find the horizontal asymtote of the equation
y=x-2/16-x^2

Question

find the horizontal asymtote of the equation
y=x-2/16-x^2

anonymous · Answer

Do you mean $$\frac{x-2}{16}$$ or $$x-\frac{2}{16}$$

anonymous · Answer

Is it $$y=\frac{x-2}{16-x^2}$$

anonymous · Answer

$$y=\frac{ x-2 }{ 16-x^2 }$$

anonymous · Answer

i think you gave the intercept of the the equation

anonymous · Answer

Divide numerator and denominator by $$x^2$$ Then, you apply limit for $$\lim_{x\rightarrow \pm \infty}$$

anonymous · Answer

horizontal asymptote means,
the value that y-approaches to as x becomes +ve or -ve infinite.
steps:
1. make the substitution x=1/u
2. simplify 
3. put u=0 (as x->±inf, u->0)
4. you will get a value of y

anonymous · Answer

i think the u in there should be the lowest degree of x isn't it?

anonymous · Answer

You end up getting same answer, but yeah, in general, you use lower degree one.

anonymous · Answer

ok i'll try it thanks

anonymous · Answer

no need. just make the substitution and it will always work.

anonymous · Answer

there is a shortcut too.... (dont tell your prof ;))

1. find highest power of x = n (in this case, 2)
2. y = ratio of the coefficients of the x^x terms on top and bottom

in this case. y = 0/(-1) = 0
so, the horiz asymptote is y=0

anonymous · Answer

ahaha thanks i'll keep it for myself^_^