Welcome to our community

Be a part of something great, join today!

Maximum value of a Quadratic Equation

thorpelizts

New member
Sep 7, 2012
6
Prove that 12x-8-3x^2 can never be greater than 4.

How do you prove? Do you find he discriminant? But isnt discriminant for roots only not equations?
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
I don't know what kind of proof you teacher or professor requires for this so can you provide some context for the course?

Otherwise I would say in general that a quadratic function with a negative leading coefficient is a parabola that faces downward and has a maximum point at the vertex.

If the formula for a parabola is given in the form \(\displaystyle y=ax^2+bx+c\) then the vertex is \(\displaystyle x=-\frac{b}{2a}\). Again this requires some context for the course.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Prove that 12x-8-3x^2 can never be greater than 4.

How do you prove? Do you find he discriminant? But isnt discriminant for roots only not equations?
This is the standard non-calculus method for finding the maximum/minimum of a quadratic expression:

You first complete the square to get:

\[-3x^2+12x-8=-3(x-2)^2+12-8=-3(x-2)^2 +4\]

Now since the largest \(-3(x-2)^2\) can be is zero the largest the whole thing can be is \(+4\).

CB
 
Last edited:

Mr Fantastic

Member
Jan 26, 2012
66
Prove that 12x-8-3x^2 can never be greater than 4.

How do you prove? Do you find he discriminant? But isnt discriminant for roots only not equations?
It should be recognized as a negative parabola so your job is to find the vertical coordinate of the maximum turning point. Several methods have already been suggested.