Question

How do you find the domain and range of `y=3x^2+2x+1`?

Answer 1

Since this is a quadratic equation, the domain is ] `-oo,oo` [

Explanation:

For the range, one possible method is to express the equation in vertex form, which gives the coordinates of the minimum point. The y-coordinate of the minimum point would give you the smallest value of the range, and the largest value y can take should be `oo` as a quadratic equation does not have asymptotes.

Answer 2

Domain: `(-oo, +oo)` Range: `[2/3, +oo)`

Explanation:

`y = 3x^2 +2x +1`

`y` is a quadratic function defined `forall x in RR`

Hence the domain of `y` is `(-oo, +oo)`

Since the coefficient of `x^2` is positive `y` will have a minimum value where `y' =0`

`y' = 6x+2` [Power rule]

`6x+2=0 -> x=-1/3`

`y_"max" = y_"(-1/3)"= 3*1/9 -2*1/3+1`

`= 1/3-2/3+1 = 2/3`

Since `y` has no upper bound, the range of `y` is `[2/3, +oo)`

This can be seen by the graph of `y` below.

graph{3x^2 +2x +1 [-5.7, 5.4, -0.846, 4.704]}