Question

list any vertical asymptotes for the function given below.
f(x)=\frac{2}{x^{2}+8 x}
a. ( y = 0 ) and ( y = -8 )
b. ( x = 0 ) and ( x = -8 )
c. ( y = 2 )
d. ( x = 0 ) and ( x = 8 )

Explanation:

Step1: Find the denominator of the function

The function is \( f(x)=\frac{2}{x^{2}+8x} \), and the denominator is \( x^{2}+8x \).

Step2: Factor the denominator

Factor \( x^{2}+8x \) as \( x(x + 8) \).

Step3: Set the denominator equal to zero

Set \( x(x + 8)=0 \). Using the zero - product property \( ab = 0\) implies \(a = 0\) or \(b = 0\). So \(x=0\) or \(x+8 = 0\) (i.e., \(x=-8\)).

Step4: Recall the definition of vertical asymptotes

For a rational function \(y=\frac{f(x)}{g(x)}\), vertical asymptotes occur at the values of \(x\) that make \(g(x)=0\) (provided \(f(x)
eq0\) at those \(x\) - values). Here, the numerator \(f(x) = 2
eq0\) for all \(x\).

Answer:

B. \(x = 0\) and \(x=-8\)