Question

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

Answer 1

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

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

Answer 2

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

How to determine the vertical asymptote?

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

Set the denominator to 0

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

Divide through by 4

$x^2-x-2 = 0$

Expand

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

Factorize

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

Factor out x + 1

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

Solve for x

x = 2 or x = -1

Hence, the vertical asymptotes are x = 2 and x = -1

Read more about vertical asymptotes at:

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