# What is the vertex of y=3x^2+6?

Dec 27, 2015

$\left(0 , 6\right)$

#### Explanation:

This is a 2nd degree quadratic function so its graph will be a parabola.
Such function of form $y = a {x}^{2} + b x + c$ has turning point at $x = - \frac{b}{2 a}$, so in this case at $x = 0$ which implies corresponding y-value is at the y-intercept itself of $6$.

Here is the graph as verification :

graph{3x^2+6 [-24.28, 40.64, -4.72, 27.74]}