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INSTRUCTORS Raffi Hovasapian John Zhu
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Lecture Comments (13)

0 answers

Post by Nolan Zhang on April 9, 2015

If the degrees are the same, what I did was to divide the coefficients instead of their degrees. It worked out for a function like (17x^2-8x)/(4X^2+9), because if you divide their degrees, you will always get y=1 which appears not to be the case with the graph of that particular function. Please correct me if I'm wrong.(im talking about ex #3)

0 answers

Post by John Michael Musaazi on August 9, 2014

for the last example why is it (x-2)(x-5) instead of (x+2)(x-5)

0 answers

Post by Julia Qiu on February 21, 2014

The graph of the first example that you draw is different from what I have done on the calculators. I think the behavior of your final function maybe have some problem.

0 answers

Post by John Michael Musaazi on January 8, 2014

How do you factor these equations and bring them down to a small number,am sorry i have not done math for a while.

0 answers

Post by Mirza Baig on December 18, 2013

I don't get it why do you guys look at your paper

2 answers

Last reply by: John Zhu
Mon Aug 12, 2013 8:52 PM

Post by Taylor Wright on August 11, 2013

In example 3:

Wouldn't there be another vertical asymptote at -4?

1 answer

Last reply by: John Zhu
Mon Aug 12, 2013 8:48 PM

Post by Taylor Wright on August 11, 2013

In example 2:


(4-x) does not equal -(x+4)

(4-x) = -(x-4)    but    -(x+4) = -4-x

2 answers

Last reply by: Angela Patrick
Sun Sep 1, 2013 2:27 PM

Post by Taylor Wright on August 11, 2013

How could there be a Horizontal Asymptote at y=0 if the point (1,0) is on the graph???

Rational Functions

  • Factor to get terms, where c is a constant
  • Numerator: zeros (x-intercepts)
  • Denominator: Asymptotes (undefined x values)
  • Plug-in values into function to get point of function

Rational Functions

Find the roots of x2 + 5x + 6
  • x2 + 5x + 6 = (x + 3)(x + 2)
  • (x + 3)(x + 2) = 0
x = −3, −2
Find the roots of 4x2 + 8x + 2
  • 4x2 + 8x + 2 = 2(2x2 + 4x + 1) = 2(2x + 1)2
Two identical roots, x = −[1/2], −[1/2]
Find the asymptote(s) of f(x) = [1/(4x2 + 8x + 2)]
We already found the roots on the previous problems. The asymptote is where the function is undefined. Here, it's when the denominator is equal to zero. We know that the denominator is equal to zero when x = −[1/2]. The function goes to positive infinity when approaching the root (−[1/2]) from the positive direction. The function also goes to positive infinity when approaching the root from the negative direction (The degree of the polynomial in the denominator is even).
Graph y = [1/(4x2 + 8x + 2)]
  • We already determined the asymptote behavior, what about the edge behavior? As x gets very large and positve, y approaches zero. As x gets very large and negative, y also approaches zero from the positive side.
Find the asymptote(s) of f(x) = [1/x]
The function is undefined at x = 0. The function goes to positive infinity when approaching zero from the positive direction. The function goes to negative infinity when approaching zero from the negative direction (Remember, the polynomial in the denominator is of odd degree).
Find the asymptote(s) of f(x) = tanx
  • tanx = [sinx/cosx]. The function is undefined when cosx = 0. cosx = 0 when x = ..., −[(π)/2], [(π)/2], [(3π)/2], [(5π)/2], ...
There are an infinite number of asymptotes for f(x) = tanx and those asymptotes are at odd integer multiples of [(π)/2]
Graph y = [x/(x2 + 10x + 25)] + [5/(x2 + 10x + 25)]
  • y = [x/(x2 + 10x + 25)] + [5/(x2 + 10x + 25)]
  • = [(x + 5)/(x2 + 10x + 25)]
  • = [(x + 5)/((x + 5)2)]
  • = [1/(x + 5)]
  • This is similar to the graph of y = [1/x], but it is shifted 5 to the left.
Graph y = [1/(x2 − 4)]
  • Even degree polynomial in the denominator. As x gets very large in either direction, y approaches zero. There are two asymptotes here, -2 and 2. That gives us 4 different cases for asymptotic behavior. Approaching -2 from the left, Approaching -2 from the right, approaching 2 from the left, and approaching 2 from the right. Plugging in values to test for these cases.
  • f(−2.001) ≈ 250
  • f(−1.999) ≈ −250
  • f(1.999) ≈ −250
  • f(2.001) ≈ 250
Graph y = [x/(x2 −4)]
  • Asymptotes are at same locations as previous problem, so we can find the asymptote behavior in a similar fashion
  • f(−2.001) ≈ −500
  • f(−1.999) ≈ 500
  • f(1.999) ≈ −500
  • f(2.001) ≈ 500
Graph f(x) = [1/lnx]
  • Remember the properties of lnx. Domain (0, ∞), ln1 = 0, and lnx is an increasing function. As x gets larger, f(x) approaches zero and f(x) has an asymptote at x = 1.
  • f(.999) ≈ −1000
  • f(1.001) ≈ 1000

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Rational Functions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Types of Functions: Rational - Definition 0:06
  • Example 1: Graph Rational Func. 0:36
  • Example 2: Find Asymptotes of Func. 7:02
  • Example 3: Find Asymptotes of Func. 8:59
  • Example 4: Graph Rational Func. 11:08