What is the vertex of # y=3x^2 + 6x + 1 #?

1 Answer
Nov 17, 2015

Answer:

#(-1, -2)#

Explanation:

Derivate the function and calculate #y'(0)# to find where the the slope is equal to #0#.

#y = 3x^2 + 6x + 1#
#y' = 2 * 3x^(2-1) + 1 * 6x^(1-0)#
#y' = 6x + 6#

Calculate #y'(0)#:
#y'(0) = 0#
#6x + 6 = 0#
#6x = -6#
#x = -1#

Put this #x# value into the original function to find the y-value.
NOTE: Put it in #y#, not #y'#.

#y = 3*(-1)^2 + 6*(-1) + 1#
#y = 3 * 1 - 6 + 1#
#y = 3 - 6 + 1 = -2#

The vertex is at #(-1, -2)#