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In this problem, we walk through a general procedure of finding horizontal and vertical asymptotes of a function. Fill in the blanks and make the appropriate selections to fill in the details of the following sketch. Let f(x) = -4z²-15z+4 (x+4)(x-4) horizontal line with equation Let's first find all horizontal asymptotes. First, we compute that lim f(x) = Number 1-X is a horizontal asymptote. Next, we evaluate lim f(x), and find that: H-18 lim f(z) exists and is not equal to _lim f(x). Thus, we have found a second horizontal asymptote. I→→∞ I→∞0 lim f(z) exists and is equal to lim f(z). Thus, there isn't a second horizontal asymptote. I→→∞ I→∞0 lim f(x) does not exist. Thus, there isn't a second horizontal asymptote. H→ ∞ Now, we try to find the vertical asymptotes, which can only occur where f(z) is discontinuous. Notice that the function has two discontinuities, at x = -4 and at z = 4. Analyzing f(z) around=-4, we find that At least one of lim f(x), 2-4 lim f(z) is infinite. Thus, z=-4 is indeed a vertical asymptote. z+-4+ Neither lim f(z) nor lim f(z) is infinite. Thus, z = -4 is not a vertical asymptote. z+-4+ I-4 Finally, we analyze f(z) around z = 4. We find that At least one of lim_ f(x), lim f(x) is infinite. Thus, z = 4 is indeed a vertical asymptote. 2+4+ I-4 This tells us that the Neither lim f(z) nor lim f(z) is infinite. Thus, z = 4 is not a vertical asymptote. I-4 2+4+

In this problem, we walk through a general procedure of finding horizontal and vertical asymptotes of a function. Fill in the blanks and make the appropriate selections to fill in the details of the following sketch. Let f(x) = -4z²-15z+4 (x+4)(x-4) horizontal line with equation Let's first find all horizontal asymptotes. First, we compute that lim f(x) = Number 1-X is a horizontal asymptote. Next, we evaluate lim f(x), and find that: H-18 lim f(z) exists and is not equal to _lim f(x). Thus, we have found a second horizontal asymptote. I→→∞ I→∞0 lim f(z) exists and is equal to lim f(z). Thus, there isn't a second horizontal asymptote. I→→∞ I→∞0 lim f(x) does not exist. Thus, there isn't a second horizontal asymptote. H→ ∞ Now, we try to find the vertical asymptotes, which can only occur where f(z) is discontinuous. Notice that the function has two discontinuities, at x = -4 and at z = 4. Analyzing f(z) around=-4, we find that At least one of lim f(x), 2-4 lim f(z) is infinite. Thus, z=-4 is indeed a vertical asymptote. z+-4+ Neither lim f(z) nor lim f(z) is infinite. Thus, z = -4 is not a vertical asymptote. z+-4+ I-4 Finally, we analyze f(z) around z = 4. We find that At least one of lim_ f(x), lim f(x) is infinite. Thus, z = 4 is indeed a vertical asymptote. 2+4+ I-4 This tells us that the Neither lim f(z) nor lim f(z) is infinite. Thus, z = 4 is not a vertical asymptote. I-4 2+4+

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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In this problem, we walk through a general procedure of finding horizontal and vertical asymptotes of a function. Fill in the blanks and make the appropriate selections to fill in the details of the following sketch. Let f(x) = -4z²-15z+4 (x+4)(x-4) horizontal line with equation Let's first find all horizontal asymptotes. First, we compute that_lim_ f(z) = Number I→∞ is a horizontal asymptote. Next, we evaluate lim f(x), and find that: I→→∞0 lim f(z) exists and is not equal to I→→∞ lim f(x) exists and is equal to I→→∞0 ∞--I lim f(z). Thus, we have found a second horizontal asymptote. I→∞ lim f(z). Thus, there isn't a second horizontal asymptote. I→∞ lim f(z) does not exist. Thus, there isn't a second horizontal asymptote. Now, we try to find the vertical asymptotes, which can only occur where f(z) is discontinuous. Notice that the function has two discontinuities, at x = -4 and at z = 4. Analyzing f(z) around + = -4, we find that At least one of lim f(z), lim f(z) is infinite. Thus, z = -4 is indeed a vertical asymptote. I→→4 2+-4+ Neither lim f(z) nor lim f(z) is infinite. Thus, z = -4 is not a vertical asymptote. z+-4+ I→-4- Finally, we analyze f(x) around = 4. We find that At least one of lim_ f(z), lim f(z) is infinite. Thus, z = 4 is indeed a vertical asymptote. I-4 2+4+ This tells us that the Neither lim f(z) nor lim f(z) is infinite. Thus, z = 4 is not a vertical asymptote. I-4 2+4+
Transcribed Image Text:In this problem, we walk through a general procedure of finding horizontal and vertical asymptotes of a function. Fill in the blanks and make the appropriate selections to fill in the details of the following sketch. Let f(x) = -4z²-15z+4 (x+4)(x-4) horizontal line with equation Let's first find all horizontal asymptotes. First, we compute that_lim_ f(z) = Number I→∞ is a horizontal asymptote. Next, we evaluate lim f(x), and find that: I→→∞0 lim f(z) exists and is not equal to I→→∞ lim f(x) exists and is equal to I→→∞0 ∞--I lim f(z). Thus, we have found a second horizontal asymptote. I→∞ lim f(z). Thus, there isn't a second horizontal asymptote. I→∞ lim f(z) does not exist. Thus, there isn't a second horizontal asymptote. Now, we try to find the vertical asymptotes, which can only occur where f(z) is discontinuous. Notice that the function has two discontinuities, at x = -4 and at z = 4. Analyzing f(z) around + = -4, we find that At least one of lim f(z), lim f(z) is infinite. Thus, z = -4 is indeed a vertical asymptote. I→→4 2+-4+ Neither lim f(z) nor lim f(z) is infinite. Thus, z = -4 is not a vertical asymptote. z+-4+ I→-4- Finally, we analyze f(x) around = 4. We find that At least one of lim_ f(z), lim f(z) is infinite. Thus, z = 4 is indeed a vertical asymptote. I-4 2+4+ This tells us that the Neither lim f(z) nor lim f(z) is infinite. Thus, z = 4 is not a vertical asymptote. I-4 2+4+
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Step 1: Define the problem

The given function is f open parentheses x close parentheses equals fraction numerator negative 4 x squared minus 15 x plus 4 over denominator open parentheses x plus 4 close parentheses open parentheses x minus 4 close parentheses end fraction

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