Question

How do you find the vertex on a porabola​

Answer 1

A parabola ALWAYS has a lowest point like unless it would be upside down.

How to solve:

You would calculate two of the numbers to get the vertex.

Answer 2

The vertex on a parabola can be generalized trivially using differential calculus. See that $f(x) = ax^2 + bx + c$ is a parabola.

Now we use the formula $\displaystyle\frac{\mathrm{d}}{\mathrm{d}x}(x^m) = mx^{m-1}$ and $\displaystyle\frac{\mathrm{d}}{\mathrm{d}x}(f(x) + g(x)) = \frac{\mathrm{d}}{\mathrm{d}x} f(x) + \frac{\mathrm{d}}{\mathrm{d}x}g(x)$ to solve this problem. We use the derivative of $f$

$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} f(x) = \frac{\mathrm{d}}{\mathrm{d}x}(ax^2) + \frac{\mathrm{d}}{\mathrm{d}x}(bx) + \frac{\mathrm{d}}{\mathrm{d}x}(c) = 2ax + b + 0$

We find where $\frac{\mathrm{d}}{\mathrm{d}x} f(x) = 0$. So this is when $\displaystyle 2ax + b = 0 \Rightarrow x = \frac{-b}{2a}$.

This gives vertex x location. To find y location you calculate ycoordinate using f(x).

$f\left( \frac{-b}{2a} \right) = a\left( \frac{-b}{2a}\right)^2 + b\left(\frac{-b}{2a}\right) + c = c - \frac{b^2}{4a}$

The vertex is $V\left( \frac{-b}{2a}, c - \frac{b^2}{4a} \right)$