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

consider the following equation: $f(x)=\frac{x^{2}+4}{4x^{2}-4x - 8}$. name the vertical asymptote(s). $x=-1$ and $x = 2$ complete because $m < n$ $m=n$ $a_{m}<b_{n}$ $a_{m}=b_{n}$ this is where the function is undefined done

consider the following equation: $f(x)=\frac{x^{2}+4}{4x^{2}-4x - 8}$. name the vertical asymptote(s). $x=-1$ and $x = 2$ complete because $m < n$ $m=n$ $a_{m}<b_{n}$ $a_{m}=b_{n}$ this is where the function is undefined done

Answer

Explanation:

Step1: Recall vertical - asymptote concept

Vertical asymptotes occur where the denominator of a rational function is zero and the numerator is non - zero.

Step2: Set the denominator equal to zero

Set $4x^{2}-4x - 8=0$. First, factor out a common factor of 4: $4(x^{2}-x - 2)=0$. Then factor the quadratic inside the parentheses: $4(x - 2)(x+1)=0$.

Step3: Solve for x

Using the zero - product property, if $4(x - 2)(x + 1)=0$, then $x-2 = 0$ or $x + 1=0$. Solving these equations gives $x=2$ and $x=-1$. At these values, the function $f(x)=\frac{x^{2}+4}{4x^{2}-4x - 8}$ is undefined since the denominator is zero while the numerator $x^{2}+4$ is non - zero for $x = 2$ and $x=-1$.

Answer:

this is where the function is undefined