chapter iii.3\nscore: 1/4 answered: 1/4\nquestion 2\nidentify all vertical and horizontal asymptotes of the…

chapter iii.3\nscore: 1/4 answered: 1/4\nquestion 2\nidentify all vertical and horizontal asymptotes of the rational function q that is given by\n$q(x)=\frac{(2x + 8)^{9}cdot(9x - 15)^{5}cdot(1x - 3)^{3}}{(3x - 1)^{5}cdot(-2x + 7)^{4}cdot(-4x - 7)^{4}cdot(13x + 11)^{5}cdot(-12x - 13)^{1}}$.\nseparate your answers with a comma.\nthe vertical asymptotes are at\n$x =$\nnone\nthe horizontal asymptotes are at\n$y =$\nnone

chapter iii.3\nscore: 1/4 answered: 1/4\nquestion 2\nidentify all vertical and horizontal asymptotes of the rational function q that is given by\n$q(x)=\frac{(2x + 8)^{9}cdot(9x - 15)^{5}cdot(1x - 3)^{3}}{(3x - 1)^{5}cdot(-2x + 7)^{4}cdot(-4x - 7)^{4}cdot(13x + 11)^{5}cdot(-12x - 13)^{1}}$.\nseparate your answers with a comma.\nthe vertical asymptotes are at\n$x =$\nnone\nthe horizontal asymptotes are at\n$y =$\nnone

Answer

Explanation:

Step1: Find vertical asymptotes

Set the denominator equal to zero and solve for $x$. $(3x - 1)^5\cdot(-2x + 7)^4\cdot(-4x - 7)^4\cdot(13x + 11)^5\cdot(-12x - 13)^1=0$ For $3x - 1=0$, we get $x=\frac{1}{3}$; for $-2x + 7 = 0$, we get $x=\frac{7}{2}$; for $-4x - 7=0$, we get $x=-\frac{7}{4}$; for $13x + 11 = 0$, we get $x=-\frac{11}{13}$; for $-12x - 13=0$, we get $x=-\frac{13}{12}$.

Step2: Find horizontal asymptotes

The degree of the numerator $n$: The degree of $(2x + 8)^9$ is $9$, of $(9x - 15)^5$ is $5$, of $(x - 3)^3$ is $3$. So $n=9 + 5+3=17$. The degree of the denominator $m$: The degree of $(3x - 1)^5$ is $5$, of $(-2x + 7)^4$ is $4$, of $(-4x - 7)^4$ is $4$, of $(13x + 11)^5$ is $5$, of $(-12x - 13)^1$ is $1$. So $m=5 + 4+4+5+1=19$. Since $n<m$, the horizontal - asymptote is $y = 0$.

Answer:

The vertical asymptotes are at $x=\frac{1}{3},\frac{7}{2},-\frac{7}{4},-\frac{11}{13},-\frac{13}{12}$ The horizontal asymptote is at $y = 0$