# What is the range of a quadratic function?

Mar 17, 2018

The range of $f \left(x\right) = a {x}^{2} + b x + c$ is:

$\left\{\left(\left[c - {b}^{2} / \left(4 a\right) , \infty\right) \text{ if " a > 0), ((-oo, c-b^2/(4a)] " if } a < 0\right)\right.$

#### Explanation:

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

We can complete the square to find:

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

For real values of $x$ the squared term ${\left(x + \frac{b}{2 a}\right)}^{2}$ is non-negative, taking its minimum value $0$ when $x = - \frac{b}{2 a}$.

Then:

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

If $a > 0$ then this is the minimum possible value of $f \left(x\right)$ and the range of $f \left(x\right)$ is $\left[c - {b}^{2} / \left(4 a\right) , \infty\right)$

If $a < 0$ then this is the maximum possible value of $f \left(x\right)$ and the range of $f \left(x\right)$ is $\left(- \infty , c - {b}^{2} / \left(4 a\right)\right]$

Another way of looking at this is to let $y = f \left(x\right)$ and see if there's a solution for $x$ in terms of $y$.

Given:

$y = a {x}^{2} + b x + c$

Subtract $y$ from both sides to find:

$a {x}^{2} + b x + \left(c - y\right) = 0$

The discriminant $\Delta$ of this quadratic equation is:

$\Delta = {b}^{2} - 4 a \left(c - y\right) = \left({b}^{2} - 4 a c\right) + 4 a y$

In order to have real solutions, we require $\Delta \ge 0$ and so:

$\left({b}^{2} - 4 a c\right) + 4 a y \ge 0$

Add $4 a c - {b}^{2}$ to both sides to find:

$4 a y \ge 4 a c - {b}^{2}$

If $a > 0$ then we can simply divide both sides by $4 a$ to get:

$y \ge c - {b}^{2} / \left(4 a\right)$

If $a < 0$ then we can divide both sides by $4 a$ and reverse the inequality to get:

$y \le c - {b}^{2} / \left(4 a\right)$