Question

Consider the following equation f(x)=(x^2+4)/(4x^2-4x-8)

Answer 1

Question:

Consider the following equation: f(x)=x^2+4\4x^2-4x-8 name the vertical asymptote(s)

Answer:

The vertical asymptotes are x = 2 or x = -1

Solution:

Given that,

$f(x)=\frac{x^2+4}{4x^2-4x-8}$

For vertical asymptotes, set the denominator to zero and then solve the quadratic equation

$4x^2-4x-8=0\\\\Divide\ by\ 2\\\\x^2 - 2x - 4 = 0$

$Split\ the\ middle\ term\\\\x^2+x-2x-2=0\\\\Group\ the\ equation\\\\(x^2 + x) - (2x + 2) = 0\\\\Factor\ the\ common\ term\\\\x(x + 1) - 2(x + 1) = 0\\\\(x + 1)(x - 2) = 0\\\\Therefore,\\\\x + 1 = 0\\\\x = -1\\\\And\\\\x - 2 = 0\\\\x = 2$

Thus the vertical asymptotes are x = - 1 and x = 2

Answer 2

Answer: x=-1 and x=2 because (this is where the function is undefined) name of horizontal asymptote (y=1/4) because (m=n)

Step-by-step explanation: on edge