# 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 \(D≥ 0:\)

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

**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
\[\begin{align}
ax^2 + bx + c - \left(\dfrac{4ac - b^2}{4a}\right) &= 0\\
4a^2x^2 + 4abx + b^2 &= 0\\
(2ax + b)^2 &= 0\\
x &= \dfrac{-b}{2a}.
\end{align}\]
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
\[\begin{align}
ax^2 + bx + c -\left(\dfrac{4ac - b^2}{4a}\right) &=0\\
4a^2x^2 + 4abx + b^2 &= 0\\
(2ax + b)^2 &= 0\\
x &= \dfrac{-b}{2a}.
\end{align}\].
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/