find the horizontal and vertical asymptotes of the curve. you may want to use a grap comma - separated…

find the horizontal and vertical asymptotes of the curve. you may want to use a grap comma - separated lists. if an answer does not exist, enter dne.)\n$y = \\frac { x ^ { 3 } - x } { x ^ { 2 } - 9 x + 8 }$\n$x =$\nfantastic work!\n$y =$\nenhanced feedback\nplease try again. for finding the horizontal asymptotes, find the limits of the function as x\nfor finding the vertical asymptotes, look for numbers at which you are dividing by zero. eva\nresources
Answer
Explanation:
Step1: Find vertical asymptotes
First, factor the denominator (x^{2}-9x + 8=(x - 1)(x - 8)). The function (y=\frac{x^{3}-x}{x^{2}-9x + 8}=\frac{x(x - 1)(x + 1)}{(x - 1)(x - 8)}). Cancel out the common factor ((x - 1)) (for (x\neq1)), the simplified function is (y=\frac{x(x + 1)}{x - 8}) ((x\neq1)). The vertical asymptote occurs when the denominator is zero (after simplification), so (x = 8).
Step2: Find horizontal asymptotes
Use the rule for rational functions. If (f(x)=\frac{a_{n}x^{n}+a_{n - 1}x^{n-1}+\cdots+a_{0}}{b_{m}x^{m}+b_{m - 1}x^{m-1}+\cdots+b_{0}}), when (n>m), we can use polynomial long - division. Divide (x^{3}-x) by (x^{2}-9x + 8). [ \begin{align*} \frac{x^{3}-x}{x^{2}-9x + 8}&=x + 9+\frac{71x-72}{x^{2}-9x + 8}\ \end{align*} ] As (x\to\pm\infty), (\lim_{x\to\pm\infty}\frac{71x-72}{x^{2}-9x + 8}=0). So (y=x + 9) is an oblique (slant) asymptote. Since the degree of the numerator ((n = 3)) is one more than the degree of the denominator ((m=2)), there is no horizontal asymptote.
Answer:
Vertical asymptote (x = 8), horizontal asymptote (DNE)