How do you graph #y=3x^2 +6x+1#?

1 Answer
bp
Apr 17, 2015

The equation represents a quadratic function, hence its graph would be a vertical parabola, opening upwards, as the coefficient of #x^2# is positive. To sketch the curve we can get the vertex and the axis of symmetry from the give equation as follows:

y= #3(x^2 +2x) +1#
= #3(x^2 +2x+1) +1-3#
= #3(x+1)^2 -2#
This shows the vertex as (-1, -2) and the axis of symmetry x= -1. The x intercepts would be#-1 +- 1/3 sqrt6#. The y intercept would be (0,1). With all these inputs, the curve can be easily sketched.

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