Answer
a. vertex $(-\frac{b}{2a}, \frac{4ac-b^2}{4a})$
b. concave up when $a\gt0$, concave down when $a\lt0$
Work Step by Step
a. A vertex represents an extreme point of the parabola where $y'=0$. For $y=ax^2+bx+c$, we have $y'=2ax+b=0$ which gives $x=-\frac{b}{2a}$ and $y=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c=\frac{b^2}{4a}-\frac{b^2}{2a}+c=\frac{4ac-b^2}{4a}$. Thus the coordinates of the vertex are $(-\frac{b}{2a}, \frac{4ac-b^2}{4a})$
b. The parabola will be concave up if $y''\gt0$ which means $y''=2a\gt0$ and this gives $a\gt0$. Similarly, the parabola will be concave down if $y''\lt0$ which means $y''=2a\lt0$ and this gives $a\lt0$
