To find the vertical asymptotes of the function
[tex]\[ f(x) = \frac{x - 6}{x^2 - 13x + 42}, \][/tex]
we need to determine where the denominator of the function is equal to zero. Vertical asymptotes occur at values of [tex]\(x\)[/tex] where the denominator is zero and the numerator is not zero.
1. Identify the denominator and set it equal to zero:
[tex]\[ x^2 - 13x + 42 = 0. \][/tex]
2. Solve the quadratic equation:
To solve [tex]\(x^2 - 13x + 42 = 0\)[/tex], we can factorize it. We are looking for two numbers that multiply to [tex]\(42\)[/tex] and add up to [tex]\(-13\)[/tex].
These numbers are [tex]\(-6\)[/tex] and [tex]\(-7\)[/tex] because:
[tex]\[ (-6) \times (-7) = 42, \][/tex]
[tex]\[ (-6) + (-7) = -13. \][/tex]
3. Factorize the quadratic equation:
[tex]\[ x^2 - 13x + 42 = (x - 6)(x - 7) = 0. \][/tex]
4. Set each factor equal to zero:
[tex]\[ x - 6 = 0 \quad \text{or} \quad x - 7 = 0. \][/tex]
5. Solve for [tex]\(x\)[/tex]:
[tex]\[ x = 6 \quad \text{or} \quad x = 7. \][/tex]
These solutions, [tex]\(x = 6\)[/tex] and [tex]\(x = 7\)[/tex], are the values of [tex]\(x\)[/tex] that make the denominator zero.
6. Verify if the numerator is non-zero at these points:
We check if the numerator [tex]\(x - 6\)[/tex] is non-zero at [tex]\(x = 6\)[/tex] and [tex]\(x = 7\)[/tex].
- At [tex]\( x = 6 \)[/tex]:
[tex]\[ x - 6 = 6 - 6 = 0. \][/tex]
Thus, [tex]\( x = 6 \)[/tex] causes both the numerator and denominator to be zero. This indicates that [tex]\( x = 6 \)[/tex] is not a vertical asymptote, but rather a removable discontinuity.
- At [tex]\( x = 7 \)[/tex]:
[tex]\[ x - 6 = 7 - 6 = 1 \neq 0. \][/tex]
Thus, [tex]\( x = 7 \)[/tex] makes the denominator zero but does not make the numerator zero.
Therefore, the vertical asymptote of the function [tex]\( f(x) \)[/tex] is:
[tex]\[ \boxed{x = 7}. \][/tex]
