Asked by Math

What is the horizontal asymptote?

y=x^2+11x+16/x+8
Answered by bobpursley
There is no horizontal asymptote.

As x>>>very large, y=x+11 so it is a sloping line it approaches.
Answered by Steve
While the asymptote is a sloping line, and it looks like it ought to be x+11, it is not.

(x^2+11x+16)/(x+8) = x+3 - 8/(x+8)

So, the asymptote is actually the line y=x+3
Answered by Damon
LOL - looks like = x to me, the 3 is tiny

Perhaps she means where the slope is zero?

dy/dx = [(x+8)(2x+11) -x^2-11x-16)]/bottom^2
where is numerator = 0?

0 =2x^2+27x+88-x^2-11x-16
0 = x^2 + 16 x + 72
complex roots :( none

Answered by Steve
The 3 is tiny, but the curve does not approach the line y=x. It gets arbitrarily close to the line y=x+3.
Answered by Damon
( 1,000,000 , 1,000,003 )
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