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Is it true to say that, for all real numbers $a$, $b$, and $c$ such that $a \not = 0$, $x = \dfrac{-b}{2a}$ is a maximiser of the function $f(x) = ax^2 + bx + c$? If not, please give a counterexample to show that the conjecture is false.

I would greatly appreciate it if someone could please take the time to confirm/deny this.

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    What happens if $a > 0$? – 2017-02-03
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    Consider $f(x)=x^2$. – 2017-02-03
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    @dxiv You're right. $x = 0$ would be the supposed maximum point according to $x = \dfrac{-b}{2a}$, but $x^2$ would actually diverge to infinity, right? – 2017-02-03
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    @ThePointer $x=-b/2a$ is a local extremum (which, in the case of quadratics, is also a global extremum). For $a \gt 0$ the extremum is a minimum. – 2017-02-03
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    @dxiv I understand. Thank you all for the assistance. :) – 2017-02-03

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