Question

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

Answer 1

Answer:

Consider the following equation:

Name the vertical asymptote(s).

A). x = -1 and x = 2

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because E). this is where the function is undefined

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                       x^2 + 4

 f (x) =         ------------------

                   4x^2 - 4x - 8

Name the horizontal asymptote(s).

D). v = 1/4

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because B). m = n

Answer 2

ANSWER

The vertical asymptotes are

$\Rightarrow x=2\:or\:x=-1$

EXPLANATION

We have

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

For vertical asymptotes we set the denominator to zero and solve the quadratic equation;

$4x^2-4x-8=0$

$\Rightarrow x^2-x-2=0$

We split the middle term to obtain,

$x^2+x-2x-2=0$

$\Rightarrow x(x+1)-2(x+1)=0$

$\Rightarrow (x-2)(x+1)=0$

$\Rightarrow x=2\:or\:x=-1$

Therefore the vertical asymptotes are

$x=2\:or\:x=-1$