Find equation of asymptotes: very basic

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Homework Statement



Find the equations of the asymptotes.

Homework Equations



y = 5x^3 - 4x^2 + 11 / 2x (x^2 - 9)

( 5x^3 - 4x^2 + 11 is the numerator, 2x (x^2 - 9) is the denominator)

The Attempt at a Solution



Vertical: x = 0, x = -3, x = 3
Horizontal: Quick question...

The degree of the numerator for the first coefficient of the numerator is 3, for first of the denominator it is 1, correct? Thus there is no horizontal asymptote?

OR

Is the degree of the denominator 3, because of 2x and then x^2, in which case it would be 5/2?
 
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The second. The degree of the denominator is 3. 2x(x^2-9)=2x^3-18x.
 
Last edited:
You're getting slightly confused, but did arrive at the answer kinda. Basically for x large, the x^3 terms on both the denominator and numerator will dominate each expression. Since they are both of degree 3 (that is the highest power of x is 3 for each) we simply look at the coeffecients of both x^3 terms and this will give us the answer. The coeffecient on top is 5, and on the bottom is 2, hence we have an asymptote at y=5/2.
 
939 said:

Homework Equations



y = 5x^3 - 4x^2 + 11 / 2x (x^2 - 9)

( 5x^3 - 4x^2 + 11 is the numerator, 2x (x^2 - 9) is the denominator)
I'm glad you clarified what you wrote. When you write fractions with multiple terms in either the numerator or denominator, put parentheses around the whole numerator or denominator.

Here is your equation as it should be written:
y = ([/color]5x^3 - 4x^2 + 11)[/color] / ([/color]2x (x^2 - 9))[/color]
 
Dick said:
The second. The degree of the denominator is 3. 2x(x^2-9)=2x^3-18x.

Gengar said:
You're getting slightly confused, but did arrive at the answer kinda. Basically for x large, the x^3 terms on both the denominator and numerator will dominate each expression. Since they are both of degree 3 (that is the highest power of x is 3 for each) we simply look at the coeffecients of both x^3 terms and this will give us the answer. The coeffecient on top is 5, and on the bottom is 2, hence we have an asymptote at y=5/2.
Thanks :).

Mark44 said:
I'm glad you clarified what you wrote. When you write fractions with multiple terms in either the numerator or denominator, put parentheses around the whole numerator or denominator.

Here is your equation as it should be written:
y = ([/color]5x^3 - 4x^2 + 11)[/color] / ([/color]2x (x^2 - 9))[/color]

Thanks, I learned something similar when using wolframalpha, but wasn't sure if I should always write them like that :).
 
When you write a fraction with inline text (as opposed to using LaTeX), you should always put parentheses around numerators and/or denominators that have multiple terms.
What you wrote -
5x^3 - 4x^2 + 11 / 2x (x^2 - 9)

would be interpreted by many as

5x3 - 4x2 + ##\frac{11}{2}## x(x2 - 9)

Pretty obviously, that's not what you intended.
 
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