Question

the graph of $f(x) = \frac{9(x - 6)}{x^2 - 7x + 6}$ has a vertical asymptote at $x = \square$.

Explanation:

Step1: Factor the denominator

To find vertical asymptotes, we first factor the denominator \(x^2 - 7x + 6\). We need two numbers that multiply to \(6\) and add to \(-7\). Those numbers are \(-1\) and \(-6\), so the denominator factors as \((x - 1)(x - 6)\). So the function becomes \(f(x)=\frac{9(x - 6)}{(x - 1)(x - 6)}\).

Step2: Simplify the function (remove common factors)

We can cancel out the common factor of \((x - 6)\) from the numerator and the denominator, but we have to note that \(x
eq6\) (since we can't divide by zero). After canceling, the function simplifies to \(f(x)=\frac{9}{x - 1}\) for \(x
eq6\).

Step3: Find where the denominator is zero (excluding holes)

Vertical asymptotes occur where the denominator is zero, but we have to exclude any values that make both the numerator and denominator zero (those are holes). We already canceled \((x - 6)\), so the denominator of the simplified function is \(x - 1\). Setting \(x - 1 = 0\) gives \(x = 1\). The value \(x = 6\) is a hole, not a vertical asymptote, because it made both the numerator and denominator zero before simplifying.

Answer:

\(1\)