# What is the focus and vertex of the parabola described by 3x^2+1x+2y+7=0 ?

Vertex is at $= \left(- \frac{1}{6} , - \frac{83}{24}\right)$ Focus is at $\left(- \frac{1}{6} , - \frac{87}{24}\right)$
$2 y = - 3 {x}^{2} - x - 7 \mathmr{and} y = - \frac{3}{2} {x}^{2} - \frac{x}{2} - \frac{7}{2} = - \frac{3}{2} \left({x}^{2} + \frac{x}{3} + \frac{1}{36}\right) + \frac{1}{24} - \frac{7}{2} = - \frac{3}{2} {\left(x + \frac{1}{6}\right)}^{2} - \frac{83}{24}$ Vertex is at $= \left(- \frac{1}{6} , - \frac{83}{24}\right)$ The parabola opens down as co -efficient of ${x}^{2}$ is negative. distance between vertex and focus is $\frac{1}{|} 4 a | = \frac{1}{4 \cdot \frac{3}{2}} = \frac{1}{6}$ Hence the focus is at $- \frac{1}{6} , \left(- \frac{83}{24} - \frac{1}{6}\right) \mathmr{and} \left(- \frac{1}{6} , - \frac{87}{24}\right)$ graph{-3/2x^2-x/2-7/2 [-20, 20, -10, 10]}[Ans]