Question

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

Answer 1

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.