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For more information, please see full course syllabus of Calculus AB

For more information, please see full course syllabus of Calculus AB

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)

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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)

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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.

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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.

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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???