# How do you find the VERTEX of a parabola y= 3/4x^2?

Apr 1, 2018

$\left(0 , 0\right)$

#### Explanation:

In general, the vertex of a parabola in the form $y = a {x}^{2}$ where $a$ is any number is $\left(0 , 0\right) ,$ as the parabola has not been shifted up, down, left, or right at all.

We can prove this by comparing $y = \frac{3}{4} {x}^{2}$ to the standard form of a quadratic, $y = a {x}^{2} + b x + c$

In this case, $a = \frac{3}{4} , b = 0 , c = 0$

The $x -$coordinate of the vertex is given by $- \frac{b}{2 a}$. In this case, it would be $\left(- \frac{0}{2 \cdot \frac{3}{4}}\right) = 0$.

The $y -$coordinate of the vertex is given by plugging in the result of $- \frac{b}{2 a}$ into the equation. In this case, $- \frac{b}{2 a} = 0 , y = \frac{3}{4} \left({0}^{2}\right) = 0$

So, the vertex is $\left(0 , 0\right)$