ExitExam+ESSLCE Support
Open in Telegram
The aim of this channel is to offer educational resources and tutorials to assist students.
Show moreThe country is not specifiedEducation42 110
4 101
Subscribers
No data24 hours
-157 days
-10130 days
Posts Archive
3️⃣9️⃣ The degree and leading coefficient of the polynomial function f(x) = 3x^4 - 2x^3 + 5x^2 - 7x +1 are:
3️⃣8️⃣What is the domain of the function f(x) = (x^2 - 4)/(x^2 - 9)?
3️⃣7️⃣What is the range of the function
f(x) = 1/(x^2 + 1)?
3️⃣6️⃣ What is the domain of the function f(x) = 1/(x^2 + 1)?
3️⃣5️⃣What is the range of the function f(x) = (x + 2)/(x^2 - 4)?
3️⃣4️⃣What is the domain of the rational function f(x) = (x + 2)/(x^2 - 4)?
✅SOLUTION3️⃣3️⃣👆👆
📚To find the equation of the oblique asymptote of the function
f(x) = (x^2 − 11x + 30)/(x − 4),
Use long division or synthetic division to divide the numerator by the denominator.
📌We will get a quotient and a remainder.
📌The quotient will be the equation of the oblique asymptote.
Long division gives:
x - 7
____
x - 4 | x^2 - 11x + 30
x^2 - 4x
----------
-7x + 30
-7x + 28
--------
2
📝Therefore, the quotient is x - 7, and the remainder is 2/(x - 4).
As x approaches infinity or negative infinity, the remainder approaches 0, so the equation of the oblique asymptote is
y = x - 7.
Therefore, the answer is (C)
📣📣
🙏HELP THE CHANNEL TO REACH TO THOSE WHO NEEDS IT🙏
🙏 Share👇👇
https://t.me/tutorialpointeth
✅SOLUTION3️⃣2️⃣👆👆
📚We are given that the graph of the function:
👉f(x) = (3x-k)/(2x+k)
passes through the point (3, 2).
📚This means that when x = 3, f(x) = 2. We use this information to find for k.
Substituting x = 3 and f(x) = 2 into the equation for f(x), we get:
2 = [3(3) - k]/[2(3) + k]
Simplifying this equation, we get:
2(2(3) + k) = 9 - k
Expanding the left side of the equation, we get:
12 + 2k = 9 - k
Collecting similar terms, we get:
3k = -3
Dividing both sides by 3, gives:
k = -1
📚Therefore, the value of k that makes the graph of f(x) pass through the point (3, 2) is k = -1.
📣📣
🙏HELP THE CHANNEL TO REACH TO THOSE WHO NEEDS IT🙏
🙏 Share👇👇
https://t.me/tutorialpointeth
✅SOLUTION3️⃣1️⃣👆👆
📚horizontal asymptote of a a rational function is the value that the function approaches as x approaches positive or negative infinity.
📚A function has a horizontal asymptote y=0 if and only if the limit of the function as x approaches infinity or negative infinity is equal to 0.
📌Choice A)
To find the horizontal asymptote of f(x) = 2x/(3-x), we can divide the numerator and denominator by x. This gives us f(x) = 2/(1 - 3/x). As x approaches infinity or negative infinity, 3/x approaches 0, so the denominator approaches 1. Therefore, the limit of f(x) as x approaches infinity or negative infinity is 2/1 = 2, which is not equal to 0. Therefore, f(x) does not have a horizontal asymptote y=0 but y=2 is its horizontalasymptote.
📌Choice B) To find the horizontal asymptote of f(x) = 1/(x-4), we can see that as x approaches infinity or negative infinity, the denominator approaches infinity or negative infinity, respectively. Therefore, the limit of f(x) as x approaches infinity or negative infinity is 0. Therefore, f(x) has a horizontal asymptote y=0.
📌Choice C) To find the horizontal asymptote of f(x) = (x^2-5x+6)/(x-3), we can use long division or synthetic division to divide the numerator by the denominator. This gives us f(x) = (x-2) + 3/(x-3). As x approaches infinity or negative infinity, 3/(x-3) approaches 0, so the limit of f(x) as x approaches infinity or negative infinity is x-2, which is not equal to 0.
👉Therefore, f(x) does not have a horizontal asymptote y=0, but y=x-2 is an oblique asymptote to f(x).
📌Choice D) To find the horizontal asymptote of f(x) = (x^2-x)/(2x^2+4), we can divide the numerator and denominator by x^2. This gives us f(x) = (1 - 1/x)/(2 + 4/x^2). As x approaches infinity or negative infinity, 1/x and 4/x^2 approach 0, so the denominator approaches 2. Therefore, the limit of f(x) as x approaches infinity or negative infinity is (1-0)/2 = 1/2, which is not equal to 0. Therefore, f(x) does not have a horizontal asymptote y=0.
📝Therefore, the only function that has a horizontal asymptote y=0 is f(x) = 1/(x-4), so the answer is (B).
3️⃣3️⃣Find the equation of the oblique asymptote of the function
f (x) = (x^2 − 11x + 30)/(x − 4)?
3️⃣2️⃣What is the value of k in the function f(x)=(3x-k)/(2x+k) if its graph passes through the point (3, 2)?
3️⃣1️⃣Which of the following functions has horizontal asymptote y=0?
3️⃣0️⃣Consider the rational function g(x) = (3x + 1)/(x^2 - 9). Which of the following statements about g(x) is NOT true?
2️⃣9️⃣What is the domain of the rational function f(x) = (x +4)/(x^2 - 9)?
2️⃣8️⃣Which of the following is the horizontal asymptote of the rational function
f(x) = (3x^2 + 2x - 1)/(2x^2 - 5x + 3)?
2️⃣6️⃣Which of the following is NOT a factor of the polynomial function
f(x) = x^3 - 4x^2 + x + 6?
2️⃣5️⃣What is the remainder when f(x) = 2x^3 - 5x^2 + 3x - 1 is divided by (x - 2)?
2️⃣4️⃣What is the degree of the polynomial function f(x) = 3x^4 - 2x^3 + 5x^2 - 7x + 1?
