To find stationary points of a function, substitute $x$ value of given point into derivative. Solve for ${dy \over dx}=0$. Integrate to find original function.
+## Rates
+
+### Related rates
+
+$${da \over db} \quad \text{change in } a \text{ with respect to } b$$
+
+#### Gradient at a point on parametric curve
+
+$${dy \over dx} = {{dy \over dt} \over {dx \over dt}} \> \vert \> {dx \over dt} \ne 0$$
+
+$${d^2 \over dx^2} = {d(y^\prime) \over dx} = {{dy^\prime \over dt} \over {dx \over dt}} \> \vert \> y^\prime = {dy \over dx}$$
+
+# Rational functions
+
+$$f(x) = {P(x) \over Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}$$
+
+## Addition of ordinates
+
+- when two graphs have the same ordinate, $y$-coordinate is double the ordinate
+- when two graphs have opposite ordinates, $y$-coordinate is 0 i.e. ($x$-intercept)
+- when one of the ordinates is 0, the resulting ordinate is equal to the other ordinate
+