Extrema of Quadratic functions

[Qube]

Homework Statement

Does every quadratic function have a relative extrema?

Homework Equations

Quadratic function: ax^2 + bx + c. Aka a polynomial.

Polynomials are continuous through all real numbers.

The Attempt at a Solution

It seems as if all quadratic functions would have a relative extrema since the basic shape of a quadratic function is U-shaped and it's graphically obvious that the second derivative changes sign; all quadratic functions have a vertex whose x coordinate is given by -b/2a and this vertex is also the location of a horizontal tangent line and always represents either the max or min of the quadratic function (depending on orientation). And taking the derivative of the general form of a quadratic function yields 2ax + b where a and b are constants and it would appear that one can easily make the derivative both positive and negative given that the domain of quadratic functions is all real numbers.

Are there any exceptions? (I'm guessing no).

 

[mfb]

It seems as if all quadratic functions would have a relative extrema since the basic shape of a quadratic function is U-shaped and it's graphically obvious that the second derivative changes sign; all quadratic functions have a vertex whose x coordinate is given by -b/2a and this vertex is also the location of a horizontal tangent line and always represents either the max or min of the quadratic function (depending on orientation). And taking the derivative of the general form of a quadratic function yields 2ax + b where a and b are constants and it would appear that one can easily make the derivative both positive and negative given that the domain of quadratic functions is all real numbers.

Right

Are there any exceptions? (I'm guessing no).

No. Just make sure that "quadratic functions" implies a!=0 in your formula (otherwise it is not a quadratic function).

 

[vela]

Homework Statement

Does every quadratic function have a relative extrema?

Homework Equations

Quadratic function: ax^2 + bx + c. Aka a polynomial.

Polynomials are continuous through all real numbers.

The Attempt at a Solution

It seems as if all quadratic functions would have a relative extrema since the basic shape of a quadratic function is U-shaped and it's graphically obvious that the second derivative changes sign

The second derivative is constant. You mean the first derivative changes sign.

By the way, extrema is the plural of extremum. You should say that quadratic functions have a relative extremum.

 

[Qube]

Right, and looking at the first derivative, it can only be positive since it's the product of three squared terms. So there can be no sign change regardless.

 

[mfb]

Right, and looking at the first derivative, it can only be positive since it's the product of three squared terms. So there can be no sign change regardless.

I guess this belongs to your other thread.

 

[Mark44]

Right, and looking at the first derivative, it can only be positive since it's the product of three squared terms. So there can be no sign change regardless.

?
"it" = what?
If y = ax² + bx + c, then y' = 2ax + b

Where are you getting the three squared terms? mfb seems to know, but I don't recall seeing that other thread.

If "it" refers to y', the derivative will always change sign in a quadratic function.

If "it" refers to y'', that's 2a, so I still don't see where the three squared terms business comes in.

 

[Qube]

I guess this belongs to your other thread.

Whoa LOL yes this belongs in the other thread. Confused. My apologies.

 

[mfb]

Here is the other thread
I don't want to move the posts, as this would give a mess now.