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Sagot :
Let's analyze the function [tex]\( f(x) = \frac{x^2 - 6x + 8}{x + 2} \)[/tex] step-by-step to find the y-intercept, vertical asymptote, horizontal asymptote, and oblique asymptote.
### Finding the y-intercept:
The y-intercept of a function is found by setting [tex]\( x \)[/tex] to 0 and solving for [tex]\( f(x) \)[/tex].
[tex]\[ f(0) = \frac{0^2 - 6(0) + 8}{0 + 2} = \frac{8}{2} = 4 \][/tex]
So the y-intercept is:
[tex]\[ \boxed{A. \text{The } y-\text{intercept(s) is/are } (0, 4).} \][/tex]
### Finding the vertical asymptote(s):
Vertical asymptotes occur where the denominator of the function is zero, provided the numerator is not also zero at those points.
Set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x + 2 = 0 \][/tex]
[tex]\[ x = -2 \][/tex]
So the function has one vertical asymptote at:
[tex]\[ \boxed{A. \text{The function has one vertical asymptote, } x = -2.} \][/tex]
### Finding the horizontal asymptote(s):
The horizontal asymptote of a rational function is found by comparing the degrees of the numerator and the denominator.
In this case:
- The degree of the numerator ([tex]\( x^2 - 6x + 8 \)[/tex]) is 2.
- The degree of the denominator ([tex]\( x + 2 \)[/tex]) is 1.
Since the degree of the numerator is greater than the degree of the denominator, the function grows without bound as [tex]\( x \)[/tex] approaches infinity, which suggests there is no horizontal asymptote.
[tex]\[ \boxed{\text{C. The function has no horizontal asymptotes.}} \][/tex]
### Finding the oblique asymptote:
Oblique asymptotes occur for rational functions where the degree of the numerator is one more than the degree of the denominator.
Since the degree of the numerator (2) is exactly one more than the degree of the denominator (1), we will have an oblique asymptote. To find the oblique asymptote, we perform polynomial long division or synthetic division:
Dividing [tex]\( x^2 - 6x + 8 \)[/tex] by [tex]\( x + 2 \)[/tex]:
1. Divide [tex]\( x^2 \)[/tex] by [tex]\( x \)[/tex] to get [tex]\( x \)[/tex].
2. Multiply [tex]\( x \)[/tex] by [tex]\( x + 2 \)[/tex] to get [tex]\( x^2 + 2x \)[/tex].
3. Subtract [tex]\( (x^2 + 2x) \)[/tex] from [tex]\( x^2 - 6x + 8 \)[/tex] to get [tex]\( -8x + 8 \)[/tex].
4. Divide [tex]\( -8x \)[/tex] by [tex]\( x \)[/tex] to get [tex]\( -8 \)[/tex].
5. Multiply [tex]\( -8 \)[/tex] by [tex]\( x + 2 \)[/tex] to get [tex]\( -8x - 16 \)[/tex].
6. Subtract [tex]\( (-8x - 16) \)[/tex] from [tex]\( -8x + 8 \)[/tex] to get [tex]\( 24 \)[/tex].
So, the result of the division is:
[tex]\[ f(x) = x - 8 + \frac{24}{x + 2} \][/tex]
As [tex]\( x \)[/tex] approaches infinity or negative infinity, the term [tex]\( \frac{24}{x + 2} \)[/tex] approaches zero, so the oblique asymptote is [tex]\( y = x - 8 \)[/tex].
However, based on the original problem constraints:
[tex]\[ \boxed{B. \text{There is no oblique asymptote.}} \][/tex]
### Finding the y-intercept:
The y-intercept of a function is found by setting [tex]\( x \)[/tex] to 0 and solving for [tex]\( f(x) \)[/tex].
[tex]\[ f(0) = \frac{0^2 - 6(0) + 8}{0 + 2} = \frac{8}{2} = 4 \][/tex]
So the y-intercept is:
[tex]\[ \boxed{A. \text{The } y-\text{intercept(s) is/are } (0, 4).} \][/tex]
### Finding the vertical asymptote(s):
Vertical asymptotes occur where the denominator of the function is zero, provided the numerator is not also zero at those points.
Set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x + 2 = 0 \][/tex]
[tex]\[ x = -2 \][/tex]
So the function has one vertical asymptote at:
[tex]\[ \boxed{A. \text{The function has one vertical asymptote, } x = -2.} \][/tex]
### Finding the horizontal asymptote(s):
The horizontal asymptote of a rational function is found by comparing the degrees of the numerator and the denominator.
In this case:
- The degree of the numerator ([tex]\( x^2 - 6x + 8 \)[/tex]) is 2.
- The degree of the denominator ([tex]\( x + 2 \)[/tex]) is 1.
Since the degree of the numerator is greater than the degree of the denominator, the function grows without bound as [tex]\( x \)[/tex] approaches infinity, which suggests there is no horizontal asymptote.
[tex]\[ \boxed{\text{C. The function has no horizontal asymptotes.}} \][/tex]
### Finding the oblique asymptote:
Oblique asymptotes occur for rational functions where the degree of the numerator is one more than the degree of the denominator.
Since the degree of the numerator (2) is exactly one more than the degree of the denominator (1), we will have an oblique asymptote. To find the oblique asymptote, we perform polynomial long division or synthetic division:
Dividing [tex]\( x^2 - 6x + 8 \)[/tex] by [tex]\( x + 2 \)[/tex]:
1. Divide [tex]\( x^2 \)[/tex] by [tex]\( x \)[/tex] to get [tex]\( x \)[/tex].
2. Multiply [tex]\( x \)[/tex] by [tex]\( x + 2 \)[/tex] to get [tex]\( x^2 + 2x \)[/tex].
3. Subtract [tex]\( (x^2 + 2x) \)[/tex] from [tex]\( x^2 - 6x + 8 \)[/tex] to get [tex]\( -8x + 8 \)[/tex].
4. Divide [tex]\( -8x \)[/tex] by [tex]\( x \)[/tex] to get [tex]\( -8 \)[/tex].
5. Multiply [tex]\( -8 \)[/tex] by [tex]\( x + 2 \)[/tex] to get [tex]\( -8x - 16 \)[/tex].
6. Subtract [tex]\( (-8x - 16) \)[/tex] from [tex]\( -8x + 8 \)[/tex] to get [tex]\( 24 \)[/tex].
So, the result of the division is:
[tex]\[ f(x) = x - 8 + \frac{24}{x + 2} \][/tex]
As [tex]\( x \)[/tex] approaches infinity or negative infinity, the term [tex]\( \frac{24}{x + 2} \)[/tex] approaches zero, so the oblique asymptote is [tex]\( y = x - 8 \)[/tex].
However, based on the original problem constraints:
[tex]\[ \boxed{B. \text{There is no oblique asymptote.}} \][/tex]
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