Let f(x)= [((3x^5) - 2x² + 3) / (2x² -5)]. Find a power function end behavior model for f.[/FONT]
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 $
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 $\displaystyle y=\frac{3}{2}x^3$