Horizontal Asymptotes

[Shoesta]

Hi all, how do I work out the horizontal asymptote? I know this is  a stupid question, but I genuinely can't work them out! Thanks.

[jamonwindeyer]

Hi all, how do I work out the horizontal asymptote? I know this is  a stupid question, but I genuinely can't work them out! Thanks.

Hey! So horizontal asymptotes look at limiting behaviour of functions, let's see what happens as \(x\to\infty\):



In the middle there, I've divided every term in the fraction by \(x^2\) to make the behaviour more clear. In the new form, clearly the 2 and 1 are unaffected, while the \(\frac{9}{x^2}\) will shrink away as \(x\) gets huge. So, really, all that will be left is \(\frac{2}{1}=2\)! Now this is the limiting behaviour, it will never actually reach the value of 2: So, \(y=2\) becomes our horizontal asymptote

Does this help a little? Horizontal asymptotes are a little tricky

[Shoesta]

Hey! So horizontal asymptotes look at limiting behaviour of functions, let's see what happens as \(x\to\infty\):



In the middle there, I've divided every term in the fraction by \(x^2\) to make the behaviour more clear. In the new form, clearly the 2 and 1 are unaffected, while the \(\frac{9}{x^2}\) will shrink away as \(x\) gets huge. So, really, all that will be left is \(\frac{2}{1}=2\)! Now this is the limiting behaviour, it will never actually reach the value of 2: So, \(y=2\) becomes our horizontal asymptote

Does this help a little? Horizontal asymptotes are a little tricky

Thanks So, I should divide all the terms in the function by the highest power of x? And then look at how the function will behave when I insert increasing values of x?

[RuiAce]

Thanks So, I should divide all the terms in the function by the highest power of x? And then look at how the function will behave when I insert increasing values of x?