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
Explanation:
Step1: Factor the denominator
$x^{2}-7x + 6=(x - 1)(x - 6)$
Step2: Find the values that make the denominator zero
Set $(x - 1)(x - 6)=0$. Then $x-1 = 0$ gives $x = 1$ and $x - 6=0$ gives $x = 6$. But when $x = 6$, the numerator $9(x - 6)=0$ too. After simplifying $f(x)=\frac{9(x - 6)}{(x - 1)(x - 6)}=\frac{9}{x - 1},x\neq6$.
Step3: Determine the vertical - asymptote
The vertical asymptote occurs when the denominator of the simplified rational function is zero. For $y=\frac{9}{x - 1}$, setting $x - 1=0$ gives $x = 1$.
Answer:
$1$