$$y=ax^2+bx+c\qquad a\ne0$$
this is a parabola.
If a>0 then it is concave up, If a<0 then it is concave down.
The y intercept is c
$$\mbox{The axis of symmetry is }\qquad x=\dfrac{-b}{2a}\\
\mbox{Substitute in this value of x to get the y co-ordinate of the vertex}$$
The roots, x intercepts are
$$x=\dfrac{-b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a}$$
That is usually enough.
Please, put your question in a more appriate and mathematically correct shape. Your question is too general.
$$y=ax^2+bx+c\qquad a\ne0$$
this is a parabola.
If a>0 then it is concave up, If a<0 then it is concave down.
The y intercept is c
$$\mbox{The axis of symmetry is }\qquad x=\dfrac{-b}{2a}\\
\mbox{Substitute in this value of x to get the y co-ordinate of the vertex}$$
The roots, x intercepts are
$$x=\dfrac{-b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a}$$
That is usually enough.