# How do you find the domain and range for y=4x^2 - 2x?

May 19, 2018

Domain: $\left(- \infty , + \infty\right)$, Range: $\left[- \frac{1}{4} , + \infty\right)$

#### Explanation:

$y = 4 {x}^{2} - 2 x$

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

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

$y$ is a quadratic function of the form: $a {x}^{2} + b x + c$
Where: $a = 4 , b = - 2 , c = 0$

Since $a > 0$, $y$ will have a minimum value where $x = - \frac{b}{2 a}$

I.e. where $x = \frac{2}{2 \times 4} = \frac{1}{4}$

$\therefore {y}_{\min} = y \left(\frac{1}{4}\right) = 4 \cdot {\left(\frac{1}{4}\right)}^{2} - 2 \cdot \left(\frac{1}{4}\right)$

$= \frac{1}{4} - \frac{1}{2} = - \frac{1}{4}$

Since $y$ has no finite upper bound the range of $y$ is $\left[- \frac{1}{4} , + \infty\right)$

We can infer these results from the graph of $y$ below.

graph{4x^2-2x [-2.35, 3.125, -0.713, 2.024]}