The horizontal asymptote of a rational function tells us the limiting value of that function as it approaches infinity.
For a rational function to have a horizontal asymptote of
![\frac{2}{9}](https://tex.z-dn.net/?f=%20%5Cfrac%7B2%7D%7B9%7D%20)
then,
![the highest degree of x in the numerator must be equal to the highest degree of x in the denominator.](https://tex.z-dn.net/?f=%3Cb%3Ethe%20highest%20degree%20of%20x%20in%20the%20numerator%20must%20be%20equal%20to%20the%20highest%20degree%20of%20x%20in%20the%20denominator.%3C%2Fb%3E)
The second condition is that,
![the coefficient of the highest degree in the numerator and the coefficient of the highest degree in the denominator should be in the ratio 2:9.](https://tex.z-dn.net/?f=%3Cb%3Ethe%20coefficient%20of%20the%20highest%20degree%20in%20the%20numerator%20and%20the%20coefficient%20of%20the%20highest%20degree%20in%20the%20denominator%20should%20be%20in%20the%20ratio%202%3A9.%3C%2Fb%3E)
Example are given in the graph above.
Here are some other examples,
![y = \frac{2x - 1}{9x + 2}](https://tex.z-dn.net/?f=%20y%20%3D%20%5Cfrac%7B2x%20-%201%7D%7B9x%20%2B%202%7D%20)
Answer:
hello :
y = –2x – 1....(1)
y = 3x–1.....(2)
by (1) and (2) : 3x-1 = -2x-1
5x =0
x =0 ....(the x-coordinates of the solutions)
Step-by-step explanation:
Answer:
To find out the first three terms of 3n - 2 substitute 1 ,2 and 3 into the equation.
so the three terms a
3(1)-2=1
3(2)-2=6-2=4
3(3)-2=9-2=7
As you can see the sequence goes up in 1,4 and 7.To find out the 10th term you also substitute 10 into the equation so 3(9)-2=18-2=16
so 16 is the tenth term
Hope this helped!
Answer:
![\large\boxed{7.\ B.\ 8s^2+16s+1}\\\\\boxed{8.\ D.\ -15t^9u^{11}}\\\\\boxed{11.\ C.\ 3,\ -2}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B7.%5C%20B.%5C%208s%5E2%2B16s%2B1%7D%5C%5C%5C%5C%5Cboxed%7B8.%5C%20D.%5C%20-15t%5E9u%5E%7B11%7D%7D%5C%5C%5C%5C%5Cboxed%7B11.%5C%20C.%5C%203%2C%5C%20-2%7D)
Step-by-step explanation:
![7.\\(3s^2+7s+2)+(5s^2+9s-1)=3s^2+7s+2+5s^2+9s-1\\\\\text{combine like terms}\\\\=(3s^2+5s^2)+(7s+9s)+(2-1)\\\\=8s^2+16s+1](https://tex.z-dn.net/?f=7.%5C%5C%283s%5E2%2B7s%2B2%29%2B%285s%5E2%2B9s-1%29%3D3s%5E2%2B7s%2B2%2B5s%5E2%2B9s-1%5C%5C%5C%5C%5Ctext%7Bcombine%20like%20terms%7D%5C%5C%5C%5C%3D%283s%5E2%2B5s%5E2%29%2B%287s%2B9s%29%2B%282-1%29%5C%5C%5C%5C%3D8s%5E2%2B16s%2B1)
![8.\\(-3t^2u^3)(5t^7u^8)=(-3\cdot5)(t^2t^7)(u^3u^8)\qquad\text{use}\ a^na^m=a^{n+m}\\\\=-15t^{2+7}u^{3+8}=-15t^9u^{11}](https://tex.z-dn.net/?f=8.%5C%5C%28-3t%5E2u%5E3%29%285t%5E7u%5E8%29%3D%28-3%5Ccdot5%29%28t%5E2t%5E7%29%28u%5E3u%5E8%29%5Cqquad%5Ctext%7Buse%7D%5C%20a%5Ena%5Em%3Da%5E%7Bn%2Bm%7D%5C%5C%5C%5C%3D-15t%5E%7B2%2B7%7Du%5E%7B3%2B8%7D%3D-15t%5E9u%5E%7B11%7D)
![11.\\n-the\ number\\\\n^2=n+6\qquad\text{subtract}\ n\ \text{and}\ 6\ \text{from both sides}\\\\n^2-n-6=0\\\\n^2+2n-3n-6=0\\\\n(n+2)-3(n+2)=0\\\\(n+2)(n-3)=0\iff n+2=0\ \vee\ n-3=0\\\\n+2=0\qquad\text{subtract 2 from both sides}\\n=-2\\\\n-3=0\qquad\text{add 3 to both sides}\\n=3](https://tex.z-dn.net/?f=11.%5C%5Cn-the%5C%20number%5C%5C%5C%5Cn%5E2%3Dn%2B6%5Cqquad%5Ctext%7Bsubtract%7D%5C%20n%5C%20%5Ctext%7Band%7D%5C%206%5C%20%5Ctext%7Bfrom%20both%20sides%7D%5C%5C%5C%5Cn%5E2-n-6%3D0%5C%5C%5C%5Cn%5E2%2B2n-3n-6%3D0%5C%5C%5C%5Cn%28n%2B2%29-3%28n%2B2%29%3D0%5C%5C%5C%5C%28n%2B2%29%28n-3%29%3D0%5Ciff%20n%2B2%3D0%5C%20%5Cvee%5C%20n-3%3D0%5C%5C%5C%5Cn%2B2%3D0%5Cqquad%5Ctext%7Bsubtract%202%20from%20both%20sides%7D%5C%5Cn%3D-2%5C%5C%5C%5Cn-3%3D0%5Cqquad%5Ctext%7Badd%203%20to%20both%20sides%7D%5C%5Cn%3D3)
Answer:
domain(-8. -2. 1 4
range( -10 0 10 0
and the the given exp is function