Thread: Curve sketching problem

The problem is: f(x)=(9/x)+x+1 where x>0

There are two asymptotes. I found x=0. But i'm stuck as I don't know how to get y=x+1 as an asymptote.

I first solved f'(x) and made that equal zero which I got x=+/- 3. But because x>0, so x=3 ---> (3, 7)

and I computed f''(x), plugged 3 in, the answer >0 so concave up and (3,7) is the minimum value.

I know there is an asymptote at x=0, but what about the other one?

Thanks!

hi

$\displaystyle f(x)=\frac{x^2+x+9}{x}$

so obviously , the vertical asymtote is x=0

the other one is an oblique/slant asymtote , divide x by x^2+x+9 (long division)

Originally Posted by fabxx

The problem is: f(x)=(9/x)+x+1 where x>0

There are two asymptotes. I found x=0. But i'm stuck as I don't know how to get y=x+1 as an asymptote.

As x goes to $\displaystyle \pm \infty$, $\displaystyle \frac{1}{x}$ goes to 0 so the graph is that of the rest of the formula, x+ 1.

I first solved f'(x) and made that equal zero which I got x=+/- 3. But because x>0, so x=3 ---> (3, 7)

and I computed f''(x), plugged 3 in, the answer >0 so concave up and (3,7) is the minimum value.

I know there is an asymptote at x=0, but what about the other one?

Thanks!