? idont understsnd be more specific
Answer:
answe is about 23. something
Step-by-step explanation:
The standard form of a quadratic equation is
![\displaystyle{ y=ax^2+bx+c](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%20y%3Dax%5E2%2Bbx%2Bc)
, while the vertex form is:
![y=a(x-h)^2+k](https://tex.z-dn.net/?f=y%3Da%28x-h%29%5E2%2Bk)
, where (h, k) is the vertex of the parabola.
What we want is to write
![\displaystyle{ y=3x^2-18x-6](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%20y%3D3x%5E2-18x-6)
as
![y=a(x-h)^2+k](https://tex.z-dn.net/?f=y%3Da%28x-h%29%5E2%2Bk)
First, we note that all the three terms have a factor of 3, so we factorize it and write:
![\displaystyle{ y=3(x^2-6x-2)](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%20y%3D3%28x%5E2-6x-2%29)
.
Second, we notice that
![x^2-6x](https://tex.z-dn.net/?f=x%5E2-6x)
are the terms produced by
![(x-3)^2=x^2-6x+9](https://tex.z-dn.net/?f=%28x-3%29%5E2%3Dx%5E2-6x%2B9)
, without the 9. So we can write:
![x^2-6x=(x-3)^2-9](https://tex.z-dn.net/?f=x%5E2-6x%3D%28x-3%29%5E2-9)
, and substituting in
![\displaystyle{ y=3(x^2-6x-2)](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%20y%3D3%28x%5E2-6x-2%29)
we have:
![\displaystyle{ y=3(x^2-6x-2)=3[(x-3)^2-9-2]=3[(x-3)^2-11]](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%20y%3D3%28x%5E2-6x-2%29%3D3%5B%28x-3%29%5E2-9-2%5D%3D3%5B%28x-3%29%5E2-11%5D)
.
Finally, distributing 3 over the two terms in the brackets we have:
![y=3[x-3]^2-33](https://tex.z-dn.net/?f=y%3D3%5Bx-3%5D%5E2-33)
.
Answer:
Answer:
I believe it would be - (n+4) x 2
Step-by-step explanation:
Answer:
its the second answer!
Step-by-step explanation: