How do you find the vertex and the intercepts for # f(x)= 3x^2 + 12x + 1#?

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Answer 1 · 2017-05-24T00:45:16

Vertex #(-2, -11)#

#f(x) = 3x^2 + 12x + 1#

x-coordinate of vertex:
#x = -b/(2a) = -12/6 = - 2#

y-coordinate of vertex: use #x=-2#
#f(-2) = 12 - 24 + 1 = - 11#

Vertex #(-2, -11)#.

To find #x#-intercepts, make #f(x) = 0,# and solve the quadratic equation:

#3x^2 + 12x + 1 = 0#

#D = d^2 = b^2 - 4ac = 144 - 12 = 132 rarr #

#d = +- 2sqrt33#

There are 2 real roots (two #x#-intercepts):

#x = - 12/6 +- (2sqrt33)/6 = - 2 +- sqrt33/3#