# Sketching the graphs of functions using differentiation.

• Jul 28th 2009, 07:16 AM
NIKKI020
Sketching the graphs of functions using differentiation.
Hi, i have got the horzontal asymptote wrong on the following question but can't work out why.

The function is f(x) = 8(5+x)/16-x^2

I have worked out that the vertical asymtotes are x=4 and x=-4 because the denominator is 0 when x = 4 and -4.

and that the stationary points are f(-8)=1/2 and f(-2) = 2

and that the x intercept is -5 and the y is 5/2

But when i divide both the numerator and the denomiator by x^2 to get the horizontal asymtote and look at the behaviour as x tend to infinity i keep getting 0. This doesn't work on the graph because of the values of the stationary point f(-8) and the x intercept.

I'm really stuck on it, if anyone can see where i'm going wrong i'd be grateful. Cheers.

Nikki
• Jul 28th 2009, 07:29 AM
songoku
I don't check the details of your work, but you got it right with the horizontal asymptote. It's y = 0

FYI, some of the graphs have intersection with horizontal asymptote
• Jul 28th 2009, 07:37 AM
earboth
Quote:

Originally Posted by NIKKI020
Hi, i have got the horzontal asymptote wrong on the following question but can't work out why.

The function is f(x) = 8(5+x)/16-x^2

I have worked out that the vertical asymtotes are x=4 and x=-4 because the denominator is 0 when x = 4 and -4.

and that the stationary points are f(-8)=1/2 and f(-2) = 2

and that the x intercept is -5 and the y is 5/2

But when i divide both the numerator and the denomiator by x^2 to get the horizontal asymtote and look at the behaviour as x tend to infinity i keep getting 0. This doesn't work on the graph because of the values of the stationary point f(-8) and the x intercept.

I'm really stuck on it, if anyone can see where i'm going wrong i'd be grateful. Cheers.

Nikki

Hmmm ...

everything is OK! Have a look on the graph!
• Jul 28th 2009, 07:51 AM
NIKKI020
Thanks for that i'll leave it how it is then.
Cheers

Nikki

Quote:

Originally Posted by songoku
I don't check the details of your work, but you got it right with the horizontal asymptote. It's y = 0

FYI, some of the graphs have intersection with horizontal asymptote

• Jul 28th 2009, 07:52 AM
NIKKI020
Thanks, that's been a big help drawing the graph as well.

Cheers
Nikki

Quote:

Originally Posted by earboth
Hmmm ...

everything is OK! Have a look on the graph!