# Maximum value of a Quadratic Equation

#### thorpelizts

##### New member
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

Staff member
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
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

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#### Mr Fantastic

##### Member
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.