Question

8.- use the following rational function: $r(x) = \frac{x^3 - 27}{x^2 - 9}$ to find
(a) domain
(b) $x$ and $y$ intercepts

(c) the horizontal (if any) asymptotes
(d) the vertical asymptotes.

(e) the behavior around the vertical asymptotes

Explanation:

Part (a): Domain

Step 1: Identify Denominator

The function is \( r(x) = \frac{x^3 - 27}{x^2 - 9} \). The denominator is \( x^2 - 9 \).

Step 2: Solve Denominator ≠ 0

Factor the denominator: \( x^2 - 9 = (x - 3)(x + 3) \). Set \( (x - 3)(x + 3) eq 0 \). So \( x eq 3 \) and \( x eq -3 \).

Step 3: Determine Domain

The domain is all real numbers except \( x = 3 \) and \( x = -3 \), so \( (-\infty, -3) \cup (-3, 3) \cup (3, \infty) \).

Part (b): x and y intercepts

x - intercepts

Step 1: Set \( r(x) = 0 \)

\( \frac{x^3 - 27}{x^2 - 9} = 0 \). The numerator must be 0 (denominator ≠ 0).

Step 2: Solve \( x^3 - 27 = 0 \)

Factor \( x^3 - 27 = (x - 3)(x^2 + 3x + 9) \). Set \( x - 3 = 0 \) (since \( x^2 + 3x + 9 \) has no real roots). But \( x = 3 \) makes denominator 0, so no x - intercept? Wait, wait: \( x^3 - 27 = 0 \) → \( x^3 = 27 \) → \( x = 3 \), but at \( x = 3 \), denominator is 0, so the function is undefined there. Wait, maybe I made a mistake. Wait, \( x^3 - 27 = (x - 3)(x^2 + 3x + 9) \), and denominator is \( (x - 3)(x + 3) \). So we can simplify the function? Wait, \( r(x) = \frac{(x - 3)(x^2 + 3x + 9)}{(x - 3)(x + 3)} \), for \( x eq 3 \). So simplify to \( r(x) = \frac{x^2 + 3x + 9}{x + 3} \), \( x eq 3, -3 \). Now, set numerator 0: \( x^2 + 3x + 9 = 0 \). Discriminant: \( 9 - 36 = -27 < 0 \), so no real x - intercepts? Wait, but original numerator at \( x = 3 \) is 0, but x = 3 is excluded. So x - intercepts: none? Wait, maybe I messed up. Wait, let's check y - intercept.

y - intercept

Step 1: Set \( x = 0 \)

\( r(0) = \frac{0 - 27}{0 - 9} = \frac{-27}{-9} = 3 \). So y - intercept is (0, 3).

Part (c): Horizontal Asymptotes

Step 1: Compare Degrees

Degree of numerator (after simplifying, original numerator is degree 3, denominator degree 2). Wait, original function: numerator degree 3, denominator degree 2. When degree of numerator > degree of denominator, there is no horizontal asymptote. Wait, but we simplified the function by canceling (x - 3), but the original function has a hole at x = 3, not a vertical asymptote. Wait, let's re - examine. The original function: \( r(x) = \frac{x^3 - 27}{x^2 - 9} = \frac{(x - 3)(x^2 + 3x + 9)}{(x - 3)(x + 3)} \), so for \( x eq 3 \), it's \( \frac{x^2 + 3x + 9}{x + 3} \). Now, the simplified function has numerator degree 2, denominator degree 1. So degree of numerator (2) > degree of denominator (1), so no horizontal asymptote. Wait, but original function: numerator degree 3, denominator degree 2. So in general, for rational functions \( \frac{N(x)}{D(x)} \), if \( \deg(N) > \deg(D) \), no horizontal asymptote. So no horizontal asymptote.

Part (d): Vertical Asymptotes

Step 1: Simplify the Function

We have \( r(x) = \frac{(x - 3)(x^2 + 3x + 9)}{(x - 3)(x + 3)} \), \( x eq 3 \). So after canceling (x - 3), the function is \( \frac{x^2 + 3x + 9}{x + 3} \), with \( x eq 3, -3 \). The vertical asymptotes occur where the denominator of the simplified function is 0 (and the numerator is not 0 there). Set denominator \( x + 3 = 0 \) → \( x = -3 \). Check numerator at \( x = -3 \): \( (-3)^2 + 3(-3) + 9 = 9 - 9 + 9 = 9 eq 0 \). So vertical asymptote at \( x = -3 \).

Part (e): Behavior around Vertical Asymptote (x = -3)

Step 1: Analyze Left - Hand Limit (\( x \to -3^- \))

\( \lim_{x \to -3^-} \frac{x^2 + 3x + 9}{x + 3} \). Let \( x = -3 - h \), where \( h \to 0^+ \). Then numerator: \( (-3 - h)^2 + 3(-3 - h) + 9 = 9 + 6h + h^2 - 9 - 3h + 9 = h^2 + 3h + 9 \), which approaches 9 as \( h \to 0 \). Denominator: \( (-3 - h)+ 3=-h \), which approaches 0 from the negative side (since \( h \to 0^+ \), -h → 0^-). So the limit is \( \frac{9}{0^-} = -\infty \).

Step 2: Analyze Right - Hand Limit (\( x \to -3^+ \))

Let \( x = -3 + h \), where \( h \to 0^+ \). Numerator: \( (-3 + h)^2 + 3(-3 + h)+ 9 = 9 - 6h + h^2 - 9 + 3h + 9 = h^2 - 3h + 9 \), approaches 9 as \( h \to 0 \). Denominator: \( (-3 + h)+ 3 = h \), approaches 0 from the positive side. So the limit is \( \frac{9}{0^+} = +\infty \).

Part (e) (Behavior around vertical asymptote x = -3)

As \( x \to -3^- \), \( r(x) \to -\infty \); as \( x \to -3^+ \), \( r(x) \to +\infty \).

Answers:

(a) Domain: \( (-\infty, -3) \cup (-3, 3) \cup (3, \infty) \)
(b) x - intercepts: None; y - intercept: \( (0, 3) \)
(c) No horizontal asymptote
(d) Vertical asymptote: \( x = -3 \)
(e) As \( x \to -3^- \), \( r(x) \to -\infty \); as \( x \to -3^+ \), \( r(x) \to +\infty \)

Answer:

s:
(a) Domain: \( (-\infty, -3) \cup (-3, 3) \cup (3, \infty) \)
(b) x - intercepts: None; y - intercept: \( (0, 3) \)
(c) No horizontal asymptote
(d) Vertical asymptote: \( x = -3 \)
(e) As \( x \to -3^- \), \( r(x) \to -\infty \); as \( x \to -3^+ \), \( r(x) \to +\infty \)