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

May 1, 2017

Since this is a quadratic equation, the domain is ] $- \infty , \infty$ [

#### 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 $\infty$ as a quadratic equation does not have asymptotes.

May 1, 2017

Domain: $\left(- \infty , + \infty\right)$ Range: $\left[\frac{2}{3} , + \infty\right)$

#### Explanation:

$y = 3 {x}^{2} + 2 x + 1$

$y$ is a quadratic function defined $\forall x \in \mathbb{R}$

Hence the domain of $y$ is $\left(- \infty , + \infty\right)$

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

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

$6 x + 2 = 0 \to x = - \frac{1}{3}$

${y}_{\text{max" = y_"(-1/3)}} = 3 \cdot \frac{1}{9} - 2 \cdot \frac{1}{3} + 1$

$= \frac{1}{3} - \frac{2}{3} + 1 = \frac{2}{3}$

Since $y$ has no upper bound, the range of $y$ is $\left[\frac{2}{3} , + \infty\right)$

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

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