#math/calculus A circle with center $(a,b)$ and radius $r$ consists of all points a distance $r$ from the center: $ (x-a)^{2}+(y-b)^{2}=r^{2}. $ A vertical parabola with vertex $(p,q)$ has the form $ y=A(x-p)^{2}+q, $ where $A>0$ opens upward and $A<0$ opens downward. For $y=ax^{2}+bx+c$ with $a\neq 0$, completing the square gives $ p=-\frac{b}{2a},\qquad q=f(p)=c-\frac{b^{2}}{4a}. $ Horizontal and vertical shifts replace $x$ by $x-p$ and add $q$, respectively. Thus $y=f(x-p)+q$ shifts the graph of $f$ right by $p$ and up by $q$.