Question

problem. 5 : consider the rational function $f(x)=\frac{x^2 + 2x}{x^2 - 4}$. identify any vertical asymptotes.
$x = \square$
problem. 6 : consider the rational function $f(x)=\frac{x^2 - 9x + 20}{x^2 - 3x + 2}$. identify any vertical asymptotes.
(note: input answers below in increasing values of $x$; ie if your answers were $x = -3$ and $x = 22$, then the left answer box would be $-3$ and the right would be $22$).
$x = \square$ $x = \square$
problem. 7 : consider the rational function $f(x)=\frac{x^2 - 7x + 12}{x^3 - 5x^2 + 3x + 9}$. identify any vertical asymptotes. (note: input answers below in increasing values of $x$; ie if your answers were $x = -3$ and $x = 22$, then the left answer box would be $-3$ and the right would be $22$).
$x = \square$ $x = \square$

Explanation:

Problem 5

Step1: Factor numerator and denominator

Factor \( f(x)=\frac{x^2 + 2x}{x^2 - 4} \). Numerator: \( x(x + 2) \), Denominator: \( (x - 2)(x + 2) \)

Step2: Cancel common factors

Cancel \( (x + 2) \), so \( f(x)=\frac{x}{x - 2} \) (for \( x
eq - 2 \))

Step3: Find vertical asymptote

Vertical asymptote where denominator is 0 (and not canceled). Solve \( x - 2 = 0 \), so \( x = 2 \)

Step1: Factor numerator and denominator

Factor \( f(x)=\frac{x^2 - 9x + 20}{x^2 - 3x + 2} \). Numerator: \( (x - 4)(x - 5) \), Denominator: \( (x - 1)(x - 2) \)

Step2: Check for common factors

No common factors, so vertical asymptotes where denominator is 0

Step3: Solve denominator

Solve \( (x - 1)(x - 2)=0 \), so \( x = 1 \) and \( x = 2 \) (in increasing order)

Step1: Factor numerator and denominator

Factor numerator: \( x^2 - 7x + 12=(x - 3)(x - 4) \)
Factor denominator: \( x^3 - 5x^2 + 3x + 9 \). Try rational roots, \( x = 3 \) is a root. Divide by \( (x - 3) \): \( x^3 - 5x^2 + 3x + 9=(x - 3)(x^2 - 2x - 3)=(x - 3)(x - 3)(x + 1) \)

Step2: Cancel common factors

Cancel \( (x - 3) \), so \( f(x)=\frac{(x - 4)}{(x - 3)(x + 1)} \) (for \( x
eq 3 \))

Step3: Find vertical asymptotes

Vertical asymptotes where denominator is 0 (and not canceled). Solve \( (x - 3)(x + 1)=0 \), so \( x=-1 \) and \( x = 3 \) (in increasing order)

Answer:

\( x = 2 \)

Problem 6