- #1

benji

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I haven't ever done one of these before so I really don't know where to start.

Here's the problem:

y=(2x^2+5x+11)/(x+1)

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- Thread starter benji
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- #1

benji

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I haven't ever done one of these before so I really don't know where to start.

Here's the problem:

y=(2x^2+5x+11)/(x+1)

- #2

Pyrrhus

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http://home.att.net/~srschmitt/precalc/precalc-12-03-00.html [Broken]

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- #3

benji

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I need to factor the numerator then, correct? But it comes out as an imaginary number?

- #4

Pyrrhus

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The Oblique Asymptote will be the quotient.

- #5

HallsofIvy

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With something like [tex]y= \frac{2x^2+5x+11}{x+1}[/tex] the simplest thing to do is to divide both numerator and denominator by the largest power of x in the denominator- here just x. [tex]y= \frac{2x+ 5+ \frac{11}{x}}{1+ \frac{1}{x}}[/tex]. Since those fractions with x in the denominator will go to 0 as x goes to either positive or negative infinity, it's easy to see that y approaches 2x+ 5.

- #6

Galileo

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- #7

dale123

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Someone pointed out earlier that for something like [tex]y= \frac{2x^2+5x+11}{x+1}[/tex] you can just divide by "x" and get that the asymptote is: [tex]y= \frac{2x+ 5+ \frac{11}{x}}{1+ \frac{1}{x}}[/tex]

However, that is not the case I believe. The asymptote is actually 2x+3, and it is found by:

= Lim x -> infinity, f(x) - (mx+b) = 0

So I would do long division, get : 2x + 3 + 8/(x+1) and then use the formula mentioned above:

= lim x-> infinity of 2x + 3 + 8/(x+1) - (2x+3)

that's equal = 0, hence the mx+b part is an oblique asymptote. Which is, 2x+3 = oblique asymptote in this case.

Hope this helps!

- #8

caddya89

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- #9

erh5060

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With something like [tex]y= \frac{2x^2+5x+11}{x+1}[/tex] the simplest thing to do is to divide both numerator and denominator by the largest power of x in the denominator- here just x. [tex]y= \frac{2x+ 5+ \frac{11}{x}}{1+ \frac{1}{x}}[/tex]. Since those fractions with x in the denominator will go to 0 as x goes to either positive or negative infinity, it's easy to see that y approaches 2x+ 5.

You're both wrong, dale 123 is correct. To find the oblique asymptote, you must use polynomial long division, and then analyze the function as it approaches infinity.

Taking the limit first, like HallsofIvy did, is wrong because 11/x and 1/x approach infinity at different rates, and therefore add to the numerator and denominator in slightly different ways.

If you are not finding the limit at infinity of the entire function, as in this problem, then you MUST write it in the form "term + term + ... + term", and analyze each individual term as x approaches infinity. This is because fractions can do unexpected things when it comes to limits.

- #10

HallsofIvy

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- #11

Bohrok

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Forgive me; you mean

- #12

HallsofIvy

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- #13

sutupidmath

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Let f(x) be sucha function and assume that L=ax+b is the oblique assymptote. Let P be any point in the graph of f and Q be a point in the line L, such that d(P,Q) is the shortest distance. Also let denote by K the point on the line L when we draw a perpendicular segment from point P to the x-axis. Now, after a little bit of geometry we find that L is an oblique assymptote iff

[tex]\lim_{x\rightarrow \infty}(f(x)-L)=lim_{x\rightarrow \infty}(f(x)-ax-b)=0[/tex]

If we divide the whole thing by x, and take the limit we find that:

[tex]a=lim_{x\rightarrow\infty}\frac{f(x)}{x}[/tex]

and

[tex]b=\lim_{x\rightarrow\infty}(f(x)-ax)[/tex]

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