Rational function question

[Sonal7]

I am unsure how to get the version i need to find the limits. I did try but I could not do it but just by looking you know its going to have a horizontal asymptote at y=-2

 

[tkhunny]

What's wrong with the "rearrange and look" method?

 

[Dr.Peterson]

You really need to clearly tell us (a) what the problem is, (b) what you did, and (c) what they give as the solution, if that is what you are asking about. I think the problem here is to find the horizontal asymptote of the given equation, and you've shown their solution and are asking how they did what they did. Am I right??

There are several ways to do this; what they show is not very clear to me, either.

The way closest to what they show is to use long division to write 4x/(1-2x) as -2 + 1/(1-2x). Then as x gets large, 1/(1-2x) approaches zero, so the function approaches -2.

The informal way is just to take the ratio of leading terms, (4x)/(-2x) = -2.

 

[Sonal7]

[MATH]y=\frac{4x}{1-2x}[/MATH][MATH]y=\frac{4x+2x-2x}{1-2x}[/MATH][MATH]y=\frac{-2(1-2x)-2x}{1-2x}[/MATH][MATH]y=-2-\frac{-2x}{1-2x}[/MATH]

 

[Sonal7]

You really need to clearly tell us (a) what the problem is, (b) what you did, and (c) what they give as the solution, if that is what you are asking about. I think the problem here is to find the horizontal asymptote of the given equation, and you've shown their solution and are asking how they did what they did. Am I right??

There are several ways to do this; what they show is not very clear to me, either.

The way closest to what they show is to use long division to write 4x/(1-2x) as -2 + 1/(1-2x). Then as x gets large, 1/(1-2x) approaches zero, so the function approaches -2.

The informal way is just to take the ratio of leading terms, (4x)/(-2x) = -2.

I have done some workings but I cant get their version. Not that i need to for the graph to be honest.

 

[Dr.Peterson]

[MATH]y=\frac{4x}{1-2x}[/MATH][MATH]y=\frac{4x+2x-2x}{1-2x}[/MATH][MATH]y=\frac{-2(1-2x)-2x}{1-2x}[/MATH][MATH]y=-2-\frac{-2x}{1-2x}[/MATH]

Are you saying this is your attempt to obtain their result? It's good, except that you have an extra negative in the last line. This is an alternate way to do the division.

EDIT: I take that back; something changed while I was answering, and I didn't look closely. Check each step to see if it makes sense.

 

[Sonal7]

Are you saying this is your attempt to obtain their result? It's good, except that you have an extra negative in the last line. This is an alternate way to do the division.

Are you still not happy with it?

Okay it should be:
[MATH]y=-2+ \frac{-2x}{1-2x}[/MATH]I am guessing you take out -2 as a factor:
[MATH]y=-2(1-\frac{x}{1-2x})[/MATH]

no it doesnt work which means they might have made a mistake.

 

[Sonal7]

Are you saying this is your attempt to obtain their result? It's good, except that you have an extra negative in the last line. This is an alternate way to do the division.

EDIT: I take that back; something changed while I was answering, and I didn't look closely. Check each step to see if it makes sense.

yup theres a mistake. I should have added 2 and -2. I will try now.

 

[Sonal7]

[MATH]y=\frac{4x}{1-2x}[/MATH][MATH]y=\frac{4x+2-2}{1-2x}[/MATH][MATH]y=-2\frac{(1-2x)-1}{1-2x}[/MATH][MATH]y=-2(1-\frac{1}{1-2x})[/MATH]

 

[Sonal7]

Are you saying this is your attempt to obtain their result? It's good, except that you have an extra negative in the last line. This is an alternate way to do the division.

EDIT: I take that back; something changed while I was answering, and I didn't look closely. Check each step to see if it makes sense.

I got it now , I find algebra a bit hard

 

[Deleted member 4993]

I am unsure how to get the version i need to find the limits. I did try but I could not do it but just by looking you know its going to have a horizontal asymptote at y=-2View attachment 21615

\(\displaystyle y \ = \ \frac{4x}{1-2x} \ = \ \frac{4x\ - 2 \ + 2}{1-2x} \ = \ \frac{2(2x-1) \ + 2}{1-2x}\ =\ \frac{2 - 2(1-2x)}{1-2x}=\ 2 \cdot \ \frac{1 - (1-2x)}{1-2x}=\ (-2) \cdot \ \left[1 - \frac{1}{1-2x}\right]\)

Take time to do the algebra "correctly" - without that Calculus will be "far away"!