Answer:
69
Step-by-step explanation:
A'(- 4, 3 )
Under a counterclockwise rotation about the origin of 90°
a point (x, y ) → (- y, x )
A(3, 4 ) → A'(- 4, 3 )
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:
Answer:
1
Step-by-step explanation:
First, we can find the equation of the parabola. The standard form of a parabola is ax^2 + bx + c,
where c is the y-intercept. The y-intercept on the graph is -5, and every option starts with x^2, so the equation must be x^2 - 5. This rules out options 3 and 4.
Next, we can find the equation of the line. The options are all given in slope-intercept form: y = mx + b, where b is the y-intercept. The y-intercept on the graph is 1, and option 1 has 1 in the place of b. Therefore, option 1 is the answer.