There is no horizontal asymptote.
As x>>>very large, y=x+11 so it is a sloping line it approaches.
What is the horizontal asymptote?
y=x^2+11x+16/x+8
5 answers
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
(x^2+11x+16)/(x+8) = x+3 - 8/(x+8)
So, the asymptote is actually the line y=x+3
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
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
The 3 is tiny, but the curve does not approach the line y=x. It gets arbitrarily close to the line y=x+3.
( 1,000,000 , 1,000,003 )
1 answer