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:
No.
When you are distributing, you are multiplying to each monomial
4(4a) = 16a
4(20b) = 80b
16a + 80b is your actual answer
hope this helps
Answer:
1 1/6
Step-by-step explanation:
im guessing you need this fast so i wont explain it
Complete question is;
Peter drew two rays, AC and AP with A as a common endpoint. Which of the following statements
might describe Peter's drawing?
I. AC and AP are parallel.
II. PAC is an angle
III. AC and AP are perpendicular
A. I and II
B. II and III
C. I and 111
D. I, II, and III
Answer:
Option B: II & III
Step-by-step explanation:
We are told that Peter drew two rays which are AC and AP.
We are told that A is a common endpoint.
If A is a common endpoint, it means the 2 rays interaction point at A is at an angle.
The angle could also be 90° which means it's possible that the rays AC and AP are perpendicular.
Thus, the correct statements that describe his drawing are: II & III
1 and 6 are not corresponding angles. They have different degrees. hope this helps :)
