# How do you find stationary and inflection points for x^2-2x-3?

The unique stationary point is $\left(1 , - 4\right)$. There are no inflection points.
When $f \left(x\right) = {x}^{2} - 2 x - 3$ we get $f ' \left(x\right) = 2 x - 2$ so that $x = 1$ is the first coordinate of the unique stationary point of this function (it also happens to be the lowest point on the graph). The second coordinate of this point is $f \left(1\right) = 1 - 2 - 3 = - 4$ so that you can think of the stationary point as being $\left(1 , f \left(1\right)\right) = \left(1 , - 4\right)$.
A point is an "inflection point" if the second derivate changes sign there. Since $f ' ' \left(x\right) = 2$ for all $x$, the second derivative never changes sign. There are no inflection points for this function (the graph never changes from concave up (smile-like) to concave down (frown-like) or vice versa...in this case, the graph is always concave up).