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'' = 2a

which is a constant, and it means that ...

(*)

y''>0

when

a>0

(so the stationary point is a local minimum), or

(**)

y''<0

when

a<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 3 months ago)

Reply 2

DFranklin

18

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+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 3 months ago)

Math help - finding the stationary points using second derivative test

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