Question

identify the vertical asymptote of the function.\\( f(x) = \frac{8}{x + 2} - 2 \\)

Explanation:

Step1: Recall vertical asymptote rule

For a rational function, vertical asymptotes occur where the denominator is zero (and numerator is non - zero at that point).
The function is \(f(x)=\frac{8}{x + 2}-2\). The rational part is \(\frac{8}{x + 2}\), and its denominator is \(x+2\).

Step2: Solve for denominator zero

Set the denominator equal to zero: \(x + 2=0\).
Solving for \(x\), we get \(x=- 2\).
We check the numerator of the rational part at \(x = - 2\). The numerator is 8, which is non - zero. Also, the subtraction of 2 does not affect the vertical asymptote (since vertical asymptotes are determined by the values that make the function undefined, and the \(-2\) is a vertical shift which does not change the x - value where the function is undefined).

Answer:

The vertical asymptote of the function \(f(x)=\frac{8}{x + 2}-2\) is \(x=-2\)