QUESTION IMAGE
Question
chapter iii.3
score: 1/4 answered: 1/4
question 2
identify all vertical and horizontal asymptotes of the rational function q that is given by
$q(x)=\frac{(2x + 8)^{9}cdot(9x - 15)^{5}cdot(1x - 3)^{3}}{(3x - 1)^{5}cdot(-2x + 7)^{4}cdot(-4x - 7)^{4}cdot(13x + 11)^{5}cdot(-12x - 13)^{1}}$.
separate your answers with a comma.
the vertical asymptotes are at
$x =$
none
the horizontal asymptotes are at
$y =$
none
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
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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 $y = 0$