(x) = \frac{x^2 - 2x}{x - 6} \)
1. Horizontal Asymptote:
Evaluating the limit as \( x \) approaches infinity:
[tex]\[ \lim_{{x \to \infty}} \frac{x^2 - 2x}{x - 6} = \infty \][/tex]
This implies there is no horizontal asymptote (or it approaches infinity).
2. Oblique Asymptote:
Performing polynomial long division of \( x^2 - 2x \) by \( x - 6 \):
[tex]\[ \frac{x^2 - 2x}{x - 6} = x + 4 + \frac{24}{x - 6} \][/tex]
As \( x \) approaches infinity, the term \(\frac{24}{x-6}\) approaches 0.
Thus, the oblique asymptote is \( y = x + 4 \).
### Summary:
- Horizontal Asymptotes:
- For \( f(x) \): \( y = -2 \)
- For \( g(x) \): \( y = 1 \)
- For \( h(x) \): No horizontal asymptote
- Oblique Asymptotes:
- For \( f(x) \): None (as it simplifies directly to a horizontal asymptote)
- For \( g(x) \): Complex form with division remainders forming regions
- For \( h(x) \): \( y = x + 4 \)
I hope this detailed analysis helps you understand the key features of these functions!
### Function \( f(x) = \frac{-2x + 4}{x - 6} \)
1. Horizontal Asymptote:
To find the horizontal asymptote, we take the limit of \( f(x) \) as \( x \) approaches infinity.
[tex]\[ \lim_{{x \to \infty}} \frac{-2x + 4}{x - 6} = -2 \][/tex]
Thus, the horizontal asymptote is \( y = -2 \).
2. Oblique Asymptote:
To identify the oblique asymptote, we perform polynomial long division of the numerator by the denominator:
[tex]\[ \frac{-2x + 4}{x - 6} = -2 + \frac{16}{x - 6} \][/tex]
As \( x \) approaches infinity, the term \(\frac{16}{x-6}\) approaches 0.
Therefore, the oblique asymptote is given by \( y = -2 \).
### Function \( g(x) = \frac{x^2 - 2x}{x^2 + x - 6} \)
1. Horizontal Asymptote:
Again, we find the limit as \( x \) approaches infinity:
[tex]\[ \lim_{{x \to \infty}} \frac{x^2 - 2x}{x^2 + x - 6} = 1 \][/tex]
Thus, the horizontal asymptote is \( y = 1 \).
2. Oblique Asymptote:
Here, we perform polynomial long division of \( x^2 - 2x \) by \( x^2 + x - 6 \):
[tex]\[ \frac{x^2 - 2x}{x^2 + x - 6} = 1 + \frac{-3x + 6}{x^2 + x - 6} \][/tex]
Thus, the oblique asymptote is represented by remainders once the polynomial division is performed, indicating unique regions for simplified form analysis.
### Function \( h(x) = \frac{x^2 - 2x}{x - 6} \)
1. Horizontal Asymptote:
Evaluating the limit as \( x \) approaches infinity:
[tex]\[ \lim_{{x \to \infty}} \frac{x^2 - 2x}{x - 6} = \infty \][/tex]
This implies there is no horizontal asymptote (or it approaches infinity).
2. Oblique Asymptote:
Performing polynomial long division of \( x^2 - 2x \) by \( x - 6 \):
[tex]\[ \frac{x^2 - 2x}{x - 6} = x + 4 + \frac{24}{x - 6} \][/tex]
As \( x \) approaches infinity, the term \(\frac{24}{x-6}\) approaches 0.
Thus, the oblique asymptote is \( y = x + 4 \).
### Summary:
- Horizontal Asymptotes:
- For \( f(x) \): \( y = -2 \)
- For \( g(x) \): \( y = 1 \)
- For \( h(x) \): No horizontal asymptote
- Oblique Asymptotes:
- For \( f(x) \): None (as it simplifies directly to a horizontal asymptote)
- For \( g(x) \): Complex form with division remainders forming regions
- For \( h(x) \): \( y = x + 4 \)
I hope this detailed analysis helps you understand the key features of these functions!
