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.
$f(x) = \frac{7x - 8}{x^3 - 16x}$

Explanation:

Step1: Factor the denominator

To find vertical asymptotes, we first factor the denominator of the rational function. The denominator is \(x^3 - 16x\). We can factor out an \(x\) first: \(x(x^2 - 16)\). Then, notice that \(x^2 - 16\) is a difference of squares, which factors as \((x - 4)(x + 4)\). So the factored form of the denominator is \(x(x - 4)(x + 4)\).

Step2: Find the values that make the denominator zero

Vertical asymptotes occur where the denominator is zero (and the numerator is not zero at those points). We set the factored denominator equal to zero: \(x(x - 4)(x + 4)=0\). Using the zero - product property, we set each factor equal to zero:

• \(x = 0\)
• \(x - 4=0\) gives \(x = 4\)
• \(x + 4=0\) gives \(x=-4\)

Now we check the numerator at these values. The numerator is \(7x - 8\).

• When \(x = 0\), the numerator is \(7(0)-8=-8

eq0\)

• When \(x = 4\), the numerator is \(7(4)-8=28 - 8=20

eq0\)

• When \(x=-4\), the numerator is \(7(-4)-8=-28 - 8=-36

eq0\)

Since the numerator is not zero at \(x = 0\), \(x = 4\), and \(x=-4\), these are the vertical asymptotes.

Answer:

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