# How do you find domain and range of a quadratic function?

Jun 28, 2018

The domain is all real numbers and the range is all the reals at or above the vertex y coordinate (if the coefficient on the squared term is positive) or all the reals at or below the vertex (if said coefficient is negative).

Jun 28, 2018

See explanation...

#### Explanation:

Suppose:

$f \left(x\right) = a {x}^{2} + b x + c \setminus$ where $a \ne 0$

First note that $f \left(x\right)$ is well defined for any value of $x$. So the domain of $f \left(x\right)$ is $\mathbb{R}$.

We can complete the square and find:

$f \left(x\right) = a {\left(x + \frac{b}{2 a}\right)}^{2} + c - {b}^{2} / \left(4 a\right)$

This is vertex form for a parabola with vertex at $\left(- \frac{b}{2 a} , c - {b}^{2} / \left(4 a\right)\right)$ and multiplier $a$.

Note that for real values of $x$, we have:

${\left(x + \frac{b}{2 a}\right)}^{2} \ge 0$

with equality when $x = - \frac{b}{2 a}$.

Hence if $a > 0$ then $f \left(x\right) \ge c - {b}^{2} / \left(4 a\right)$.

In fact for any $y \ge c - \frac{b}{2 a}$ we have:

$f \left(\frac{- b \pm \sqrt{{b}^{2} - 4 a \left(c - y\right)}}{2 a}\right) = y$

So the range of $f \left(x\right)$ is $\left[c - {b}^{2} / \left(4 a\right) , \infty\right)$

Similarly if $a < 0$ then $f \left(x\right) \le c - {b}^{2} / \left(4 a\right)$ and for any $y < c - \frac{b}{2 a}$ we have:

$f \left(\frac{- b \pm \sqrt{{b}^{2} - 4 a \left(c - y\right)}}{2 a}\right) = y$

So the range of $f \left(x\right)$ is $\left(- \infty , c - {b}^{2} / \left(4 a\right)\right]$