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Mathematics 9 Online
OpenStudy (anonymous):

find the HORIZONTAL asymptote, if any, of the graph of the rational function f(x)= 25x/5x^2+1 A) y = 5 B) y = 1/5 C) y = 0 D) no horizontal asymptote?

OpenStudy (callisto):

For horizontal asymptotes, the degree of polynomial of the numerator of the fraction should be the same as the degree of polynomial of the denominator of the fraction. Can you work out the answer now?

OpenStudy (anonymous):

so it should be no horizontal asymptote cause 5x^2+1 = 0 5x^2 = -1?

OpenStudy (anonymous):

or 25 (x)/ 5 (x^2 +1) = 5 (x) / (x^2 +1) right there?

OpenStudy (callisto):

No. If you're doing in that way, you're trying to find the VERTICAL asymptotes. To find horizontal asymptotes, you actually need to take limit. But for this question, it's quite obvious because my little trick works. :S

OpenStudy (anonymous):

I know how to get the vertical what about the horizantal? is this correct? 25(x)/ 5(x^2+1) = 5x/x2+1?

OpenStudy (callisto):

No, it is not. 5 is not the common factor for denominator. You cannot do it like that.

OpenStudy (anonymous):

is it 5?

OpenStudy (callisto):

No.

OpenStudy (callisto):

Take limit for the function, x tends to infinity, to find the horizontal asymptotes, or arrange the function in the form \(f(x) = C + \frac{P(x)}{Q(x)}\). Horizontal asymptotes is at y = c.

OpenStudy (callisto):

*Horizontal asymptotes is y = c.

OpenStudy (anonymous):

I still didn't understand it

OpenStudy (anonymous):

I still don't get it.

OpenStudy (callisto):

Have you learnt limit?

OpenStudy (anonymous):

Nope

OpenStudy (callisto):

Alright. Then use the second method.. Example: \(f(x)= \frac{5x^2+3}{5x^2+1}\) It can be rearranged in the form C + P(x)/Q(x) \[f(x)= \frac{5x^2+3}{5x^2+1}\]\[=\frac{5x^2+1+2}{5x^2+1}\]\[=\frac{5x^2+1}{5x^2+1}+\frac{2}{5x^2+1}\]\[=1+\frac{2}{5x^2+1}\] So, the horizontal asymptotes is y=1. Got it?

OpenStudy (anonymous):

I'm gonna take note of that GOT IT SIR

OpenStudy (callisto):

Miss, not sir :)

OpenStudy (anonymous):

Wait don't go yet I want to make sure my answer is correct

OpenStudy (callisto):

I'll leave in 5 minutes.. I'm sorry!

OpenStudy (anonymous):

so this is it. I followed what you did. 5x(5) / 5x (x+1) is that correct?

OpenStudy (callisto):

Nope :|

OpenStudy (anonymous):

Oh. I'm dumb flutter

OpenStudy (callisto):

Calm... Observe the denominator and the numerator.. and compare..

OpenStudy (anonymous):

they have both x. and multiples of 5

OpenStudy (callisto):

No. of terms?

OpenStudy (callisto):

Sorry, have to go now...

OpenStudy (anonymous):

alright thanks for your help though.

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