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Rational Functions Quiz

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This quiz tests the work covered in Lecture 6 and corresponds to the second half of Section 1.6 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).

There is a written explanation at http://id.mind.net/ zona/mmts/functionInstitute/rationalFunctions/rationalFunctions.html which covers the material in the text but it also has links to a variety of applets which are all very useful.

Question 1

Consider 5x3 - x2 - 4 f(x) =--3----2---. 2x + x - 1
Which of the following set of statements is correct?

a)

f(x) → 5 2 asx → - ∞
f(x) → 5 2 asx → +∞
  b)
f(x) → - 52 asx → - ∞
f(x) → 52 asx → + ∞
c)
f(x) → - ∞ asx → - ∞
f(x) → ∞ asx → +∞
  d)
f (x) → ∞ asx → - ∞
f (x) → ∞ asx → +∞

 

Your answer is correct.
5x3 5 f(x)  2x3 = 2 for large positive x and large negative x.
Not correct. Choice (b) is false.
Try again, f(x)  5x3= 5 2x3 2 for large positive x and large negative x.
Not correct. Choice (c) is false.
Try putting some large positive numbers and large negative numbers in to the function and see what happens.
Not correct. Choice (d) is false.
Try putting some large positive numbers and large negative numbers in to the function and see what happens.

Question 2

Which of the following are the x -intercepts for the function below?
  3x2 - 12x+ 9 y = --2--------. x + 5x+ 6

a)
x = - 3, x = - 2.
  b)
x = - 3, x = - 1.
c)
x = 1, x = 9.
  d)
x = 3, x = 1

 

Not correct. Choice (a) is false.
Try again, you have factorized the denominator, not the numerator.
Not correct. Choice (b) is false.
Try again, you may not have factorized the numerator correctly.
Not correct. Choice (c) is false.
Try again, you have not factorized the numerator correctly.
Your answer is correct.
3(x2 - 4x+ 3) 3(x - 3)(x - 1) y = -x2 +-5x-+-6-= -x2-+-5x+-6--.
Hence y = 0 when x = 3 or x = 1 and these are the x -intercepts.

Question 3

Which of the following are the vertical asymptotes for the function below?
  y = 3x2 --12x+-9. x2 + 5x+ 6

a)
x = - 3, x = - 2.
  b)
x = 3, x = 2.
c)
x = 3, x = 1.
  d)
x = - 6, x = 1.

 

Your answer is correct.
y = 3(x2 --4x-+-3)= 3(x---3)(x--1). x2 + 5x+ 6 (x+ 3)(x + 2)
Hence y is undefined when x = - 3 or x = - 2 and these are the vertical asymptotes.
Not correct. Choice (b) is false.
Try again, you may not have factorized the denominator correctly.
Not correct. Choice (c) is false.
Try again, you may have factorized the numerator, not the denominator.
Not correct. Choice (d) is false.
Try again, you may not have factorized the denominator correctly.
Note that x2 + 5x + 6 = (x+ 3)(x+ 2).

Question 4

Which of the following is the horizontal asymptote for   5x2 - 8x+ 3 y = --2---------? 3x + 13x + 4

a)
x = - 1
b)
y = 5
c)
y = 53
d)
Since the numerator and the denominator cannot be factorized we cannot determine the horizontal asymptote.

 

Not correct. Choice (a) is false.
Try again, this is one of the vertical asymptotes.
Remember horizontal asymptotes are of the form y = b.
Not correct. Choice (b) is false.
Try again, look at what happens for large values of x.
Your answer is correct.
5x2 5 f(x)  3x2 = 3 for large positive x and large negative x.
Not correct. Choice (d) is false.
Try again, the numerator and the denominator can be factorized,
but even if they couldn’t be factorized it would make no difference,
we could still find the horizontal asymptote.
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