Where is the vertex of y=3(x-26)^2+14?
1 answer:
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The standard form of a quadratic equation is 
, while the vertex form is:
, where (h, k) is the vertex of the parabola.
What we want is to write
as
First, we note that all the three terms have a factor of 3, so we factorize it and write:
.
Second, we notice that
are the terms produced by 
, without the 9. So we can write:
, and substituting in 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:
What we want is to write
First, we note that all the three terms have a factor of 3, so we factorize it and write:
Second, we notice that
Finally, distributing 3 over the two terms in the brackets we have:
Answer:
You know that BC is congruent to x so you need to solve for x using the ratio:
So then we need to find BC.
We know:
Therefore BC =8
Then:
So then we need to find BC.
We know:
Therefore BC =8
Then: