# How do you find the intercept and vertex of x^2+6y=0?

Vertex is at$\left(0 , 0\right)$ and Intercept is at origin$\left(0 , 0\right)$
$6 \cdot y = - {x}^{2} \mathmr{and} y = - \frac{1}{6} \cdot {x}^{2}$ or $y = - \frac{1}{6} \cdot {\left(x - 0\right)}^{2} + 0$: Vertex is at $0 , 0$