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ra1l [238]
3 years ago
15

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

Mathematics
2 answers:
klemol [59]3 years ago
8 0

Answer:

Consider the following equation:

Name the vertical asymptote(s).

A). x = -1 and x = 2

__________________________________________________________

because E). this is where the function is undefined

__________________________________________________________

                       x^2 + 4

 f (x) =         ------------------

                   4x^2 - 4x - 8

Name the horizontal asymptote(s).

D). v = 1/4

__________________________________________________________

because B). m = n

zhenek [66]3 years ago
3 0

ANSWER

The vertical asymptotes are


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

<u>EXPLANATION</u>

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





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