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
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.