Find the vertical asymptotes of the function f(x) = \frac{x^4 - 4x^3 -x^2 - 4}{2x^3-2}

Question:

Find the vertical asymptotes of the function {eq}f(x) = \frac{x^4 - 4x^3 -x^2 - 4}{2x^3-2} {/eq}

Answer and Explanation:

We are given the function

{eq}\displaystyle f(x) = \frac{x^4 - 4x^3 -x^2 - 4}{2x^3-2} = \frac{x^4 - 4x^3 -x^2 - 4}{2(x^3-1)} {/eq}

The vertical asymtptotes of the function are defined by the left and right limits of the function

at points where the denominator is zero

{eq}\displaystyle 2(x^3-1) = 0 \Rightarrow x=1. {/eq}

Specifically, we have that

{eq}\displaystyle \lim{ x\to 1^- } f(x) = \lim{ x\to 1^- } \frac{x^4 - 4x^3 -x^2 - 4}{2(x^3-1)} = \frac{1^{-} - 4^{-} -1^{-} -4}{0^-} = \infty \\ \displaystyle \lim{ x\to 1^+ } f(x) = \lim{ x\to 1^+ } \frac{x^4 - 4x^3 -x^2 - 4}{2(x^3-1)} = \frac{1^{+} - 4^{+} -1^{+} -4}{0^+} = -\infty \\ {/eq}


Learn more about this topic:

Vertical Asymptotes: Definition & Rules

from High School Algebra II: Help and Review

Chapter 16 / Lesson 14
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