Maximum value of a Quadratic Equation

[thorpelizts]

Prove that 12x-8-3x^2 can never be greater than 4.

How do you prove? Do you find he discriminant? But isn't discriminant for roots only not equations?

 

[Jameson]

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 [math]y=ax^2+bx+c[/math] then the vertex is [math]x=-\frac{b}{2a}[/math]. Again this requires some context for the course.

 

[CaptainBlack]

Prove that 12x-8-3x^2 can never be greater than 4.

How do you prove? Do you find he discriminant? But isn't 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

 

[Mr Fantastic]

Prove that 12x-8-3x^2 can never be greater than 4.

How do you prove? Do you find he discriminant? But isn't 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.