The Student Room Group

Math help - finding the stationary points using second derivative test

y = ax^2 +bx + c
I know that this function has a minimum turning point when a>0 and a maximum when a<0, but is there a way to actually show this through derivatives? I'm told I need to find the 1st derivative then equal it to 0...is that right? What do I do next?
Original post by Nice_100
y = ax^2 +bx + c
I know that this function has a minimum turning point when a>0 and a maximum when a<0, but is there a way to actually show this through derivatives? I'm told I need to find the 1st derivative then equal it to 0...is that right? What do I do next?

The second derivative is y=2ay'' = 2a which is a constant, and it means that ...

(*) y>0y''>0 when a>0a>0 (so the stationary point is a local minimum), or

(**) y<0y''<0 when a<0a<0 (so the stationary point is a local maximum).


A quadratic is the only function which will differentiate twice to give a non-zero constant. In this sense, there is no need to set the first derivative to zero as we do not use the x-coordinate of the stationary point in the second derivative.

For all other functions; yes, differentiate once, set to zero, obtain x values of stationary points, then differentiate once more and substitute these x values into the second derivative. Observe sign to deduce the nature (maximum/minimum/point of inflection).
(edited 10 months ago)
Reply 2
In terms of understanding why a > 0 gives a minimum, you would do better to complete the square (https://en.wikipedia.org/wiki/Completing_the_square).

This allows you to rewrite ax^2+bx+c in the form a(x+B)2+Ca(x+B)^2+C for suitable choices of B and C. Hopefully at this point it's obvious if a > 0 the minimum value is C, obtained when x = -B (and similarly C is the max value if a < 0).
(edited 10 months ago)

Quick Reply