Prove that, if $x^∗ = \dfrac{−b}{2a}$ is a maximizer of the function $f(x) = ax^2 + bx + c$, then a < 0.

Question

Proposition: Suppose that $a$, $b$, and $c$ are real numbers with $a \not = 0$. Prove that, if $x^∗ = \dfrac{−b}{2a}$ is a maximizer of the function $f(x) = ax^2 + bx + c$, then a < 0.

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A (hypothesis): $x^∗ = \dfrac{−b}{2a}$ is a maximizer of the function $f(x) = ax^2 + bx + c$ where $a \not = 0$, $b$, and $c$ are real numbers.

B (conclusion): $a < 0$

A1: For all real numbers $x$, $x^* = \dfrac{−b}{2a}$ is a maximiser of the function $f(x) = ax^2 + bx + c$.

A1 rephrases A using the universal quantifier "for all".

A2: Let $x \in \mathbf{R}$ and $x = \dfrac{-b}{2a}$.

A3: $f(x) = ax^2 + bx + c$

$\implies f\left(\dfrac{−b}{2a}\right) = a\left(\dfrac{−b}{2a}\right)^2 + b\left(\dfrac{−b}{2a}\right) + c$

$ = \dfrac{ab^2}{4a^2} - \dfrac{b^2}{2a} + c$

$ = \dfrac{b^2}{4a} - \dfrac{b^2}{2a} + c$

$ = \dfrac{b^2 - 2b^2}{4a} + c$

$ = \dfrac{-b^2}{4a} + c$

A4: $\dfrac{-b^2}{4a} + c \ge ax^2 + bx + c$

$\implies \dfrac{-b^2}{4a} \ge ax^2 + bx$

$\therefore a < 0$

$Q.E.D.$

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I would greatly appreciate it if people could please take the time to look over my proof and provide feedback.

Answer 1

$$ax^2+bx+c=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}\right)\leq-\frac{b^2-4ac}{4a}$$ The equality occurs for $x=-\frac{b}{2a}$.

Answer 2

In your (A4) part, there is a big jump to reach the “therefore” part.

Following the post by @MichaelRozenberg, we have $ax^2+bx+c= … = a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}\right)$

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

$\leq-\frac{b^2-4ac}{4a}$ for all x ONLY when $a\left(x+\frac{b}{2a}\right)^2$ is negative

This further means 'a' must be negative since $\left(x+\frac{b}{2a}\right)^2$ is always positive for all x.