×

# 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
3 years, 7 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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

- 3 years, 7 months ago

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

- 3 years, 7 months ago

If you find the roots of the first derivative, it will yield the x coordinate of the minima/maxima of a function, in the

- 3 years, 7 months ago