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Parabola vertex proof

Yay, my first note!! In this note, I will discuss the rather simple proof of a parabola's vertex equation being \(x=\frac{-b}{2a}\), I couldn't find it anywhere on brilliant so I figured I may as well post it.

We start with the standard parabola equation \(y=ax^2+bx+c\). Next, we take the derivative of this equation \(\dfrac{\text{d}}{\text{d}x}y=ax^2+bx+c\Rightarrow2ax+b\).

Now, a special property of the first derivative is that its roots occur at the same x coordinate as its antiderivatives's relative maxima/minima. Thus the vertex of a parabola will occur when \(2ax+b=0\). Solving for x, we get \(x=\dfrac{-b}{2a}\).

Note by Trevor Arashiro
2 years, 11 months ago

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why did u finded roots of the equation from which this general equation is made to come out from ? reply please M Siddiqui · 2 years, 11 months ago

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@M Siddiqui In the case of a second degree polynomial,c it will yield the vertex. Trevor Arashiro · 2 years, 11 months ago

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@M Siddiqui If you find the roots of the first derivative, it will yield the x coordinate of the minima/maxima of a function, in the Trevor Arashiro · 2 years, 11 months ago

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