## Related questions with answers

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

Solution

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

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

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

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

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

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

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

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

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