Extreme point of quadratic equation

Question

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Write a function quadratic that returns the interval of all values $f(t)$ such that $t$ is in the argument interval $x$ and $f(t)$ is a quadratic function:

$$f(t) = at^2 + bt + c$$ Make sure that your implementation returns the smallest such interval, one that does not suffer from the multiple references problem.

Hint: the derivative $f'(t) = 2at + b$, and so the extreme point of the quadratic is $-\frac{b}{2a}$.

I would like to understand,

What is extreme point of quadratic equation? If I imagine a graph of that function>

why the extreme point of quadratic equation is $-\frac{b}{2a}$?

Answer 1

The extreme point is a maximum or minimum, in the quadratic function case it will be global maximum of minimum.

If you know calculus, this is easy to understand: at the extreme point, the first derivative at that point is $0$. Differentiate $ax^2+bx+c$ you get $2ax+b=0$, thus $x = -\frac{b}{2a}$

If you don't know calculus, then this is the way you may understand: notice that quadratic function is symmetrical by its extreme. The quadratic formula tells you the two roots are $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$. They are symmetric by the extreme, so the extreme is at $x=-\frac{b}{2a}$.

Answer 2

Extreme point occurs where derivative vanishes. Setting derivative to zero $$ x_{ExtremePoint}= \dfrac{-b}{2 a}$$and corresponding $$ y_{ExtremePoint}= c-\frac{b^2}{4 a}$$.