Question

list any vertical asymptotes for the function given below.

$f(x)=\frac{2(x - 7)}{x^{2}-8x + 7}$

$\bigcirc$a. $x = 1$

$\bigcirc$b. $x = 1$ and $x = 7$

$\bigcirc$c. $x = 0$

$\bigcirc$d. $y = 1$

Explanation:

Step1: Factor the denominator

Factor \(x^{2}-8x + 7\) using the formula \(x^{2}+(a + b)x+ab=(x + a)(x + b)\). Here \(a=-1\) and \(b=-7\), so \(x^{2}-8x + 7=(x - 1)(x - 7)\). The function becomes \(f(x)=\frac{2(x - 7)}{(x - 1)(x - 7)}\).

Step2: Simplify the function (if possible)

Cancel out the common factor \((x - 7)\) (for \(x
eq7\)). The simplified function is \(f(x)=\frac{2}{x - 1}\) (with a hole at \(x = 7\) since the original function is undefined at \(x=7\) but the factor cancels).

Step3: Find the vertical asymptote

Set the denominator of the simplified function \(x-1\) equal to zero. Solving \(x - 1=0\) gives \(x = 1\).

Answer:

A. \(x = 1\)