X^2 + 8x + 6
(x^2 + 8x + ) + 6
erin was completing the square of the quadratic function in order to find the extreme value. what is the next step in the process, and what is the extreme value?
a) (x + 8)2 - 58; the extreme minimum is -8
b) (x + 4)2 - 10; the extreme minimum is -4
c) (x2 + 8x + 16) + 6 - 16; the extreme minimum is -10
d) (x2 + 8x + 64) + 6 - 64; the extreme minimum is -58

Ax^2+bx+c=0
a=leading term

ok so if the leading term is positive then opens up and has a min
if leading term is negative then opens down and has a max

leading term is positive
1x^2+8x
has a max

to complete the square, move c aside take 1/2 of b and square it
b=8
8/2=4
4^2=16
now add that to both sides
x^2+8x+16+6=0+16
factor perfect square
(x+4)^2+6=16
subtract 6
(x+4)^2=10
subtract 10
(x+4)^2-10=0
vertex aka min or max is (h,k) when ou have
y=a(x-h)+k
h=-4
k=-10

C

(-14+3/2b) - (1+8/2b) = -5 -5/2b(-10+b) + (7b-5) = 8b-15(7/2b-3) - (8+6b) = -5/2b-11(5+2b) + (2b+3/2) = 4b+ 13/2