1. what is the horizontal asymptote for the function? f(x) = (4x² - 6x + 3)/(2x³ + 5) 2. find the horizontal…

Explanation:

Step1: Recall horizontal - asymptote rules

For a rational function $y=\frac{f(x)}{g(x)}=\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots + b_0}$, if $n < m$, the horizontal asymptote is $y = 0$; if $n=m$, the horizontal asymptote is $y=\frac{a_n}{b_m}$; if $n>m$, there is no horizontal asymptote.

Step2: Solve for $f(x)=\frac{4x^2 - 6x + 3}{2x^3+5}$

Here $n = 2$ (degree of numerator) and $m = 3$ (degree of denominator). Since $n<m$, the horizontal asymptote is $y = 0$.

Step3: Solve for $y=\frac{8x^3+4x + 2}{12x^3-x^2-10}$

Here $n = 3$ and $m = 3$. Then $a_n = 8$ and $b_m=12$. So the horizontal asymptote is $y=\frac{8}{12}=\frac{2}{3}$.

Step4: Solve for $g(x)=\frac{3x^4-3x^3 + 1}{x^3-3x^2+3x - 3}$

Here $n = 4$ and $m = 3$. Since $n>m$, there is no horizontal asymptote.

Step5: Solve for $h(x)=\frac{12x^3-4x^2+3}{3x^3+9x + 8}$

Here $n = 3$ and $m = 3$. Then $a_n = 12$ and $b_m = 3$. So the horizontal asymptote is $y=\frac{12}{3}=4$.

Step6: Solve for $f(x)=\frac{x^2+5x + 6}{x^2+7x + 12}$

Here $n = 2$ and $m = 2$. Then $a_n = 1$ and $b_m = 1$. So the horizontal asymptote is $y=\frac{1}{1}=1$.

Answer:

1. $y = 0$
2. $y=\frac{2}{3}$
3. No horizontal asymptote
4. $y = 4$
5. $y = 1$