Question

determine the vertical asymptote(s) of the following function. enter each asymptote as a separate equation, separating multiple asymptotes with a comma. if there are no vertical asymptotes, enter none.

$$ f ( x ) = \frac { 5 x - 6 } { x ^ { 3 } - 16 x } $$

Explanation:

Step1: Factor the denominator

Factor \(x^{3}-16x\) as \(x(x^{2} - 16)=x(x - 4)(x + 4)\) using the difference - of - squares formula \(a^{2}-b^{2}=(a - b)(a + b)\) (here \(a=x\) and \(b = 4\)). So the function becomes \(f(x)=\frac{5x - 6}{x(x - 4)(x + 4)}\).

Step2: Find the vertical asymptotes

Vertical asymptotes occur at the values of \(x\) that make the denominator equal to zero (while the numerator is not zero at those values).
Set \(x(x - 4)(x + 4)=0\). Solving \(x=0\), \(x - 4=0\) (i.e., \(x = 4\)), and \(x+4=0\) (i.e., \(x=-4\)).
Check the numerator at these values:
When \(x = 0\), \(5x-6=5\times0 - 6=-6
eq0\).
When \(x = 4\), \(5x-6=5\times4-6 = 14
eq0\).
When \(x=-4\), \(5x-6=5\times(-4)-6=-26
eq0\).

Answer:

\(x = 0,x = 4,x=-4\)