PLEASE HELP ASAP 30 POINTS MY BRAINS TO FRIED TO THINK. I ALREADY DID A PART OF THE ANSWER, PLEASE SOLVE THE REST.

An expression is shown below:

f(x) = −16x2 + 22x + 3

What are the coordinates of the vertex? Justify your answers and show your work.

ANSWER:
To find the vertex: the x value = f(x)=ax^2+bx+c= is -b/(2a), and y value = f(-b/(2a)).
Given: f(x)= −16x2 + 22x + 3
a= -16 and b= 22
b-/2a= -(22)/(-16*2)=11/16
PLEASE SOLVE THE REST

Vertex is at (11/16, 169/16)

Step-by-step explanation:

f(x) = −16x^2 + 22x + 3

The vertex is at x = -b/2a

x = -22/(2)(-16) = -22/-32 = 11/16

f(11/16) = -16 (11/16)^2 +22(11/16) +3

           =-16 *11/16 *11/16 +22*11/16 +3

            = -121/16 +242/16 +3

           -121/16 +242/16 +48/16

              169/16

Vertex is at (11/16, 169/16)

Vertex: (11/16,169/16)

Step-by-step explanation:

You plug in the x value to find the y value so

-16(11/16)^2+22(11/16)+3=

-16(121/256)+121/8+3=

-121/16+145/8=

Y=169/16

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'.

slope= 1

y-inter slope= 0

x       y

o     0

1       1

2     2

ect.

step-by-step explanation: