# Power Function

• Jan 15th 2009, 01:36 PM
h4hv4hd4si4n
Power Function
Let f(x)= [((3x^5) - 2x² + 3) / (2x² -5)]. Find a power function end behavior model for f.[/FONT]
• Jan 15th 2009, 01:55 PM
Jester
Quote:

Originally Posted by h4hv4hd4si4n
Let f(x)= [((3x^5) - 2x² + 3) / (2x² -5)]. Find a power function end behavior model for f.[/font]

When you say "end behaviour" do you mean as $x \rightarrow \pm \infty$?
• Jan 15th 2009, 02:01 PM
h4hv4hd4si4n
Yes
• Jan 15th 2009, 02:11 PM
Jester
Do you know how to divide polynominals by long division?
• Jan 15th 2009, 02:18 PM
h4hv4hd4si4n
yes, but i'm confused by what the question is asking for.
How do you find the power function and end behavior function?
• Jan 15th 2009, 02:23 PM
chabmgph
Quote:

Originally Posted by h4hv4hd4si4n
Let f(x)= [((3x^5) - 2x² + 3) / (2x² -5)]. Find a power function end behavior model for f.[/font]

Can you figure out what power functions can be the representatives for the polynomials for the numerator and denominator in the long run? Then the end behavior of a rational function can be found by looking at the ratio of the power functions you found.
• Jan 15th 2009, 03:11 PM
Jester
If you perform long division you get

$\frac{3x^5-2x^2+3}{2x^2-5} = \frac{3x^3}{2} + \frac{15 x}{4} - 1 + \frac{75x-8}{4(2x^2-5)}$

As $x \rightarrow \pm \infty$ then the last term drops and the end behaviour looks like the function

$\frac{3x^5-2x^2+3}{2x^2-5} \approx \frac{3x^3}{2} + \frac{15 x}{4} - 1$
• Jan 15th 2009, 03:43 PM
chabmgph
Quote:

Originally Posted by danny arrigo
If you perform long division you get

$\frac{3x^5-2x^2+3}{2x^2-5} = \frac{3x^3}{2} + \frac{15 x}{4} - 1 + \frac{75x-8}{4(2x^2-5)}$

As $x \rightarrow \pm \infty$ then the last term drops and the end behaviour looks like the function

$\frac{3x^5-2x^2+3}{2x^2-5} \approx \frac{3x^3}{2} + \frac{15 x}{4} - 1$

Your approach is right, however the problem calls for a power function.
Actually $y=\frac{3}{2}x^3$ would be the answer. Another way to look at this is by looking at the leading terms of the polynomials on the numerator and denominator (they each describe the end behavior of the two polynomials), which in this case is $3x^5$ and $2x^2$ and by looking at its ratio, we get $y=\frac{3}{2}x^3$
• Jan 15th 2009, 03:54 PM
Jester
Quote:

Originally Posted by chabmgph
Your approach is right, however the problem calls for a power function.
Actually $y=\frac{3}{2}x^3$ would be the answer. Another way to look at this is by looking at the leading terms of the polynomials on the numerator and denominator (they each describe the end behavior of the two polynomials), which in this case is $3x^5$ and $2x^2$ and by looking at its ratio, we get $y=\frac{3}{2}x^3$

Yes - I was hoping that the poster would come to this conclusion. Thanks for the clarifaction though. :)