- #1

JohnnyPhysics

I the curve (1-2x)/(3x+5). I have been asked to find the verticle and horizontal asymptotes. Can anyone help me with a strategy?

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JohnnyPhysics

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- #2

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Here are some hints.

Vertical asymptote: check the demominator, see which point doesn't exist.

horizontal asymptote: take limit of f(x) as x tends to infinity. What is the relation between the limit value you find and the horizontal asymptote?

Vertical asymptote: check the demominator, see which point doesn't exist.

horizontal asymptote: take limit of f(x) as x tends to infinity. What is the relation between the limit value you find and the horizontal asymptote?

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- #3

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I dunno but that answer looks a little too complicated...horizontal asymptote: take limit of f(x) as x tends to infinity. What is the relation between the limit value you find and the horizontal asymptote?

y = (1-2x)/(3x+5)

3xy + 5y = 1 - 2x

x = (1 - 5y )/(3y + 2)

Now, find the value y can't have....

but beware, I have been wrong before (intentionally of course! )

- #4

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That's interesting. I never thought of it like that. That suggests to me that the horizontal asymptote of a function should be equivalent to the vertical asymptote of its inverse (if the inverse exists). Let's see. That might be too general.Originally posted by FZ+

I dunno but that answer looks a little too complicated...

y = (1-2x)/(3x+5)

3xy + 5y = 1 - 2x

x = (1 - 5y )/(3y + 2)

Now, find the value y can't have....

but beware, I have been wrong before (intentionally of course! )

Say y(x)= (Ax^n+B)/(Cx^n+D)

Then y=A/C is its horizontal asymptote.

It's inverse:

yCx^n+yD=Ax^n+B

x^n(yC-A)=B-yD

x=[(B-yD)/(yC-A)]^(1/n)

Has as its vertical asymptote y=A/C

YEAH!

(For the second graph, x is a function of y - so y=A/C is a vertical line).

Of course, the converse should also be true.

I like that. It shows how arbitrary our placement of the axes and the definition of our variables are.

My algebra skills break down from there. I tried having nonzero coefficients for other powers of x [eg. x^(n-1)] but I'm not sure I can solve for x in that case.

- #5

Matt

3x + 5

So the horizontal asymtote is (-2/3)

Right?

- #6

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right

To find horizontal asymptotes (when dealing with rational functions of polynomials), divide both top and bottom of the fraction by the highest power of x and take the limit as x->[oo]

Essentially, all terms which have an x that is less than the highest power will tend to zero, and that is a shortcut that most people use.

For example, if the highest power of x is in the denominator, all the terms in the numerator will tend to zero and that asymptote is y=0.

btw, an asymptote is a line (or some y-value) that the graph approaches, so it should be written as y=k, rather than just k, although I'm sure anyone would know what you mean if you said the horizontal asymptote is -2/3

To find horizontal asymptotes (when dealing with rational functions of polynomials), divide both top and bottom of the fraction by the highest power of x and take the limit as x->[oo]

Essentially, all terms which have an x that is less than the highest power will tend to zero, and that is a shortcut that most people use.

For example, if the highest power of x is in the denominator, all the terms in the numerator will tend to zero and that asymptote is y=0.

btw, an asymptote is a line (or some y-value) that the graph approaches, so it should be written as y=k, rather than just k, although I'm sure anyone would know what you mean if you said the horizontal asymptote is -2/3

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