consider the following equation: $f(x)=\frac{x^{2}+4}{4x^{2}-4x - 8}$\nname the vertical asymptote(s).\n$x=-1…

Explanation:

Step1: Find vertical asymptotes

Vertical asymptotes occur where the denominator of a rational - function is zero. First, factor the denominator of $f(x)=\frac{x^{2}+4}{4x^{2}-4x - 8}$. Factor out a 4 from the denominator: $4x^{2}-4x - 8 = 4(x^{2}-x - 2)$. Then factor $x^{2}-x - 2=(x + 1)(x - 2)$. Set the denominator equal to zero: $4(x + 1)(x - 2)=0$. Solving for $x$, we get $x=-1$ and $x = 2$. These are the values where the function is undefined, so they are the vertical asymptotes.

Step2: Find horizontal asymptotes

For a rational function $y=\frac{f(x)}{g(x)}$ where $f(x)=a_mx^m+\cdots$ and $g(x)=b_nx^n+\cdots$, when $m = n$ (the degree of the numerator and the denominator are the same), the horizontal asymptote is $y=\frac{a_m}{b_n}$. In $f(x)=\frac{x^{2}+4}{4x^{2}-4x - 8}$, the degree of the numerator $m = 2$ and the degree of the denominator $n = 2$, $a_m = 1$ (the coefficient of $x^{2}$ in the numerator) and $b_n = 4$ (the coefficient of $x^{2}$ in the denominator). So the horizontal asymptote is $y=\frac{1}{4}$.

Answer:

Vertical asymptotes: $x=-1$ and $x = 2$; Horizontal asymptote: $y=\frac{1}{4}$