How do you find vertical, horizontal and oblique asymptotes for $\frac{x^{4} - 8 x^{2} + 16}{x^{2}}$ $(x^4 - 8x^2 +16) / x^2$ ?

Question

1587 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-03T12:54:26

The vertical asymptote is $x = 0$ $x=0$
No horizontal asymptote
No slant asymptote

Let's rewrite the expression

$\frac{x^{4} - 8 x^{2} + 16}{x^{2}}$ $(x^4-8x^2+16)/x^2$

Let $f (x) = \frac{x^{4} - 8 x^{2} + 16}{x^{2}}$ $f(x)=(x^4-8x^2+16)/x^2$

The domain of $f (x)$ $f(x)$ is $D_f(x)=RR-{0}$

As you canot divide by $0$ , $x!=0$

$lim_(x->+-oo)f(x)=lim_(x->+-oo)x^4/x^2=lim_(x->+-oo)x^2=+oo$

graph{(y-(x^4-8x^2+16)/(x^2))(y-1000x)=0 [-10, 10, -5, 5]}

How do you find vertical, horizontal and oblique asymptotes for (x^4 - 8x^2 +16) / x^2x4−8x2+16x2 (x^4 - 8x^2 +16) / x^2?