f(x)=(x^2 - 6x + 6)/(x - 9)\nthe (choose one) asymptote is y =
f(x)=(x^2 - 6x + 6)/(x - 9)\nthe (choose one) asymptote is y =
Answer
Explanation:
Step1: Perform polynomial long - division
Divide $x^{2}-6x + 6$ by $x - 9$. We know that $x^{2}-6x + 6=(x - 9)(x+3)+33$. So, $f(x)=\frac{x^{2}-6x + 6}{x - 9}=x + 3+\frac{33}{x - 9}$.
Step2: Determine the oblique asymptote
As $x\to\pm\infty$, the term $\frac{33}{x - 9}\to0$. The non - rational part of the function $y=x + 3$ is the oblique asymptote. Since the degree of the numerator is one more than the degree of the denominator, there is no horizontal asymptote.
Answer:
The oblique asymptote is $y=x + 3$