What is the range of a quadratic function?

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Answer 1 · 2018-03-17T10:55:57

The range of $f(x) = ax^2+bx+c$ is:

${ ([c-b^2/(4a), oo) " if " a > 0), ((-oo, c-b^2/(4a)] " if " a < 0) :}$

Given a quadratic function:

$f(x) = ax^2+bx+c" "$ with $a != 0$

We can complete the square to find:

$f(x) = a(x+b/(2a))^2+(c-b^2/(4a))$

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

Then:

$f(-b/(2a)) = c - b^2/(4a)$

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

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

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

Given:

$y = ax^2+bx+c$

Subtract $y$ from both sides to find:

$ax^2+bx+(c-y) = 0$

The discriminant $Delta$ of this quadratic equation is:

$Delta = b^2-4a(c-y) = (b^2-4ac)+4ay$

In order to have real solutions, we require $Delta >= 0$ and so:

$(b^2-4ac)+4ay >= 0$

Add $4ac-b^2$ to both sides to find:

$4ay >= 4ac-b^2$

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

$y >= c-b^2/(4a)$

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

$y <= c-b^2/(4a)$