Answer :
To graph the function [tex]\( f(x) = x^2 + \frac{8}{x} \)[/tex] step by step and find its domain, we follow these procedures:
1. Write [tex]\( f(x) \)[/tex] in its lowest terms:
The given function is [tex]\( f(x) = x^2 + \frac{8}{x} \)[/tex]. We observe that it cannot be further simplified. Therefore, the function is already in its lowest terms.
Answer: B. [tex]\( f(x) \)[/tex] is already in lowest terms.
2. Determine the domain of the function.
For the function [tex]\( f(x) = x^2 + \frac{8}{x} \)[/tex], the domain consists of all real numbers except where the denominator equals zero, which would make the function undefined. The term [tex]\(\frac{8}{x}\)[/tex] suggests [tex]\( x \neq 0 \)[/tex].
The domain is: All real numbers [tex]\( x \)[/tex] except [tex]\( x \neq 0 \)[/tex].
3. Find asymptotes:
- Vertical asymptote: Occurs where the function is undefined. As noted, [tex]\( f(x) \)[/tex] is undefined when [tex]\( x = 0 \)[/tex]. Hence, there is a vertical asymptote at [tex]\( x = 0 \)[/tex].
- Horizontal/Oblique asymptote: As [tex]\( x \to \infty \)[/tex] or [tex]\( x \to -\infty \)[/tex], the term [tex]\( \frac{8}{x} \)[/tex] approaches 0. Thus, the function [tex]\( f(x) \approx x^2 \)[/tex]. Since [tex]\( x^2 \)[/tex] grows without bound, there is no horizontal asymptote. However, we should note that for large [tex]\( |x| \)[/tex], the function behaves similarly to [tex]\( y = x^2 \)[/tex].
4. Find intercepts:
- x-intercept: Set [tex]\( f(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\( x^2 + \frac{8}{x} = 0 \)[/tex] implies [tex]\( x^3 = -8 \)[/tex]. Solving for [tex]\( x \)[/tex], we get [tex]\( x = -2 \)[/tex].
So, the x-intercept is [tex]\( (-2, 0) \)[/tex].
- y-intercept: This is found by evaluating [tex]\( f(0) \)[/tex]. However, [tex]\( f(0) \)[/tex] is undefined since [tex]\( x = 0 \)[/tex] makes the function undefined. Therefore, there is no y-intercept.
5. Critical points and behavior:
We analyze the behavior as [tex]\( x \to 0 \)[/tex] from both the positive and negative sides, and as [tex]\( |x| \to \infty \)[/tex]:
- As [tex]\( x \to 0^+ \)[/tex], [tex]\( \frac{8}{x} \to \infty \)[/tex], so [tex]\( f(x) \to \infty \)[/tex].
- As [tex]\( x \to 0^- \)[/tex], [tex]\( \frac{8}{x} \to -\infty \)[/tex], so [tex]\( f(x) \to -\infty \)[/tex].
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \approx x^2 \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \approx x^2 \)[/tex].
Using this information, we can graph the function and mark important features such as vertical asymptote at [tex]\( x = 0 \)[/tex], x-intercept at [tex]\( (-2, 0) \)[/tex], and the overall behavior described above. The graph will show that as [tex]\( |x| \)[/tex] gets large, [tex]\( f(x) \)[/tex] behaves like [tex]\( x^2 \)[/tex], and a vertical asymptote illustrates the undefined nature at [tex]\( x = 0 \)[/tex].
By linking all these points and behaviors, you should be able to sketch an accurate graph of the function [tex]\( f(x) = x^2 + \frac{8}{x} \)[/tex].
1. Write [tex]\( f(x) \)[/tex] in its lowest terms:
The given function is [tex]\( f(x) = x^2 + \frac{8}{x} \)[/tex]. We observe that it cannot be further simplified. Therefore, the function is already in its lowest terms.
Answer: B. [tex]\( f(x) \)[/tex] is already in lowest terms.
2. Determine the domain of the function.
For the function [tex]\( f(x) = x^2 + \frac{8}{x} \)[/tex], the domain consists of all real numbers except where the denominator equals zero, which would make the function undefined. The term [tex]\(\frac{8}{x}\)[/tex] suggests [tex]\( x \neq 0 \)[/tex].
The domain is: All real numbers [tex]\( x \)[/tex] except [tex]\( x \neq 0 \)[/tex].
3. Find asymptotes:
- Vertical asymptote: Occurs where the function is undefined. As noted, [tex]\( f(x) \)[/tex] is undefined when [tex]\( x = 0 \)[/tex]. Hence, there is a vertical asymptote at [tex]\( x = 0 \)[/tex].
- Horizontal/Oblique asymptote: As [tex]\( x \to \infty \)[/tex] or [tex]\( x \to -\infty \)[/tex], the term [tex]\( \frac{8}{x} \)[/tex] approaches 0. Thus, the function [tex]\( f(x) \approx x^2 \)[/tex]. Since [tex]\( x^2 \)[/tex] grows without bound, there is no horizontal asymptote. However, we should note that for large [tex]\( |x| \)[/tex], the function behaves similarly to [tex]\( y = x^2 \)[/tex].
4. Find intercepts:
- x-intercept: Set [tex]\( f(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\( x^2 + \frac{8}{x} = 0 \)[/tex] implies [tex]\( x^3 = -8 \)[/tex]. Solving for [tex]\( x \)[/tex], we get [tex]\( x = -2 \)[/tex].
So, the x-intercept is [tex]\( (-2, 0) \)[/tex].
- y-intercept: This is found by evaluating [tex]\( f(0) \)[/tex]. However, [tex]\( f(0) \)[/tex] is undefined since [tex]\( x = 0 \)[/tex] makes the function undefined. Therefore, there is no y-intercept.
5. Critical points and behavior:
We analyze the behavior as [tex]\( x \to 0 \)[/tex] from both the positive and negative sides, and as [tex]\( |x| \to \infty \)[/tex]:
- As [tex]\( x \to 0^+ \)[/tex], [tex]\( \frac{8}{x} \to \infty \)[/tex], so [tex]\( f(x) \to \infty \)[/tex].
- As [tex]\( x \to 0^- \)[/tex], [tex]\( \frac{8}{x} \to -\infty \)[/tex], so [tex]\( f(x) \to -\infty \)[/tex].
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \approx x^2 \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \approx x^2 \)[/tex].
Using this information, we can graph the function and mark important features such as vertical asymptote at [tex]\( x = 0 \)[/tex], x-intercept at [tex]\( (-2, 0) \)[/tex], and the overall behavior described above. The graph will show that as [tex]\( |x| \)[/tex] gets large, [tex]\( f(x) \)[/tex] behaves like [tex]\( x^2 \)[/tex], and a vertical asymptote illustrates the undefined nature at [tex]\( x = 0 \)[/tex].
By linking all these points and behaviors, you should be able to sketch an accurate graph of the function [tex]\( f(x) = x^2 + \frac{8}{x} \)[/tex].
