Maximum value of a quadratic expression

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 \(x = \dfrac{-b}{2a}.\)