Y = (x + 2) ^ 2 + 3x + 4의 정점은 무엇입니까?

대답:

# "정점 {-3.5", "-4.25} #

설명:

#y = (x + 2) ^ 2 + 3x + 4 #

# y = x ^ 2 + 4x + 4 + 3x + 4 #

# y = x ^ 2 + 7x + 8 "(1)"#

# (dy) / (dx) = 0 #

# (dy) / (dx) = 2x + 7 = 0 #

# 2x + 7 = 0 #

# 2x = -7 #

# x = -7 / 2 = -3.5 #

# "use (1)"#

#y = (- 7/2) ^ 2-7 (7/2) + 8 #

# y = 49 / 4-49 / 2 + 8 #

# y = (49-98 + 32) / 4 #

# y = -17 / 4 = -4.25 #

# "정점 {-3.5", "-4.25} #