What are the coordinates of the vertex for the below quadratic function? would this quadratic function open up or down? provide two additional points (not the vertex) that would fall on the function. f(x) = (x-8)2+15

1. (8, 15)
What are the coordinates of the vertex for the below quadratic function? 
Differentiate the function

f(x) = (x-8)^2+15
f'(x)= 2(x-8)

The vertex would be when f'(x)= 0. The calculation would be: 
f'(x)= 2(x-8) =0
2x-16=0
2x=16
x=8

If you put x=8 into the equation you will get:
f(x) = (x-8)^2+15
f(8) = (8-8)^2+15
f(8)= 15
The coordinate would be: (8, 15)

2. up
Would this quadratic function open up or down? 
f(x) = (x-8)^2+15
f(x) = (x-8)(x-8)+15
f(x) = x^2-8x-8x+64+15
f(x) = x^2-16x + 79
If you look at the function, the biggest exponent would be the x^2 and the coefficient is +1. The quadratic function would open up because the biggest exponent is positive. The vertex would be the lowest point

3.  (0, 79) and (1, 62)
Provide two additional points (not the vertex) that would fall on the function

The easiest point would be x=0 and x=1
f(x) = x^2-16x + 79
f(0) = 0^2-160 + 79= 79

f(x) = x^2-16x + 79
f(1) = 1^2-16(1) + 79
f(1) = 1 -16 + 79= 62

The point would be: (0, 79) and (1, 62)

Igot the answer of 15.8 years.