Question

I'm trying to sketch a curve can someone check what I have so far:
Domain: all real numbers except 8
y intercept: 8
x intercept: none
Vertical A: 8
horizontal A: 0
function: f(x)= x-8/x^2
first derivative: 1+16/x^3
critcal number = 8
second derivative: -48/x^4

Answer 1

try facebook maths nerds group they anser ur questions

Answer 2

Is the function:
\(\displaystyle f(x)=x-\frac{8}{x^2}\) or \(\displaystyle f(x)=\frac{x-8}{x^2}\)

Also the asymptotes and intercepts are wrong

Answer 3

\[f(x)= x-8/x^2\]

Answer 4

The function explodes at zero. So you will have a vertical asymptote there. You will also have an oblique asymptote because the limiting behaviour of the function is \(g(x)=x\)

Answer 5

\[\lim_{x \to \infty} x-\frac{8}{x^2} = x\]

Answer 6

I take it thats the horixonal asymptote right?

Answer 7

It's an oblique asymptote, not a horizontal asymptote.
http://en.wikipedia.org/wiki/Asymptote#Oblique_asymptotes

Answer 8

oh okay

Answer 9

You are considering what happens when x gets very large (negative or positive). Think about what happens.

Answer 10

The term \(\displaystyle\frac{8}{x^2}\) collapses

Answer 11

All you are left with is \(x\)

Answer 12

oh ok so for example this function:

\[f(x)=x^2+1/x\] would have a slant asmpotote as well because its undefined at 0?

Answer 13

You will have a vertical asymptote at zero, since the term \(\displaystyle\frac{1}{x}\) explodes at 0. You will have a quadratic asymptote for large x.