Question

find the vertical asymptotes of the function f(x)=\frac{3x(x + 2)}{x^{2}-1}

Explanation:

Step1: Factor the denominator

$$x^{2}-1=(x + 1)(x - 1)$$

Step2: Set the denominator equal to zero

$$(x + 1)(x - 1)=0$$
$$x+1 = 0\quad\text{or}\quad x - 1=0$$
$$x=-1\quad\text{or}\quad x = 1$$

Step3: Check if numerator is zero at these \(x\) - values

For \(x=-1\): \(3\times(-1)\times(-1 + 2)=-3
eq0\)
For \(x = 1\): \(3\times1\times(1 + 2)=9
eq0\)

Answer:

The vertical asymptotes of the function \(y = f(x)\) are \(x=-1\) and \(x = 1\)