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Multiple Choice Identify the choice that best completes the statement or answers the question. Find any points of discontinuity for the rational function. 1. \( y=\frac{(x-7)(x+2)(x-9)}{(x-5)(x-2)} \) a. \( x=-5, x=-2 \) b. \( x=5, x=2 \) c. \( x=-7, x=2, x=-9 \) d. \( x=7, x=-2, x=9 \) 2. \( y=\frac{(x+7)(x+4)(x+2)}{(x+5)(x-3)} \) a. \( x=-5, x=3 \) b. \( x=7, x=4, x=2 \) c. \( x=-7, x=-4, x=-2 \) d. \( x=5, x=-3 \) 3. \( y=\frac{x+4}{x^{2}+8 x+15} \) a. \( x=-5, x=-3 \) b. \( x=-4 \) c. \( x=-5, x=3 \) d. \( x=5, x=3 \) 4. \( y=\frac{x-3}{x^{2}+3 x-10} \) a. \( x=-5, x=2 \) b. \( x=5, x=-2 \) c. \( x=3 \) d. \( x \) \( =-5, x=-2 \) 6. What are the points of discontinuity? Are they all removable? \[ y=\frac{(x-4)}{x^{2}-13 x+36} \] a. \( x=-9, x=-4, x=8 \); yes b. \( x=1, x=8, x= \) -8; no c. \( x=9, x=4 \); no d. \( x=-9, x=-4 \); no 7. Describe the vertical asymptote(s) and hole(s) for the graph of \( y=\frac{(x-2)(x-5)}{(x-5)(x+2)} \). a. asymptote: \( x=2 \) and hole: \( x=-5 \) b. asymptotes: \( x=-2 \) and hole: \( x=-5 \) c. asymptote: \( x=-2 \) and hole: \( x=5 \) d. asymptote: \( x=-2 \) and hole: \( x=-2 \) a. \( x=-3, x=-8 \); no b. \( x=5, x=-7, x=1 \); no c. \( x=-5, x=7, x=-1 \); yes d. \( x=3, x=8 \); yes 8. Describe the vertical asymptote(s) and hole(s) for the graph of \( y=\frac{x+1}{x^{2}+6 x+5} \). a. asymptote: \( x=-1 \) and hole: \( x=-1 \) b. asymptote: \( x=-1 \) and hole: \( x=-1 \) c. asymptotes: \( x=-1,-1 \) and hole: \( x=-1 \) d. asymptote: -5 and hole: \( x=-1 \), 9. Find the horizontal asymptote of the graph of \( y=\frac{7 x^{6}+7 x+3}{9 x^{5}+7 x+3} \). a. \( y=0 \) b. \( y=\frac{7}{9} \) c. no horizontal asymptote d. \( y=\frac{6}{5} \)

Ask by Pena Carter. in the United States
Jan 27,2025

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Answer

**Points of Discontinuity and Asymptotes:** 1. **Question 1:** - **Discontinuities:** \( x = 5, x = 2 \) - **Answer:** b. \( x = 5, x = 2 \) 2. **Question 2:** - **Discontinuities:** \( x = -5, x = 3 \) - **Answer:** a. \( x = -5, x = 3 \) 3. **Question 3:** - **Discontinuities:** \( x = -5, x = -3 \) - **Answer:** a. \( x = -5, x = -3 \) 4. **Question 4:** - **Discontinuities:** \( x = -5, x = 2 \) - **Answer:** a. \( x = -5, x = 2 \) 5. **Question 6:** - **Discontinuities:** \( x = 9, x = 4 \) (Non-removable at \( x = 9 \)) - **Answer:** c. \( x = 9, x = 4 \); no 6. **Question 7:** - **Asymptote:** \( x = -2 \) - **Hole:** \( x = 5 \) - **Answer:** c. asymptote: \( x = -2 \) and hole: \( x = 5 \) 7. **Question 8:** - **Asymptote:** \( x = -5 \) - **Hole:** \( x = -1 \) - **Answer:** d. asymptote: \( x = -5 \) and hole: \( x = -1 \) 8. **Question 9:** - **No Horizontal Asymptote** - **Answer:** c. no horizontal asymptote **Summary:** - **Discontinuities:** Identify where the denominator is zero. - **Asymptotes:** Vertical asymptotes occur where the denominator is zero and the numerator is not, while holes occur where both numerator and denominator have common factors that cancel out. - **Horizontal Asymptotes:** Determine based on the degrees of the numerator and denominator. If you need further clarification on any of these points, feel free to ask!

Solution

Sure, let's go through each question one by one to identify the correct answers. --- ### **1.** \( y = \frac{(x-7)(x+2)(x-9)}{(x-5)(x-2)} \) **Points of Discontinuity:** - **Denominator Zeros:** \( x = 5 \) and \( x = 2 \) - **Numerator Factors:** \( x = 7, -2, 9 \) (No cancellation with denominator) **Answer:** **b. \( x = 5, x = 2 \)** --- ### **2.** \( y = \frac{(x+7)(x+4)(x+2)}{(x+5)(x-3)} \) **Points of Discontinuity:** - **Denominator Zeros:** \( x = -5 \) and \( x = 3 \) - **Numerator Factors:** \( x = -7, -4, -2 \) (No cancellation with denominator) **Answer:** **a. \( x = -5, x = 3 \)** --- ### **3.** \( y = \frac{x+4}{x^{2} + 8x + 15} \) **Points of Discontinuity:** - **Simplify Denominator:** \( x^{2} + 8x + 15 = (x+5)(x+3) \) - **Denominator Zeros:** \( x = -5 \) and \( x = -3 \) - **Numerator Factor:** \( x = -4 \) (No cancellation with denominator) **Answer:** **a. \( x = -5, x = -3 \)** --- ### **4.** \( y = \frac{x-3}{x^{2} + 3x - 10} \) **Points of Discontinuity:** - **Simplify Denominator:** \( x^{2} + 3x - 10 = (x+5)(x-2) \) - **Denominator Zeros:** \( x = -5 \) and \( x = 2 \) - **Numerator Factor:** \( x = 3 \) (No cancellation with denominator) **Answer:** **a. \( x = -5, x = 2 \)** --- ### **6.** \( y = \frac{(x-4)}{x^{2} - 13x + 36} \) **Points of Discontinuity:** - **Simplify Denominator:** \( x^{2} - 13x + 36 = (x-9)(x-4) \) - **Denominator Zeros:** \( x = 9 \) and \( x = 4 \) - **Numerator Factor:** \( x = 4 \) (Cancels with denominator, creating a hole) - **Non-Removable Discontinuity:** \( x = 9 \) **Are All Discontinuities Removable?** No, \( x = 9 \) is not removable. **Answer:** **c. \( x = 9, x = 4 \); no** --- ### **7.** \( y = \frac{(x-2)(x-5)}{(x-5)(x+2)} \) **Vertical Asymptotes and Holes:** - **Simplify:** \( y = \frac{(x-2)\cancel{(x-5)}}{\cancel{(x-5)}(x+2)} = \frac{x-2}{x+2} \) - **Hole:** At \( x = 5 \) (where \( x-5 \) was canceled) - **Vertical Asymptote:** At \( x = -2 \) (denominator zero) **Answer:** **c. asymptote: \( x = -2 \) and hole: \( x = 5 \)** --- ### **8.** \( y = \frac{x+1}{x^{2} + 6x + 5} \) **Vertical Asymptotes and Holes:** - **Simplify Denominator:** \( x^{2} + 6x + 5 = (x+1)(x+5) \) - **Simplify Expression:** \( y = \frac{\cancel{(x+1)}}{\cancel{(x+1)}(x+5)} = \frac{1}{x+5} \) - **Hole:** At \( x = -1 \) - **Vertical Asymptote:** At \( x = -5 \) **Answer:** **d. asymptote: \( x = -5 \) and hole: \( x = -1 \)** --- ### **9.** \( y = \frac{7x^{6} + 7x + 3}{9x^{5} + 7x + 3} \) **Horizontal Asymptote:** - **Degree of Numerator:** 6 - **Degree of Denominator:** 5 - **Rule:** If the degree of the numerator is greater than the degree of the denominator, **no horizontal asymptote** exists. **Answer:** **c. no horizontal asymptote** --- ## **Summary of Answers:** 1. **b.** \( x = 5, x = 2 \) 2. **a.** \( x = -5, x = 3 \) 3. **a.** \( x = -5, x = -3 \) 4. **a.** \( x = -5, x = 2 \) 6. **c.** \( x = 9, x = 4 \); no 7. **c.** asymptote: \( x = -2 \) and hole: \( x = 5 \) 8. **d.** asymptote: \( x = -5 \) and hole: \( x = -1 \) 9. **c.** no horizontal asymptote --- If you have any further questions or need explanations for any of the problems, feel free to ask!

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To find points of discontinuity in rational functions, first identify values of \( x \) that make the denominator equal to zero, since these are where the function cannot be defined. For option 1, when you set \( (x-5)(x-2)=0 \), you get \( x=5 \) and \( x=2 \) as points of discontinuity. Similarly, apply this method to other functions to arrive at the correct answers. Vertical asymptotes occur at those values of \( x \) that make the denominator zero but are not cancelled out by the numerator. If both the numerator and denominator have a common factor that cancels, it creates a removable discontinuity (or a hole) at that x-value. For instance, in the function \( y=\frac{(x-2)(x-5)}{(x-5)(x+2)} \), \( x=5 \) is a hole while \( x=-2 \) is a vertical asymptote. Stay playful with these concepts, and soon you'll be spotting discontinuities like a pro!

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