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At what value of x does the quadratic function f(x)=ax2+bx+c,a0,f(x)=a x^{2}+b x+c, a \neq 0, have a stationary point? Under what conditions is the stationary point a local maximum or a local minimum?

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Let f(x)=ax2+bx+c, a0f(x) = ax^2 + bx +c,\ a\neq 0 be an arbitrary quadratic function.

By definition, a stationary point is a point xx where f(x)=0f'(x) = 0. We calculate the derivative:

f(x)=2ax+bf(x)=02ax+b=02ax=bx=b2a\begin{align*} f'(x) &= 2ax + b\\ \Rightarrow f'(x) = 0 \Leftrightarrow 2ax+b &= 0\\ \Leftrightarrow 2ax &= -b\\ \Leftrightarrow x &= \frac{-b}{2a}\\ \end{align*}

To find out the value of ff at that point, we must substitute the calculated point into the formula:

f(b2a)=a(b2a)2+b(b2a)+c=ab24a2b22a+c=b22b2+4ac4a=b2+4ac4a\begin{align*} f\qty(\frac{-b}{2a}) &= a \qty(\frac{-b}{2a})^2 + b\qty(\frac{-b}{2a})+c\\ &= \frac{ab^2}{4a^2} - \frac{b^2}{2a} +c\\ &= \frac{b^2-2b^2+4ac}{4a}\\ &= \frac{-b^2+4ac}{4a}\\ \end{align*}

Hence, the value of the local minimum or local maximum of the function ff is located at the coordinates

(b2a,b2+4ac4a).\qty(\frac{-b}{2a},\frac{-b^2+4ac}{4a}).

If a<0a<0, that point will be a local maximum and if a>0a>0 that point will be a local minimum.

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