Question

consider the following equation: \\( f(x) = \frac{x^2 + 4}{4x^2 - 4x - 8} \\) name the vertical asymptote(s). \\( x = -1 \\) and \\( x = 2 \\) because \\( m < n \\) \\( m = n \\) \\( a_m < b_n \\) \\( a_m = b_n \\) this is where the function is undefined name the horizontal asymptote(s). \\( y = 1/4 \\) because \\( m < n \\) \\( m = n \\) \\( a_m < b_n \\) \\( a_m = b_n \\) this is where the function is undefined

Explanation:

Step1: Analyze the degrees of numerator and denominator

For a rational function \( f(x)=\frac{a_mx^m +\cdots}{b_nx^n+\cdots}\), the degree of the numerator \(m\) (for \(x^2 + 4\), \(m = 2\)) and the degree of the denominator \(n\) (for \(4x^2-4x - 8\), \(n=2\)).

Step2: Use the horizontal asymptote rule

When \(m = n\), the horizontal asymptote is \(y=\frac{a_m}{b_n}\). Here \(a_m = 1\) (coefficient of \(x^2\) in numerator) and \(b_n=4\) (coefficient of \(x^2\) in denominator). So \(y=\frac{1}{4}\).

For vertical asymptotes:

Step1: Factor the denominator

Factor \(4x^2-4x - 8=4(x^2 - x - 2)=4(x + 1)(x - 2)\).

Step2: Set the denominator equal to zero

Set \(4(x + 1)(x - 2)=0\). Solving \(x+1 = 0\) gives \(x=-1\) and solving \(x - 2=0\) gives \(x = 2\). The function \(f(x)=\frac{x^2+4}{4x^2-4x - 8}\) is undefined at \(x=-1\) and \(x = 2\) (since denominator is zero at these points and numerator is non - zero at these points).

Answer:

Vertical asymptotes: \(x=-1\) and \(x = 2\) (because the function is undefined at these points). Horizontal asymptote: \(y=\frac{1}{4}\) (because \(m = n\) where \(m\) is the degree of the numerator and \(n\) is the degree of the denominator)