Pregunta
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B.
is the square of
INSTRUCTION III Match the statements in column ’ ’ with the statemen A
21. Domain of
B
22. Vertical aympaptote of
A.
22. Harimntal ay mptote of
B.
24. Oblique asymptote of
and
25. -intercepts of
D.
26. -intercept of
E.
27. . The graph of crosses
F.
its horizontal asymptote at
G.
H. or the point
28. Range of the function
L.
29. The hole of the graph of is
J.
30. The graph off crosses
K. or the point
Its oblique asymptote at
DIRECTION V:- Work out the following problems clearly and correctly

Ask by Fowler Kelley. in Ethiopia
Jan 31,2025

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21 - D
22 - A
23 - E
24 - B
25 - L
26 - H
27 - F
28 - G
29 - K
30 - J

Solución

To solve the matching problem, we need to analyze each statement in column B and match it with the appropriate statement in column A. Let’s go through each item step by step.

21. Domain of

To find the domain, we need to determine where the denominator is not equal to zero.
  1. Factor the denominator:
  2. Set the denominator equal to zero:
  3. The domain is all real numbers except and :
Match: 21 - D.

22. Vertical asymptote of

  1. Factor the denominator:
  2. Set the denominator equal to zero:
  3. The vertical asymptote occurs at and .
Match: 22 - A.

23. Horizontal asymptote of

  1. The degrees of the numerator and denominator are the same (both are 3).
  2. The horizontal asymptote is given by the ratio of the leading coefficients:
Match: 23 - E.

24. Oblique asymptote of

  1. Perform polynomial long division of by .
  2. The result will give us the oblique asymptote.
Match: 24 - B.

25. -intercepts of

  1. Set the numerator equal to zero:
  2. The -intercept is at and the roots of .
Match: 25 - L.

26. -intercept of

  1. Substitute :
Match: 26 - H. or the point

27. The graph of crosses its horizontal asymptote at

  1. The horizontal asymptote is since the degree of the numerator is less than the degree of the denominator.
  2. Set :
Match: 27 - F.

28. Range of the function

  1. Analyze the function to find the range, which can be complex and may require calculus.
Match: 28 - G.

29. The hole of the graph of

  1. Factor the numerator:
  2. The hole occurs at .
Match: 29 - K.

30. The graph of crosses its oblique asymptote at

  1. Perform polynomial long division to find the oblique asymptote.
Match: 30 - J.

Summary of Matches:

  • 21 - D
  • 22 - A
  • 23 - E
  • 24 - B
  • 25 - L
  • 26 - H
  • 27 - F
  • 28 - G
  • 29 - K
  • 30 - J
This concludes the matching of statements from column B with those in column A.

Respondido por UpStudy AI y revisado por un tutor profesional

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Bonus Knowledge

The function has a domain which excludes points where the denominator equals zero. Factoring the denominator gives , thus leading to a domain of . This is helpful to ensure the function remains defined where it’s being analyzed!
When examining vertical asymptotes for the function , we first find the values that cause the denominator to be zero. Thus, the vertical asymptotes occur at and . Keep in mind that understanding asymptotic behavior can provide insight into the overarching behavior of the function!

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