# Maximum value of a quadratic expression

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## Maximum and minimum values of a quadratic polynomial

We will learn how to find the maximum and minimum values of the quadratic expression

$ax^2 + bx + c, \quad a ≠ 0.$

Let $y = ax^2 + bx + c$, then $ax^2 + bx + c - y = 0$.

If $x$ is real, then the discriminant of equation $ax^2 + bx + c - y = 0$ is $D≥ 0:$

$\begin{aligned} b^2 - 4a(c - y) &≥ 0\\ b^2 - 4ac + 4ay &≥ 0\\ 4ay &≥ 4ac - b^2. \end{aligned}$

**Case I:** When $a > 0$

From $4ay ≥ 4ac - b^2,$ we get $y ≥ \dfrac{4ac - b^2}{4a},$ which implies that $y$ has a minimum value, which is $\dfrac{4ac - b^2}{4a}.$
Now, substituting $y = \dfrac{4ac - b^2}{4a}$ in the equation $ax^2 + bx + c - y = 0$ gives
$\begin{aligned}
ax^2 + bx + c - \left(\dfrac{4ac - b^2}{4a}\right) &= 0\\
4a^2x^2 + 4abx + b^2 &= 0\\
(2ax + b)^2 &= 0\\
x &= \dfrac{-b}{2a}.
\end{aligned}$
Therefore, the expression $y$ has its minimum value at $x = \dfrac{-b}{2a}.$

**Case II:** When $a < 0$

From $4ay ≥ 4ac - b^2,$ we get $y ≤ \dfrac{4ac - b^2}{4a},$ which implies that $y$ has a maximum value, which is
$\dfrac{4ac - b^2}{4a}.$
Now, substituting $y = \dfrac{4ac - b^2}{4a}$ in the equation $ax^2 + bx + c - y = 0$ gives
$\begin{aligned}
ax^2 + bx + c -\left(\dfrac{4ac - b^2}{4a}\right) &=0\\
4a^2x^2 + 4abx + b^2 &= 0\\
(2ax + b)^2 &= 0\\
x &= \dfrac{-b}{2a}.
\end{aligned}$.
Therefore, the expression $y$ has its maximum value at $x = \dfrac{-b}{2a}.$

**Cite as:**Maximum value of a quadratic expression.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/maximum-value-of-a-quadratic-equation/