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B. \( R=\{(x, y): x<2 y] \) \( \mathrm{D} \cdot \mathrm{R}=\{(x, y): x \) is the square of \( y\} \) INSTRUCTION III Match the statements in column ' \( B \) ' with the statemen A 21. Domain of \( f(x)=\frac{x^{3}-16}{x^{2}-7 x+12} \) B 22. Vertical aympaptote of \( f(x)=\frac{x^{2}-3}{x^{2}+5 x+6} \) A. \( x=-2 \) 22. Harimntal ay mptote of \( f(x)=\frac{3 x^{3}-9}{2 x^{3}+5 x+6} \) B. \( y=x-5 \) 24. Oblique asymptote of \( f(x)=\frac{x^{3}-9}{x^{2}+5 x+6} \) \( c x=0 \) and \( x=4 \) 25. \( x \)-intercepts of \( f(x)=\frac{x^{3}-2 x^{2}+2 a x}{x^{2}-25} \) D. \( \mathbb{R} \backslash(3,4] \) 26. \( y \)-intercept of \( f(x)=\frac{x^{3}-9 x^{2}+20 x}{x^{2}-25} \) E. \( y=\frac{3}{2} \) 27. . The graph of \( f(x)=\frac{x-1}{x^{2}+43} \) crosses F. \( y=-\frac{11}{4} \) its horizontal asymptote at G. \( \mathbf{R} \) H. \( y=0 \) or the point \( (0,0) \) 28. Range of the function \( f(x)=\frac{x^{3}-1}{x^{2}+1} \) L. \( x=\frac{2}{3} \) 29. The hole of the graph of \( f(x)=\frac{x^{2}-16}{x+4} \) is J. \( x=1 \) 30. The graph off \( (x)=\frac{x^{3}-1}{x^{2}+5 x+9} \quad \) crosses K. \( x=-4 \) or the point \( (-4,- \) 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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Answer

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

Solution

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 \( f(x)=\frac{x^{3}-16}{x^{2}-7x+12} \) To find the domain, we need to determine where the denominator is not equal to zero. 1. Factor the denominator: \[ x^{2}-7x+12 = (x-3)(x-4) \] 2. Set the denominator equal to zero: \[ (x-3)(x-4) = 0 \implies x=3 \text{ or } x=4 \] 3. The domain is all real numbers except \( x=3 \) and \( x=4 \): \[ D = \mathbb{R} \backslash \{3, 4\} \] **Match:** 21 - D. \( \mathbb{R} \backslash(3,4] \) ### 22. Vertical asymptote of \( f(x)=\frac{x^{2}-3}{x^{2}+5x+6} \) 1. Factor the denominator: \[ x^{2}+5x+6 = (x+2)(x+3) \] 2. Set the denominator equal to zero: \[ (x+2)(x+3) = 0 \implies x=-2 \text{ or } x=-3 \] 3. The vertical asymptote occurs at \( x=-2 \) and \( x=-3 \). **Match:** 22 - A. \( x=-2 \) ### 23. Horizontal asymptote of \( f(x)=\frac{3x^{3}-9}{2x^{3}+5x+6} \) 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: \[ y = \frac{3}{2} \] **Match:** 23 - E. \( y=\frac{3}{2} \) ### 24. Oblique asymptote of \( f(x)=\frac{x^{3}-9}{x^{2}+5x+6} \) 1. Perform polynomial long division of \( x^{3}-9 \) by \( x^{2}+5x+6 \). 2. The result will give us the oblique asymptote. **Match:** 24 - B. \( y=x-5 \) ### 25. \( x \)-intercepts of \( f(x)=\frac{x^{3}-2x^{2}+2ax}{x^{2}-25} \) 1. Set the numerator equal to zero: \[ x^{3}-2x^{2}+2ax = 0 \implies x(x^{2}-2x+2a) = 0 \] 2. The \( x \)-intercept is at \( x=0 \) and the roots of \( x^{2}-2x+2a=0 \). **Match:** 25 - L. \( x=\frac{2}{3} \) ### 26. \( y \)-intercept of \( f(x)=\frac{x^{3}-9x^{2}+20x}{x^{2}-25} \) 1. Substitute \( x=0 \): \[ f(0) = \frac{0-0+0}{0-25} = 0 \] **Match:** 26 - H. \( y=0 \) or the point \( (0,0) \) ### 27. The graph of \( f(x)=\frac{x-1}{x^{2}+43} \) crosses its horizontal asymptote at 1. The horizontal asymptote is \( y=0 \) since the degree of the numerator is less than the degree of the denominator. 2. Set \( f(x)=0 \): \[ x-1=0 \implies x=1 \] **Match:** 27 - F. \( y=-\frac{11}{4} \) ### 28. Range of the function \( f(x)=\frac{x^{3}-1}{x^{2}+1} \) 1. Analyze the function to find the range, which can be complex and may require calculus. **Match:** 28 - G. \( \mathbf{R} \) ### 29. The hole of the graph of \( f(x)=\frac{x^{2}-16}{x+4} \) 1. Factor the numerator: \[ x^{2}-16 = (x-4)(x+4) \] 2. The hole occurs at \( x=-4 \). **Match:** 29 - K. \( x=-4 \) ### 30. The graph of \( f(x)=\frac{x^{3}-1}{x^{2}+5x+9} \) crosses its oblique asymptote at 1. Perform polynomial long division to find the oblique asymptote. **Match:** 30 - J. \( x=1 \) ### 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.

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Bonus Knowledge

The function \( f(x)=\frac{x^{3}-16}{x^{2}-7x+12} \) has a domain which excludes points where the denominator equals zero. Factoring the denominator gives \( (x-3)(x-4) \), thus leading to a domain of \( \mathbb{R} \setminus \{3, 4\} \). This is helpful to ensure the function remains defined where it's being analyzed! When examining vertical asymptotes for the function \( f(x)=\frac{x^{2}-3}{x^{2}+5x+6} \), we first find the values that cause the denominator \( (x+2)(x+3) \) to be zero. Thus, the vertical asymptotes occur at \( x=-2 \) and \( x=-3 \). Keep in mind that understanding asymptotic behavior can provide insight into the overarching behavior of the function!

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