Asymptotes curve (1-2x)/(3x+5)

[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?

 

[KLscilevothma]

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?

 

[FZ+]

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?

I don't know 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! )

 

[StephenPrivitera]

Originally posted by FZ+
I don't know 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! )

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.

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.

 

[Matt]

In this case the largest degrees of the variables are the same (to the first). So for horizontal asymtotes don't u just take the ratio:

-2x + 1
3x + 5

So the horizontal asymtote is (-2/3)

Right?

 

[StephenPrivitera]

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