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Sagot :
Let's analyze the function [tex]\( f(x) = \frac{1}{(x-4)^2} \)[/tex] and address each part of the question one by one.
### 1. Domain of the function
The function is defined for all [tex]\( x \)[/tex] except where the denominator is zero. The denominator [tex]\((x-4)^2\)[/tex] is zero when [tex]\( x = 4 \)[/tex]. Therefore, the function is not defined at [tex]\( x = 4 \)[/tex].
Domain: [tex]\( (-\infty, 4) \cup (4, \infty) \)[/tex]
### 2. x-intercepts of the function
The x-intercepts are the points where the function crosses the x-axis, i.e., where [tex]\( f(x) = 0 \)[/tex].
For [tex]\( \frac{1}{(x-4)^2} = 0 \)[/tex], there is no value of [tex]\( x \)[/tex] that will make the equation true because the numerator is a constant (1) and the denominator is squared, so it is never zero, thus the function has no x-intercepts.
Answer: B. The function has no x-intercepts.
### 3. y-intercepts of the function
The y-intercept is found by evaluating the function at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{1}{(0-4)^2} = \frac{1}{16} \][/tex]
So the y-intercept is [tex]\( (0, \frac{1}{16}) \)[/tex].
Answer: A. The y-intercept of the function is [tex]\( \left( 0, \frac{1}{16} \right) \)[/tex].
### 4. Vertical asymptotes
Vertical asymptotes occur where the function goes to infinity, which happens where the denominator equals zero. For [tex]\( \frac{1}{(x-4)^2} \)[/tex], the denominator is zero at [tex]\( x = 4 \)[/tex].
Answer: A. The vertical asymptote is [tex]\( x = 4 \)[/tex].
### 5. Horizontal asymptotes
Horizontal asymptotes describe the behavior of the function as [tex]\( x \)[/tex] approaches [tex]\( +\infty \)[/tex] or [tex]\( -\infty \)[/tex]. For [tex]\( f(x) \)[/tex], as [tex]\( x \)[/tex] approaches either [tex]\( +\infty \)[/tex] or [tex]\( -\infty \)[/tex], the denominator [tex]\((x-4)^2\)[/tex] becomes very large, making the function value approach 0.
Answer: A. The horizontal asymptote is [tex]\( y = 0 \)[/tex].
In summary:
- Domain: [tex]\( (-\infty, 4) \cup (4, \infty) \)[/tex]
- x-intercept(s): B. The function has no x-intercepts.
- y-intercept: A. The y-intercept of the function is [tex]\( \left( 0, \frac{1}{16} \right) \)[/tex].
- Vertical asymptote(s): A. The vertical asymptote is [tex]\( x = 4 \)[/tex].
- Horizontal asymptote(s): A. The horizontal asymptote is [tex]\( y = 0 \)[/tex].
### 1. Domain of the function
The function is defined for all [tex]\( x \)[/tex] except where the denominator is zero. The denominator [tex]\((x-4)^2\)[/tex] is zero when [tex]\( x = 4 \)[/tex]. Therefore, the function is not defined at [tex]\( x = 4 \)[/tex].
Domain: [tex]\( (-\infty, 4) \cup (4, \infty) \)[/tex]
### 2. x-intercepts of the function
The x-intercepts are the points where the function crosses the x-axis, i.e., where [tex]\( f(x) = 0 \)[/tex].
For [tex]\( \frac{1}{(x-4)^2} = 0 \)[/tex], there is no value of [tex]\( x \)[/tex] that will make the equation true because the numerator is a constant (1) and the denominator is squared, so it is never zero, thus the function has no x-intercepts.
Answer: B. The function has no x-intercepts.
### 3. y-intercepts of the function
The y-intercept is found by evaluating the function at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{1}{(0-4)^2} = \frac{1}{16} \][/tex]
So the y-intercept is [tex]\( (0, \frac{1}{16}) \)[/tex].
Answer: A. The y-intercept of the function is [tex]\( \left( 0, \frac{1}{16} \right) \)[/tex].
### 4. Vertical asymptotes
Vertical asymptotes occur where the function goes to infinity, which happens where the denominator equals zero. For [tex]\( \frac{1}{(x-4)^2} \)[/tex], the denominator is zero at [tex]\( x = 4 \)[/tex].
Answer: A. The vertical asymptote is [tex]\( x = 4 \)[/tex].
### 5. Horizontal asymptotes
Horizontal asymptotes describe the behavior of the function as [tex]\( x \)[/tex] approaches [tex]\( +\infty \)[/tex] or [tex]\( -\infty \)[/tex]. For [tex]\( f(x) \)[/tex], as [tex]\( x \)[/tex] approaches either [tex]\( +\infty \)[/tex] or [tex]\( -\infty \)[/tex], the denominator [tex]\((x-4)^2\)[/tex] becomes very large, making the function value approach 0.
Answer: A. The horizontal asymptote is [tex]\( y = 0 \)[/tex].
In summary:
- Domain: [tex]\( (-\infty, 4) \cup (4, \infty) \)[/tex]
- x-intercept(s): B. The function has no x-intercepts.
- y-intercept: A. The y-intercept of the function is [tex]\( \left( 0, \frac{1}{16} \right) \)[/tex].
- Vertical asymptote(s): A. The vertical asymptote is [tex]\( x = 4 \)[/tex].
- Horizontal asymptote(s): A. The horizontal asymptote is [tex]\( y = 0 \)[/tex].
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