How do you find the Vertical, Horizontal, and Oblique Asymptote given $f(x)= (x-4)/(x^2-8x+16)$ ?

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Answer 1 · 2016-11-02T07:47:31

$x=4$ is a Vertical Asymptote

$y=0$ is a Horizontal Asymptote

There is no Slant Asymptote.

This function is defined for all $x in RR$ except $x=4$

$f(x)=(x-4)/(x^2-2(4)x-4^2)$

$f(x)=(x-4)/(x-4)^2$

Domain of $f$ is: $(-oo,4)U(4,+oo)$

The vertical Asymptote is computed by setting the denominator to zero

$x-4=0rArrcolor(blue)(x=4)$ is a $color(blue)( V.A )$

Let us start computing the asymptote:

The degree of numerator is greater than that of the denominator

Therefore , $color(blue)(y=0)$ is a $color(blue)(H.A)$

There is no Slant asymptote.

How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x)= (x-4)/(x^2-8x+16)f(x)= (x-4)/(x^2-8x+16)?