Precalculus Textbook 8th Edition Section 2.6 Exercise 19 Solution
Question:
Find the Horizontal and Vertical asymptotes of the function f(x) = (2x2 – 1) / (x2 + 3). Use the limits to describe the corresponding behavior.
Online Precalculus Course |
One to One Student Class Fee: 20 USD Per Hour |
One to Five Student Class Fee: 15 USD Per Hour |
Batch 1 Timing (IST-Indian Standard Time): 3:00 AM to 4:00 AM |
Batch 2 Timing (IST-Indian Standard Time): 4:00 AM to 5:00 AM |
Batch 3 Timing (IST-Indian Standard Time): 5:00 AM to 6:00 AM |
Batch 4 Timing (IST-Indian Standard Time): 6:00 AM to 7:00 AM |
Batch 5 Timing (IST-Indian Standard Time): 7:00 AM to 8:00 AM |
Batch 6 Timing (IST-Indian Standard Time): 8:00 AM to 9:00 AM |
For More Details:Click Here to WhatsApp |
Solution:
Given polynomial is f(x) = (2x2 – 1) / (x2 + 3)
Horizontal asymptote of a function is line y = k where k = Value of f(x) at x tends to infinity.
Vertical asymptote of a function is line x = t where t = Values of x at which function f(x) is undefined.
Now,
Thus, the horizontal asymptote of the given function is y = 2.
Hence, the end behaviors of the given function are .
Since the given function f(x) = (2x2 – 1) / (x2 + 3), it is defined everywhere as its denominator (x2 + 3)
is non-zero for any real value of x.
Thus, there is NO vertical asymptote of the given function.
Precalculus Algebra Tutors
For More details about the AP Precalculus Course Details, Click Here.
For more update about AMBIPi Tutors, Join WhatsApp Channel.