Let f(x)= [((3x^5) - 2x² + 3) / (2x² -5)]. Find a power function end behavior model forf.[/FONT]

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- Jan 15th 2009, 01:36 PMh4hv4hd4si4nPower FunctionLet f(x)= [((3x^5) - 2x² + 3) / (2x² -5)]. Find a power function end behavior model for
*f.*[/FONT] - Jan 15th 2009, 01:55 PMJester
- Jan 15th 2009, 02:01 PMh4hv4hd4si4n
Yes

- Jan 15th 2009, 02:11 PMJester
Do you know how to divide polynominals by long division?

- Jan 15th 2009, 02:18 PMh4hv4hd4si4n
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 PMchabmgph
- Jan 15th 2009, 03:11 PMJester
If you perform long division you get

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

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

$\displaystyle \frac{3x^5-2x^2+3}{2x^2-5} \approx \frac{3x^3}{2} + \frac{15 x}{4} - 1 $ - Jan 15th 2009, 03:43 PMchabmgph

Your approach is right, however the problem calls for a power function.

Actually $\displaystyle y=\frac{3}{2}x^3$ 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 $\displaystyle 3x^5$ and $\displaystyle 2x^2$ and by looking at its ratio, we get $\displaystyle y=\frac{3}{2}x^3$ - Jan 15th 2009, 03:54 PMJester