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Math 111 Quiz Solutions - Week 7, Quizzes of Algebra

Solutions to the math 111 quiz for week 7. It covers topics such as determining if one polynomial is a factor of another, finding horizontal and vertical asymptotes, analyzing the graph of a polynomial function, and solving inequalities.

Typology: Quizzes

Pre 2010

Uploaded on 07/23/2009

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Quiz for Math 111 (CRN 13936), Week 7
Note: Show your work!
Name
1. f(x) = x3+ 7x2+ 4xโˆ’12, g(x) = xโˆ’1, is g(x) a factor of f(x)? Why? (20 points)
Solution:
Recall that โ€xโˆ’ais a factor of polynomial f(x)โ€ just means โ€f(a) = 0โ€. To judge
whether xโˆ’1 is a factor of f(x), just plug 1 into f(x), we have f(1) = 13+7ยท12+4ยท1โˆ’12 = 0,
so g(x) = xโˆ’1 is a factor of f(x).
In this example, f(x)is relatively simple, you can also use polynomial divi-
sion to verify that g(x)is a factor of f(x)(using brute force)
2. Find the horizontal and vertical asymptotes algebraically for rational function f(x)
a) f(x) = x(xโˆ’1)
(xโˆ’2)3
b) f(x) = x(xโˆ’4)
x(xโˆ’5)
Is x= 0 a pole of function fin a)? Is x= 0 a pole of function fin b)? (30 points)
Solution:
Plug x= 0 into f(x) in a), the numerator is 0 and the denominator is not 0, that is
f(0) = 0
4= 0. So x= 0 is not a pole in a) (because 0 is in the domain of f(x) in a) ).
Plug x= 0 into f(x) in b), the numerator is 0 and the denominator is also 0, so x= 0
is not in the domain of f(x) in b), besides, for all the x6= 0, we can cancel xfrom the
numerator and denominator to get f(x) = xโˆ’4
(xโˆ’5) , x 6= 0. So x= 0 is a pole of f(x) in b).
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Quiz for Math 111 (CRN 13936), Week 7

Note: Show your work!

Name

  1. f (x) = x^3 + 7x^2 + 4x โˆ’ 12, g(x) = x โˆ’ 1, is g(x) a factor of f (x)? Why? (20 points)

Solution: Recall that โ€x โˆ’ a is a factor of polynomial f (x)โ€ just means โ€f (a) = 0โ€. To judge whether xโˆ’1 is a factor of f (x), just plug 1 into f (x), we have f (1) = 1^3 +7ยท 12 +4ยท 1 โˆ’12 = 0, so g(x) = x โˆ’ 1 is a factor of f (x). In this example, f (x) is relatively simple, you can also use polynomial divi- sion to verify that g(x) is a factor of f (x) (using brute force)

  1. Find the horizontal and vertical asymptotes algebraically for rational function f (x) a) f (x) =

x(x โˆ’ 1) (x โˆ’ 2)^3 b) f (x) =

x(x โˆ’ 4) x(x โˆ’ 5) Is x = 0 a pole of function f in a)? Is x = 0 a pole of function f in b)? (30 points)

Solution: Plug x = 0 into f (x) in a), the numerator is 0 and the denominator is not 0, that is f (0) = 04 = 0. So x = 0 is not a pole in a) (because 0 is in the domain of f (x) in a) ). Plug x = 0 into f (x) in b), the numerator is 0 and the denominator is also 0, so x = 0 is not in the domain of f (x) in b), besides, for all the x 6 = 0, we can cancel x from the numerator and denominator to get f (x) = (^) (xxโˆ’โˆ’^4 5) , x 6 = 0. So x = 0 is a pole of f (x) in b).

  1. The graph of polynomial function f (x) is as below, answer the following questions (30 points) a) Is the degree of f (x) even or odd? b) Is the leading coefficient of f (x) positive or negative? c) What are the real roots of f (x)? For each root, is the multiplicity even or odd? d) What can you tell about the degree of f (x)? Whatโ€™s the smallest possible degree of f (x)?

-4 (^) -2 2 4

20

40

Solution: a) The two ends of the graph goes in opposite direction, so the degree of f (x) is odd. b) Positive. Because the right end of the graph goes upward. c) -1 (odd), 0 (odd), 2 (even), 4 (odd). d) The degree of f (x) is the sum of the multiplicity of every root. As the smallest odd number is 1 and the smallest even number is 2, the smallest possible degree of f (x) is 1 + 1 + 2 + 1 = 5.

  1. Solve the following inequalities. (20 points) a) | 2 x + 3| < 9 b) |x^2 โˆ’ 2 x โˆ’ 5 | > 2

Solution: a) โˆ’ 9 < 2 x + 3 < 9 โˆ’ 12 < 2 x < 6