Question

1. find the vertical asymptote(s) of $h(x) = \frac{(x - 6)(x + 1)}{x^2 - 4}$.

Explanation:

Step1: Factor the denominator

To find vertical asymptotes, we first factor the denominator. The denominator is \(x^2 - 4\), which is a difference of squares. Using the formula \(a^2 - b^2=(a - b)(a + b)\), we get \(x^2 - 4=(x - 2)(x + 2)\). So the function becomes \(h(x)=\frac{(x - 6)(x + 1)}{(x - 2)(x + 2)}\).

Step2: Check for common factors

Now we check if the numerator and the denominator have any common factors. The numerator is \((x - 6)(x + 1)\) and the denominator is \((x - 2)(x + 2)\). There are no common factors between the numerator and the denominator.

Step3: Find the values that make the denominator zero

Vertical asymptotes occur where the denominator is zero (and the numerator is not zero, since we already checked for common factors). So we set the denominator equal to zero: \((x - 2)(x + 2)=0\). Solving this equation, we use the zero - product property. If \(ab = 0\), then either \(a = 0\) or \(b = 0\). So \(x - 2=0\) gives \(x = 2\) and \(x+2 = 0\) gives \(x=-2\).

Answer:

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