What is the horizontal asymptote?
y=x^2+11x+16/x+8
There is no horizontal asymptote.
As x>>>very large, y=x+11 so it is a sloping line it approaches.
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
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
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 )
To find the horizontal asymptote of a function, you need to examine its behavior as x approaches positive or negative infinity. In the given function, y = (x^2 + 11x + 16) / (x + 8). To determine the horizontal asymptote, follow these steps:
1. Divide the degree of the numerator (2) by the degree of the denominator (1).
- In this case, the numerator has a degree of 2, and the denominator has a degree of 1.
2. If the numerator's degree is greater than the denominator's degree, there is no horizontal asymptote.
- Since 2 > 1, there is no horizontal asymptote.
However, it is still possible to have an oblique or slant asymptote. To determine if one exists, perform long division or synthetic division on the function. In this case, dividing (x^2 + 11x + 16) by (x + 8) yields:
x + 3 - remainder 40
____________________
x + 8 | x^2 + 11x + 16
3. If the remainder is 0 when dividing, there is a slant asymptote.
- In this case, the remainder is not 0, so there is no slant asymptote either.
Therefore, the function y = (x^2 + 11x + 16) / (x + 8) does not have a horizontal or slant asymptote.