How does the change in $b$ in the quadratic formula $ax^2+bx+c$ move the parabola in an inverted version of the quadratic function?

I will just complete my answer to your previous question for this. 

By completing the square we have 
\[ax^2+bx+c=a(x^2+\frac{b}{a}x)+c=a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c,\]
which is a parabola with the vertex $(-\frac{b}{2a},-\frac{b^2}{4a}+c)$. Note that if $a>0$ then the parabola faces up and if $a<0$ then the parabola would face down. The coefficient $b$ could also be negative or positive number. We need to look at four different cases. 

Case I. $a>0, b>0.$ Then the parabola is facing up and increasing $b$ moves the vertex of the parabola $(-\frac{b}{2a},-\frac{b^2}{4a}+c)$ to the left and down.

Case II. $a>0,  b<0$. Then the parabola is facing up and increasing $|b|$ (making $b$ larger and larger negative numbers) moves the vertex of the parabola $(-\frac{b}{2a},-\frac{b^2}{4a}+c)$ to the right and down.

Case III. $a<0,  b>0$. Then the parabola is facing down and increasing $b$ moves the vertex of the parabola $(-\frac{b}{2a},-\frac{b^2}{4a}+c)$ to the left and up.

Case IV. $a<0,  b<0$. Then the parabola is facing down and increasing $|b|$ (making $b$ larger and larger negative numbers) moves the vertex of the parabola $(-\frac{b}{2a},-\frac{b^2}{4a}+c)$ to the right and up.

To understand why these happen, think about how the numbers $-\frac{b}{2a}$ and $-\frac{b^2}{4a}+c$ change when you change $b$ in each of the above 4 cases.