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=b2ax=\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=ax2+bx+cy=ax^2+bx+c. Next, we take the derivative of this equation ddxy=ax2+bx+c2ax+b\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=02ax+b=0. Solving for x, we get x=b2ax=\dfrac{-b}{2a}.

#Algebra #Calculus #Parabola #Proofs #Easymoney

Note by Trevor Arashiro
6 years, 10 months ago

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Comments

why did u finded roots of the equation from which this general equation is made to come out from ? reply please

M Siddiqui - 6 years, 10 months ago

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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 - 6 years, 10 months ago

In the case of a second degree polynomial,c it will yield the vertex.

Trevor Arashiro - 6 years, 10 months ago
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