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

1 Answer
bp
Apr 17, 2015

A quadratic function represents a parabola, hence its vertex and the axis of symmetry can be found from the given equation as follows

y= #3(x^2 +2x)-1#
= #3 (x^2 +2x +1) +1-3#
= #3(x+1)^2 -2#

Since coefficient of#x^2# is positive, the parabola would open up, its vertex would be at (-1,-2), the axis of symmetry would be line x= -1. X-intercepts would be #-1 +-2/3 sqrt3#
The parabola would also cross y axis at point (0, -1). With these inputs the curve can be easily sketched.

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