Question

Could someone explain how to solve this?

Answer 1

?

Answer 2

Find the horizontal asymptote, if any, of the graph of the rational function.

f(x)= 14x over 6x^2 +5

Answer 3

@jim_thompson5910

Answer 4

you should find the roots of bottom

6x^2+5=0

Answer 5

? My example has some long equation in a fraction. The degree of numerator is n and denominator is m.

Answer 6

\[f(x)= {14x \over 6x^2 +5}\]??

Answer 7

Yes.

Answer 8

so if \(x \rightarrow \pm \infty\), what happens to f(x)

Answer 9

https://www.mathsisfun.com/algebra/asymptote.html

Answer 10

My example says this..

Answer 11

I have no clue what your talking about..

Answer 12

using just maths
\[f(x)= {14x \over 6x^2 +5} \\= {14{x\over x^2} \over 6{x^2\over x^2} +{5\over x^2}}\\= {{14\over x} \over 6 +{5\over x^2}}\]

that's a lot simpler than your posted solution, esp as we now know what happens as \(x \to \pm \infty\)

again, if you are trying to find asymptotes.



you are asking for a horizontal asymptote. you need to know what an asymptote is first.

good luck :p

Answer 13

I honestly didn't understand any of that.

Answer 14

to fine asymptotes, youre making the denominator = 0

Answer 15

so by equalling the denominator to 0, you're solving it and therefore finding the roots to it

Answer 16

Okay? But that's not what my example does.

Answer 17

arent you trying to find the asymptote?

Answer 18

Im trying to find the horizontal asymptote..

Answer 19

That would be trying to find the vertical asymptotes @09chunga

Answer 20

so...hmmm what's the question again?

Answer 21

If I don't solve it like the example then it will mess me up later on..

Answer 22

oh hmmm

Answer 23

That cant be right. I clicked no horizontal asymptote on my practice problem and it says thats wrong.

Answer 24

hmmm wait a second.. that went off the beaten path =(

Answer 25

lemme redo that bit hehe

Answer 26

you mean the x-axis is the asymptote

Answer 27

\(\bf \cfrac{14x^{\color{red}{ 1}}}{6x^{\color{red}{ 2}}+5}\) the denominator's degree is higher, thus the asymptote is y = 0

Answer 28

I have no clue.

Answer 29

eh? so... hmmm what are you expected do it anyway?

Answer 30

@Destinyyyy We can find the Horizontal asymptotes of a function by seeing what that function is like when we approach infinity/-infinity

That is basically what your "example" is saying without actually saying it

Let me try to relate it to that

\[\large f(x) = \frac{14x}{6x^2 + 5}\]

As your example states: lets rewite that as

\[\large f(x) = \frac{a_nx^n}{b_mx^m}\]

*The +5 goes away because when you are approaching infinity...adding 5 doesn't do anything

From here...as your example states

"If n<m, the x-axis or y = 0 is the asymptote
n=m, a/b is the asymptote
n>m, there is no asymptote

So here with your problem

\[\large f(x) = \frac{ax^1}{bx^2}\]

Since n < m *1 < 2* then the asymptote is y = 0

Answer 31

I posted pictures of the example.

Answer 32

well, the pictures are saying the same thing we are saying, so.... tis ok then
unless what you're saying is, that you don't understand the examples

Answer 33

YES!! That makes sense

Answer 34

Thank you @johnweldon1993

Answer 35

And why? Well think about it

if we have \(\large \frac{x}{x^2}\) and we approach infinity...we basically have a number over a BIGGER number...and what does that do?

example of a number / bigger number:

\[\large \frac{1}{100000} \rightarrow 0\]

It is basically 0...or a VERY small number

Hence, using the infinity logic...if n < m...we approach 0 and the asymptote is y = 0

Hope that makes sense :)

Answer 36

It does! Could you help me with another one? I can post in a new question.

Answer 37

And no problem! @Destinyyyy