7. $p(x)=\frac{(3x - 1)^2}{2x^2+3x + 5}$\nhorizontal asymptote: y or n\nif yes, equation: ________

Explanation:

Step1: Expand the numerator

First, expand $(3x - 1)^2=(3x - 1)(3x - 1)=9x^{2}-6x + 1$. So $p(x)=\frac{9x^{2}-6x + 1}{2x^{2}+3x + 5}$.

Step2: Determine the degrees of numerator and denominator

The degree of the numerator $n = 2$ (highest - power of $x$ is $2$ in $9x^{2}-6x + 1$) and the degree of the denominator $m = 2$ (highest - power of $x$ is $2$ in $2x^{2}+3x + 5$).

Step3: Find the horizontal asymptote

When $n=m$, the horizontal asymptote is given by $y=\frac{a_{n}}{b_{m}}$, where $a_{n}$ is the leading coefficient of the numerator and $b_{m}$ is the leading coefficient of the denominator. Here, $a_{n}=9$ and $b_{m}=2$. So the horizontal asymptote is $y = \frac{9}{2}$.

Answer:

Horizontal Asymptote: Y If Yes, Equation: $y=\frac{9}{2}$