Question

find all vertical asymptotes of the following function.

f(x)=\frac{6 x-20}{x^{2}-4}

answer attempt 1 out of 2

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Explanation:

Step1: Find the denominator's zeros

Set \(x^{2}-4 = 0\). Using the difference - of - squares formula \(a^{2}-b^{2}=(a + b)(a - b)\), where \(a=x\) and \(b = 2\), we have \((x + 2)(x-2)=0\).
Solving \((x + 2)(x - 2)=0\) gives \(x=-2\) or \(x = 2\).

Step2: Check if the numerator is non - zero at these points

For \(x=-2\): Substitute \(x=-2\) into the numerator \(6x-20\), we get \(6\times(-2)-20=-12 - 20=-32
eq0\).
For \(x = 2\): Substitute \(x = 2\) into the numerator \(6x-20\), we get \(6\times2-20=12-20=-8
eq0\).

Since the denominator is zero at \(x=-2\) and \(x = 2\) and the numerator is non - zero at these points, the function \(y=\frac{6x - 20}{x^{2}-4}\) has vertical asymptotes.

Answer:

The vertical asymptotes are \(x=-2\) and \(x = 2\).