[MOVED] Find vertical, horizontal asymptotes for 2x+3/x+2,
- Thread starter tonsuz0
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\(\displaystyle \L f(x) = \frac{2x+3}{x+2}\)
Vertical asymptote at the zeros of the denominator: x = -2
horizontal asymptote take the limit as x--> infinity: you can disregard everything but the highest terms.
\(\displaystyle \L lim_{x\to\infty}\frac{2x + 3}{x + 2}\)
divide all terms by the highest x term
= \(\displaystyle \L lim_{x\to\infty}\frac{\frac{2x}{x} + \frac{3}{x}}{\frac{x}{x} + \frac{2}{x}}\)
let x goto infinity:
= \(\displaystyle \L \frac{2 + 0}{1 + 0}\) = 2
Therefor you have a vertical asymptote at x = -2 and a horizontal asymptote at x = 2
John.
Vertical asymptote at the zeros of the denominator: x = -2
horizontal asymptote take the limit as x--> infinity: you can disregard everything but the highest terms.
\(\displaystyle \L lim_{x\to\infty}\frac{2x + 3}{x + 2}\)
divide all terms by the highest x term
= \(\displaystyle \L lim_{x\to\infty}\frac{\frac{2x}{x} + \frac{3}{x}}{\frac{x}{x} + \frac{2}{x}}\)
let x goto infinity:
= \(\displaystyle \L \frac{2 + 0}{1 + 0}\) = 2
Therefor you have a vertical asymptote at x = -2 and a horizontal asymptote at x = 2
John.
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- Feb 4, 2004
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You've been provided with great explanations for the first of the posted exercises (assuming the tutor guessed your meaning correctly). For general information and instruction, try the following. :idea:tonsuz0 said:
. . . . .Google results for "asymptotes vertical horizontal"
Also, please be careful with your notation in future: "2x + 3/x + 2" means "2x + (3/x) + 2", but you almost certainly meant "(2x + 3)/(x + 2)". :wink:
Thank you!
Eliz.
Re: can you walk me through this
As has already been explained to you, a vertical asymptote occurs at any value of x which makes the denominator of the fraction become 0.
Ok....when is x<SUP>2</SUP> + 1 = 0
No matter what value you choose for x, x<SUP>2</SUP> can never be less than 0. So the denominator of the fraction can't ever be 0. There is NO vertical asymptote.
For the horizontal asymptote, follow the suggestion given earlier. Divide each term in the numerator and denominator by the highest power of x:
g(x) = (5x) / (x<SUP>2</SUP> + 1)
The highest power of x is x<SUP>2</SUP>. Divide each term in the numerator and denominator by that:
g(x) = (5x / x<SUP>2</SUP>) / [ (x<SUP>2</SUP> / x<SUP>2</SUP> + 1 / x<SUP>2</SUP>)
g(x) = ( 5 / x) / [ 1 + (1 / x<SUP>2</SUP>)]
Now, think about what happens as x gets VERY large (approaches infinity) or VERY small (approaches negative infinity).
5/x gets very close to 0. 1/x<SUP>2</SUP> also gets very close to 0.
So,
g(x) gets very close to 0/(1 + 0), or 0 / 1, or 0.
The horizontal asymptote is y = 0.
I hope this helps you.
tonsuz0 said:
As has already been explained to you, a vertical asymptote occurs at any value of x which makes the denominator of the fraction become 0.
Ok....when is x<SUP>2</SUP> + 1 = 0
No matter what value you choose for x, x<SUP>2</SUP> can never be less than 0. So the denominator of the fraction can't ever be 0. There is NO vertical asymptote.
For the horizontal asymptote, follow the suggestion given earlier. Divide each term in the numerator and denominator by the highest power of x:
g(x) = (5x) / (x<SUP>2</SUP> + 1)
The highest power of x is x<SUP>2</SUP>. Divide each term in the numerator and denominator by that:
g(x) = (5x / x<SUP>2</SUP>) / [ (x<SUP>2</SUP> / x<SUP>2</SUP> + 1 / x<SUP>2</SUP>)
g(x) = ( 5 / x) / [ 1 + (1 / x<SUP>2</SUP>)]
Now, think about what happens as x gets VERY large (approaches infinity) or VERY small (approaches negative infinity).
5/x gets very close to 0. 1/x<SUP>2</SUP> also gets very close to 0.
So,
g(x) gets very close to 0/(1 + 0), or 0 / 1, or 0.
The horizontal asymptote is y = 0.
I hope this helps you.
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- Feb 4, 2004
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Yes, we got that. Which is why one was done for you, and you were provided with explanations and a link to lessons. If, after having worked through the example provided here and having read two or more online lessons, you still have absolutely no idea even how to get started, then I'm afraid you likely need much more help than we can provide.tonsuz0 said:
Please consider hiring a qualified local tutor, and setting aside a few hours every week for intensive face-to-face re-teaching. :idea:
Good luck!
Eliz.
Thank you! Together with the first explanation, the second one answered my existing questions from the first post. I am a stay at home mom and taking college courses online. I am really bad at math, but am hanging in there to better my education. It has been 12 years since I have been in school. Once again thanks to those who walked me through these problems.