finding an oblique asymptote

[Guest]

I need help clearing things up for this questions:

Find the equation of the oblique asymptote:

f(x)= 2x^2+9x+2/ (2x+3)

I did long division and got2x+3)(x+3)-7, so to get the equation: [(2x+3)(x+3)-7]/(2x+3)
x+3-(7/2x+3)
and since -7/2x+3 becomes less and less significant when a large number is substituted in, the equation is x+3


I'm confused on the next part of the question:
It says to determine whether the curve approaches the asymptote from above or below, we did an example on how to find this: so I followed it:

f(100)= -7/2x+3
=-7/203
= -0.03 <0, therefore approaches from below
f(-100)= -7/-197= 0.03 >0 therefore approaches from above

But for the next questions it asks you to determine the direction from which the curve approaches the asymptote. How do you find that?

the back of the book reads for that questions" As x--> -00 f(x) is below the line, but I thought you were suppose to figure that out for the previous question, ( "whether the curve approaches the asymptote from above or below" )

thanks for the help in advance

 

[skeeter]

I don't understand how you are doing long division ... here's how it's supposed to be done.

          x   + 3
        --------------------
(2x+3) | 2x^2 + 9x + 2
         2x^2 + 3x
         ------------------
                6x + 2
                6x + 9
                ---------
                     -7

the equation of the oblique asymptote is y = x + 3

f(100) = [2(100)^2 + 9(100) + 2]/[2(100) + 3] = approx 102. 9655517...

since y = x + 3 ...
y(100) = 103

since f(100) < y(100), f(x) approaches the oblique asymptote from below.

 

[Guest]

I didnt show how I did the long division, (I did it on paper) but I gave you the answer to it.

How come you're replacing a large number into the origional function instead of -7/2x+3?

 

[skeeter]

How come you're replacing a large number into the origional function instead of -7/2x+3?
because the graph of the original function approaches the graph of the asymptote ... that is what is being compared.

 

[JeffM]

I didnt show how I did the long division, (I did it on paper) but I gave you the answer to it.

How come you're replacing a large number into the origional function instead of -7/2x+3?

Because there are frequently a number of ways to solve a problem

\(\displaystyle \dfrac{2x^2 + 9x + 2}{2x + 3} = \dfrac{2x^2 + 3x}{2x + 3} + \dfrac{6x + 9}{2x + 3} + \dfrac{-7}{2x + 3} = x + 3 + \dfrac{-7}{2x + 3}.\)

You did this through long division; well done.

Furthermore, you saw that \(\displaystyle \displaystyle \lim_{x \rightarrow \infty}\dfrac{-7}{2x + 3} = 0.\)

Again well done

\(\displaystyle So\ x + 3\ is\ an\ asymptote.\) You are fine to there.

Skeeter took a large value of x and compared f(x) to x + 3 at that value.

\(\displaystyle \dfrac{2 * 100^2 + 9 * 100 + 2}{2 * 100 + 3} = \dfrac{20902}{203} \approx 102.97 < 100 + 3.\)

He concludes that the function approaches the asymptote from below. Make sense?

Alternative way

\(\displaystyle x > 0 \implies \dfrac{-7}{2x + 3} < 0 \implies \dfrac{2x^2 + 9x + 2}{2x + 3} = x + 3 + \dfrac{-7}{2x + 3} < x + 3 + 0 = x + 3.\)

Conclusion: the function approaches x + 3 from below.

Same result.