Question 136530
a) subtracted the constant term (+6)


b) completed the square by adding the square of half of the coefficient of the x term
__ (8/2)^2=16


c) the equation is now in the "vertex" form __ y-k=(x-h)^2
__ where the vertex is (h,k)


the domain is what values x can be
__ in this case, any real number value for x gives a value for y


the range is what values y can be
__ since the x-h is squared, it can never be less than zero
__ so y+10 can never be less than zero __ so y can't be less than -10


the y value of the vertex is the minimum value for the function (and the lower bound of the range)