Answer :
Let's solve the problem step-by-step.
1. Given Function:
[tex]\[ f(x) = \frac{x+1}{x-2} \][/tex]
2. Understanding the Function:
The function [tex]\( f(x) = \frac{x+1}{x-2} \)[/tex] is a rational function, which means it is a ratio of two polynomials. The numerator is [tex]\( x + 1 \)[/tex] and the denominator is [tex]\( x - 2 \)[/tex].
3. Domain of the Function:
The function [tex]\( f(x) \)[/tex] is defined for all real numbers except where the denominator is zero because division by zero is undefined.
Setting the denominator equal to zero gives:
[tex]\[ x - 2 = 0 \implies x = 2 \][/tex]
Therefore, the domain of [tex]\( f(x) \)[/tex] is all real numbers except [tex]\( x = 2 \)[/tex]:
[tex]\[ \text{Domain: } (-\infty, 2) \cup (2, \infty) \][/tex]
4. Vertical Asymptote:
A vertical asymptote occurs where the function is undefined due to division by zero. From our domain analysis, we have:
[tex]\[ x = 2 \][/tex]
Thus, the vertical asymptote is at [tex]\( x = 2 \)[/tex].
5. Horizontal Asymptote:
To find the horizontal asymptote, we compare the degrees of the polynomials in the numerator and the denominator.
- Both the numerator [tex]\( (x+1) \)[/tex] and the denominator [tex]\( (x-2) \)[/tex] are linear polynomials (degree 1).
When the degrees of the numerator and denominator are equal, the horizontal asymptote is given by the ratio of the leading coefficients. The leading coefficients of [tex]\( x+1 \)[/tex] and [tex]\( x-2 \)[/tex] are both 1.
Thus, the horizontal asymptote is:
[tex]\[ y = \frac{1}{1} = 1 \][/tex]
6. Intercepts:
- y-intercept: Set [tex]\( x = 0 \)[/tex] in the function:
[tex]\[ f(0) = \frac{0+1}{0-2} = \frac{1}{-2} = -\frac{1}{2} \][/tex]
So, the y-intercept is [tex]\( (0, -\frac{1}{2}) \)[/tex].
- x-intercept: Set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ \frac{x+1}{x-2} = 0 \implies x+1 = 0 \implies x = -1 \][/tex]
So, the x-intercept is [tex]\( (-1, 0) \)[/tex].
By following these steps, we have thoroughly analyzed and solved the given mathematical function [tex]\( f(x) = \frac{x+1}{x-2} \)[/tex].
1. Given Function:
[tex]\[ f(x) = \frac{x+1}{x-2} \][/tex]
2. Understanding the Function:
The function [tex]\( f(x) = \frac{x+1}{x-2} \)[/tex] is a rational function, which means it is a ratio of two polynomials. The numerator is [tex]\( x + 1 \)[/tex] and the denominator is [tex]\( x - 2 \)[/tex].
3. Domain of the Function:
The function [tex]\( f(x) \)[/tex] is defined for all real numbers except where the denominator is zero because division by zero is undefined.
Setting the denominator equal to zero gives:
[tex]\[ x - 2 = 0 \implies x = 2 \][/tex]
Therefore, the domain of [tex]\( f(x) \)[/tex] is all real numbers except [tex]\( x = 2 \)[/tex]:
[tex]\[ \text{Domain: } (-\infty, 2) \cup (2, \infty) \][/tex]
4. Vertical Asymptote:
A vertical asymptote occurs where the function is undefined due to division by zero. From our domain analysis, we have:
[tex]\[ x = 2 \][/tex]
Thus, the vertical asymptote is at [tex]\( x = 2 \)[/tex].
5. Horizontal Asymptote:
To find the horizontal asymptote, we compare the degrees of the polynomials in the numerator and the denominator.
- Both the numerator [tex]\( (x+1) \)[/tex] and the denominator [tex]\( (x-2) \)[/tex] are linear polynomials (degree 1).
When the degrees of the numerator and denominator are equal, the horizontal asymptote is given by the ratio of the leading coefficients. The leading coefficients of [tex]\( x+1 \)[/tex] and [tex]\( x-2 \)[/tex] are both 1.
Thus, the horizontal asymptote is:
[tex]\[ y = \frac{1}{1} = 1 \][/tex]
6. Intercepts:
- y-intercept: Set [tex]\( x = 0 \)[/tex] in the function:
[tex]\[ f(0) = \frac{0+1}{0-2} = \frac{1}{-2} = -\frac{1}{2} \][/tex]
So, the y-intercept is [tex]\( (0, -\frac{1}{2}) \)[/tex].
- x-intercept: Set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ \frac{x+1}{x-2} = 0 \implies x+1 = 0 \implies x = -1 \][/tex]
So, the x-intercept is [tex]\( (-1, 0) \)[/tex].
By following these steps, we have thoroughly analyzed and solved the given mathematical function [tex]\( f(x) = \frac{x+1}{x-2} \)[/tex].
