Answer :
To determine the vertical asymptotes of the function [tex]\( f(x) = \frac{x^2 - 9x + 14}{x^2 - 5x + 6} \)[/tex], and to evaluate the limits from both sides of the vertical asymptotes, follow these steps:
### Step 1: Factor the Denominator
First, factorize the denominator [tex]\( x^2 - 5x + 6 \)[/tex]:
[tex]\[ x^2 - 5x + 6 = (x - 2)(x - 3) \][/tex]
### Step 2: Identify Vertical Asymptotes
The function will have vertical asymptotes at the values of [tex]\( x \)[/tex] that make the denominator zero, as the numerator doesn't cancel these factors:
[tex]\[ (x - 2)(x - 3) = 0 \][/tex]
[tex]\[ x = 2 \quad \text{and} \quad x = 3 \][/tex]
So, the vertical asymptotes are at [tex]\( x = 2 \)[/tex] and [tex]\( x = 3 \)[/tex].
### Step 3: Evaluate Limits as [tex]\( x \)[/tex] Approaches the Vertical Asymptotes
#### For [tex]\( x = 2 \)[/tex]:
- From the right ([tex]\( x \to 2^+ \)[/tex]):
[tex]\[ \lim_{x \to 2^+} \frac{x^2 - 9x + 14}{x^2 - 5x + 6} = 5 \][/tex]
- From the left ([tex]\( x \to 2^- \)[/tex]):
[tex]\[ \lim_{x \to 2^-} \frac{x^2 - 9x + 14}{x^2 - 5x + 6} = 5 \][/tex]
- As [tex]\( x \)[/tex] approaches [tex]\( 2 \)[/tex] from either side:
[tex]\[ \lim_{x \to 2} \frac{x^2 - 9x + 14}{x^2 - 5x + 6} = 5 \][/tex]
#### For [tex]\( x = 3 \)[/tex]:
- From the right ([tex]\( x \to 3^+ \)[/tex]):
[tex]\[ \lim_{x \to 3^+} \frac{x^2 - 9x + 14}{x^2 - 5x + 6} = -\infty \][/tex]
- From the left ([tex]\( x \to 3^- \)[/tex]):
[tex]\[ \lim_{x \to 3^-} \frac{x^2 - 9x + 14}{x^2 - 5x + 6} = \infty \][/tex]
- As [tex]\( x \)[/tex] approaches [tex]\( 3 \)[/tex] from either side:
[tex]\[ \lim_{x \to 3} \frac{x^2 - 9x + 14}{x^2 - 5x + 6} = -\infty \][/tex]
### Summary
- The function [tex]\( f(x) = \frac{x^2 - 9x + 14}{x^2 - 5x + 6} \)[/tex] has vertical asymptotes at [tex]\( x = 2 \)[/tex] and [tex]\( x = 3 \)[/tex].
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ \lim_{x \to 2^+} f(x) = 5, \quad \lim_{x \to 2^-} f(x) = 5, \quad \lim_{x \to 2} f(x) = 5 \][/tex]
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ \lim_{x \to 3^+} f(x) = -\infty, \quad \lim_{x \to 3^-} f(x) = \infty, \quad \lim_{x \to 3} f(x) = -\infty \][/tex]
### Step 1: Factor the Denominator
First, factorize the denominator [tex]\( x^2 - 5x + 6 \)[/tex]:
[tex]\[ x^2 - 5x + 6 = (x - 2)(x - 3) \][/tex]
### Step 2: Identify Vertical Asymptotes
The function will have vertical asymptotes at the values of [tex]\( x \)[/tex] that make the denominator zero, as the numerator doesn't cancel these factors:
[tex]\[ (x - 2)(x - 3) = 0 \][/tex]
[tex]\[ x = 2 \quad \text{and} \quad x = 3 \][/tex]
So, the vertical asymptotes are at [tex]\( x = 2 \)[/tex] and [tex]\( x = 3 \)[/tex].
### Step 3: Evaluate Limits as [tex]\( x \)[/tex] Approaches the Vertical Asymptotes
#### For [tex]\( x = 2 \)[/tex]:
- From the right ([tex]\( x \to 2^+ \)[/tex]):
[tex]\[ \lim_{x \to 2^+} \frac{x^2 - 9x + 14}{x^2 - 5x + 6} = 5 \][/tex]
- From the left ([tex]\( x \to 2^- \)[/tex]):
[tex]\[ \lim_{x \to 2^-} \frac{x^2 - 9x + 14}{x^2 - 5x + 6} = 5 \][/tex]
- As [tex]\( x \)[/tex] approaches [tex]\( 2 \)[/tex] from either side:
[tex]\[ \lim_{x \to 2} \frac{x^2 - 9x + 14}{x^2 - 5x + 6} = 5 \][/tex]
#### For [tex]\( x = 3 \)[/tex]:
- From the right ([tex]\( x \to 3^+ \)[/tex]):
[tex]\[ \lim_{x \to 3^+} \frac{x^2 - 9x + 14}{x^2 - 5x + 6} = -\infty \][/tex]
- From the left ([tex]\( x \to 3^- \)[/tex]):
[tex]\[ \lim_{x \to 3^-} \frac{x^2 - 9x + 14}{x^2 - 5x + 6} = \infty \][/tex]
- As [tex]\( x \)[/tex] approaches [tex]\( 3 \)[/tex] from either side:
[tex]\[ \lim_{x \to 3} \frac{x^2 - 9x + 14}{x^2 - 5x + 6} = -\infty \][/tex]
### Summary
- The function [tex]\( f(x) = \frac{x^2 - 9x + 14}{x^2 - 5x + 6} \)[/tex] has vertical asymptotes at [tex]\( x = 2 \)[/tex] and [tex]\( x = 3 \)[/tex].
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ \lim_{x \to 2^+} f(x) = 5, \quad \lim_{x \to 2^-} f(x) = 5, \quad \lim_{x \to 2} f(x) = 5 \][/tex]
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ \lim_{x \to 3^+} f(x) = -\infty, \quad \lim_{x \to 3^-} f(x) = \infty, \quad \lim_{x \to 3} f(x) = -\infty \][/tex]