Question

show all work to identify the asymptotes and state the end behavior of the function $f(x)=\frac{3x}{x - 9}$. question 8 (essay worth 10 points) (mc)

Explanation:

Step1: Find Vertical Asymptote

For a rational function \( f(x)=\frac{N(x)}{D(x)} \), vertical asymptotes occur where \( D(x) = 0 \) (and \( N(x)
eq0 \) at those points).
Here, \( D(x)=x - 9 \). Set \( x - 9 = 0 \), so \( x = 9 \).
Check \( N(9)=3(9)=27
eq0 \), so \( x = 9 \) is a vertical asymptote.

Step2: Find Horizontal Asymptote

For rational functions, if the degrees of \( N(x) \) and \( D(x) \) are equal (both degree 1 here, since \( N(x)=3x \) and \( D(x)=x - 9 \)), the horizontal asymptote is the ratio of the leading coefficients.
Leading coefficient of \( N(x) \) is 3, leading coefficient of \( D(x) \) is 1. So horizontal asymptote is \( y=\frac{3}{1}=3 \).

Step3: Analyze End Behavior

• As \( x \to \infty \): Divide numerator and denominator by \( x \): \( f(x)=\frac{3x/x}{(x - 9)/x}=\frac{3}{1 - \frac{9}{x}} \). As \( x \to \infty \), \( \frac{9}{x}\to0 \), so \( f(x)\to\frac{3}{1 - 0}=3 \).
• As \( x \to -\infty \): Similarly, \( \frac{9}{x}\to0 \), so \( f(x)\to3 \).
• As \( x \to 9^+ \) (approaching 9 from the right): \( D(x)=x - 9\to0^+ \), \( N(x)=3x\to27>0 \), so \( f(x)=\frac{3x}{x - 9}\to\frac{27}{0^+}=\infty \).
• As \( x \to 9^- \) (approaching 9 from the left): \( D(x)=x - 9\to0^- \), \( N(x)=3x\to27>0 \), so \( f(x)=\frac{3x}{x - 9}\to\frac{27}{0^-}=-\infty \).

Answer:

Vertical Asymptote: \( x = 9 \); Horizontal Asymptote: \( y = 3 \); End Behavior: As \( x \to \infty \), \( f(x) \to 3 \); As \( x \to -\infty \), \( f(x) \to 3 \); As \( x \to 9^+ \), \( f(x) \to \infty \); As \( x \to 9^- \), \( f(x) \to -\infty \)