Question

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

Answer 1

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.