1. Answer: (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. Answer: 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. Answer: (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)
