the graph of $f(x)=\frac{9(x - 6)}{x^{2}-7x + 6}$ has a vertical asymptote at $x = square$.
the graph of $f(x)=\frac{9(x - 6)}{x^{2}-7x + 6}$ has a vertical asymptote at $x = square$.
Answer
Answer:
1
Explanation:
Step1: Factor the denominator
$x^{2}-7x + 6=(x - 1)(x - 6)$
Step2: Find values for vertical asymptote
Set the denominator equal to 0: $(x - 1)(x - 6)=0$. Solving gives $x=1$ or $x = 6$. But when $x = 6$, the numerator is also 0 and we can cancel out the $(x - 6)$ factor. So the vertical - asymptote is at $x = 1$.