What are the important points needed to graph #y=3x^2+6x+1#?

1 Answer
Edric Y.
Nov 5, 2015

The vertex: #(-1,-2)#
The y-intercept: #(0,1)#
The y-intercept reflected over axis of symmetry: #(-2,1)#

Explanation:

#(-b)/(2a) = (-6)/(2*3) = -1# This is the x-coordinate of the vertex.
#y=3(-1)^2+6(-1)+1=-2# This is the y-coordinate of the vertex.

The vertex: #(-1,-2)#

Now plug in #0# for x:
#y = 3(0)^2 + 6(0) + 1 = 1#

The y-intercept: #(0,1)#

Now reflect that point over the axis of symmetry (#x=-1#) to get
#(-2,1)#
to get this, you take #-1 - (0 - (-1))#

Related questions
See all questions in Quadratic Functions and Their Graphs
Impact of this question
1390 views around the world
You can reuse this answer
Creative Commons License