Mikä on symmetrian akseli ja piste graafin y = -3x ^ 2 + 12x + 4 osalta?

Vastaus:

aos = 2

kärki = (2,16)

#y = -3x ^ 2 + 12x + 4 #

#f (x) = -3x ^ 2 + 12x + 4 #

Muodossa # Y = ax ^ 2 + bx + c # sinulla on:

# A = -3 #

# B = 12 #

# C = 4 #

Symmetria-akseli (aos) on: #aos = (- b) / (2a) = (-12) / (2 * -3) = 2 #

Muistaa # Y = f (x) #

Vertex on: # (aos, f (aos)) = (2, f (2)) #:

#f (x) = -3x ^ 2 + 12x + 4 #

#f (2) = -3 (2) ^ 2 + 12 * 2 + 4 = 16 #

kärki #=(2, 16)#

kaavio {-3x ^ 2 + 12x + 4 -16.71, 23.29, -1.6, 18.4}

Vertex -

#(2,16)#

Symmetria-akseli

# X = 2 #

Annettu -

# Y = 3x ^ 2 + 12x + 4 #

Vertex -

#X = (- b) / (2a) = (- 12) / (2xx-3) = (- 12) / (- 6) = 2 #

at # X = 2; y = -3 (2 ^ 2) +12 (2) + 4 #

# Y = -12 + 24 + 4 = 16 #

#(2,16)#

Symmetria-akseli

# X = 2 #