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

1 Answer
Jul 22, 2015

The unique stationary point is #(1,-4)#. There are no inflection points.

Explanation:

A point is "stationary" if the derivative is zero there (such points are also called critical points and are usually local extreme, or "turning", points...though not always).

When #f(x)=x^2-2x-3# we get #f'(x)=2x-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(1)=1-2-3=-4# so that you can think of the stationary point as being #(1,f(1))=(1,-4)#.

A point is an "inflection point" if the second derivate changes sign there. Since #f''(x)=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).