F(x) = x^2 + 6x + 8
= b^2 - 4ac
= (6)^2 - 4(1)(8)
= 36 - 4(8)
= 36 - 32
= 4
g(x) = x2 + 4x + 8
= b^2 - 4ac
= (4)^2 - 4(1)(8)
= 16 - 4(8)
= 16 - 32
= -16
h(x) = x2 – 12x + 32
= b^2 - 4ac
= (-12)^2 - 4(1)(32)
= 144 - 4(32)
= 144 - 128
= 16
k(x) = x2 + 4x – 1
= (4)^2 - 4(1)(-1)
= 16 - 4(-1)
= 16 + 4
= 20
p(x) = 5x2 + 5x + 4
= b^2 - 4ac
= (5)^2 - 4(5)(4)
= 25 - 4(20)
= 25 - 80
= -55
t(x) = x2 – 2x – 15
= b^2 - 4ac
= (-2)^2 - 4(1)(-15)
= 4 - 4(-15)
= 4 + 60
= 64
Answer: D [-1,1]
is the inverse of function g(x).
the inverse equation of g(x) = sin(x):
sin(y) = x
y = 
(x)
using a trig circle with radius 1, we can find out that the max values of sin(x) are 1 and -1, so the domain of the inverse of sin(x) must be [-1,1]
**Note: if allowed during AP test, test from school. or assignment, you can use your graphing calculator to graph (x)
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