Question

consider the following rational function.
$f(x) = \frac{x^2 + 10}{x^3 - 64}$
step 1 of 3: find equations for the vertical asymptotes, if any, for the function.
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Explanation:

Step1: Factor Denominator

Factor \(x^3 - 64\) using difference of cubes: \(a^3 - b^3=(a - b)(a^2+ab + b^2)\). Here, \(a = x\), \(b = 4\), so \(x^3 - 64=(x - 4)(x^2+4x + 16)\).

Step2: Find Vertical Asymptotes

Vertical asymptotes occur where denominator is zero (and numerator non - zero). Solve \(x^3 - 64 = 0\), i.e., \((x - 4)(x^2+4x + 16)=0\).

• For \(x - 4=0\), \(x = 4\).
• For \(x^2+4x + 16 = 0\), discriminant \(\Delta=b^2 - 4ac=16-64=-48<0\), so no real roots.

Check numerator at \(x = 4\): \(4^2+10 = 16 + 10=26
eq0\). So vertical asymptote is \(x = 4\).

Answer:

\(x = 4\)