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 \(\displaystyle y=ax^2+bx+c\) then the vertex is \(\displaystyle x=-\frac{b}{2a}\). 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.

 

[CellCoree]

Yes, the discriminant is typically used to find the roots of a quadratic equation. However, it can also be used to determine the maximum value of a quadratic equation. In this case, the discriminant is not needed to prove that 12x-8-3x^2 can never be greater than 4.

To prove this statement, we can use the fact that the maximum value of a quadratic equation occurs at the vertex of the parabola. The vertex can be found by using the formula x = -b/2a, where a and b are the coefficients of the quadratic equation in standard form (ax^2 + bx + c = 0).

In the given equation, 12x-8-3x^2, the coefficient of x^2 is -3 and the coefficient of x is 12. Plugging these values into the formula, we get x = -12/2(-3) = 2. Therefore, the x-coordinate of the vertex is 2.

To find the y-coordinate of the vertex, we can plug in the x-coordinate (2) into the given equation. This gives us a value of -8 for the y-coordinate. Therefore, the vertex of the parabola is at the point (2, -8).

Since the vertex is the maximum point of the parabola, we can conclude that the maximum value of the given equation is -8. This means that the equation can never be greater than 4, as 4 is a larger value than -8.

In summary, by using the formula for finding the vertex of a parabola, we can prove that the given quadratic equation can never have a value greater than 4.