Please help Part A
Which is the vertex form of the function f(x) = 3x2 – 30x + 71?
O
A. 3(x - 5)2 - 4
B. 2(x – 6)2 - 1
C. 3(x – 4)2 + 23
D. 5(x – 4)2 – 9
Part B.
What is the vertex of the function?
vertex = (

Option A:

y = 3*(x - 5)^2 - 4

Step-by-step explanation:

For a quadratic equation:

y = a*x^2 + b*x + c

with the vertex (h, k), we can rewrite the function as:

such that:

h = -b/2*a

y = a*(x - h)^2 + k

Here we have the function:

y = 3*x^2 - 30*x + 71

the x-value of the vertex will be:

h = -(-30)/(2*3) = 30/6 = 5

And k is given by:

k = y(5) = 3*(5)^2 - 30*5 + 71 = -4

Then the vertex is:

(5, - 4)

And we can rewrite the equation in the vertex form as:

y = 3*(x - 5)^2 + (-4)

y = 3*(x - 5)^2 - 4

Then the correct option is A.

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.