1. Answer: (8, 15) What are the coordinates of the vertex for the below quadratic function? <span> 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: </span>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)
<span>2. Answer: up Would this quadratic function open up or down? </span>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)<span> Provide two additional points (not the vertex) that would fall on the function </span> The easiest point would be x=0 and x=1 f(x) = x^2-16x + 79 f(0) = 0^2-160 + 79= 79