haleyelizabeth2017
  • haleyelizabeth2017
Determine the equation of the horizontal asymptotes, if any, of the function. f(x)=(2x+1)/(x+1)
Mathematics
schrodinger
  • schrodinger
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haleyelizabeth2017
  • haleyelizabeth2017
\[f(x)=\frac{ 2x+1 }{ x+1 }\]
Rizags
  • Rizags
are you allowed to use limits?
haleyelizabeth2017
  • haleyelizabeth2017
I'm not sure.

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Rizags
  • Rizags
what class is this for?
haleyelizabeth2017
  • haleyelizabeth2017
PreCalculus
Rizags
  • Rizags
ok then maybe some limits are allowed
Rizags
  • Rizags
are you familiar with limits?
haleyelizabeth2017
  • haleyelizabeth2017
Nope. But...the book gives us two methods we can use. Solving for x in terms of y is one method, and divide the numerator and denominator by the highest power of x is the second.
Rizags
  • Rizags
yes use the second
Rizags
  • Rizags
multiply the whole thing by \[\large \frac{\frac{1}{x}}{\frac{1}{x}}\]
anonymous
  • anonymous
the numerator and denominator are both polynomials the degrees are the same (they are both of degree 1) the horizontal asymptotes is the ratio of the leading coefficients
anonymous
  • anonymous
in your example it is \[y=\frac{2}{1}=2\]
anonymous
  • anonymous
it always works this way if the degrees are the same if they are not, then it is different
haleyelizabeth2017
  • haleyelizabeth2017
Okay, sorry...was afk for a moment there
anonymous
  • anonymous
if the degree of the denominator is larger, then it is \(y=0\) if the degree of the numerator is larger, then there is no horizontal asymptote
haleyelizabeth2017
  • haleyelizabeth2017
Okay, good to know. Thank you

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