Finding the Maximum Value of a Quadratic Function

Introduction to Quadratic Functions and Their Maximum Values

Quadratic functions are a fundamental topic in algebra, often encountered in various fields such as physics, engineering, and economics. These functions are graphically represented as parabolas, which can open upwards or downwards depending on the sign of the leading coefficient. In this article, we will explore how to find the maximum value of a quadratic function, specifically the function y 3 - 4x - 5x^2 for integer values of x. We will use both the standard quadratic form and the vertex theorem to solve this problem.

The Standard Quadratic Form and Parabolas

A quadratic function is generally written in the form:

y ax^2 bx c

For the function we are considering, y 3 - 4x - 5x^2, we can rewrite it in standard form as follows:

y -5x^2 - 4x 3

In this form, we can easily identify the coefficient of x^2 (which is a -5) and the coefficient of x (which is b -4). Since the coefficient of x^2 is negative, the parabola opens downwards, indicating a maximum point at its vertex.

Applying the Vertex Theorem

The vertex theorem states that the x-coordinate of the vertex of a parabola given by y ax^2 bx c is:

x -frac{b}{2a}

For the given function, this yields:

x -frac{-4}{2(-5)} frac{4}{10} 0.4

Since x must be an integer, we will evaluate the function at the nearest integer values to 0.4, which are x 0 and x 1.

Evaluating the Function at Integer Values

Let's evaluate the function at x 0 and x 1:

y(0) 3 - 4(0) - 5(0)^2 3

y(1) 3 - 4(1) - 5(1)^2 3 - 4 - 5 -6

Additionally, we should also consider the value of the function at x -1:

y(-1) 3 - 4(-1) - 5(-1)^2 3 4 - 5 2

Lastly, we check the value of the function at x 2:

y(2) 3 - 4(2) - 5(2)^2 3 - 8 - 20 -25

From the above evaluations, the maximum value of y for integer x values is 3, which occurs at x 0.

Additional Insights

We can also find the vertex of the quadratic function using calculus. The derivative of the function with respect to x is:

dy/dx -4 - 1

Setting this derivative to zero gives us the vertex:

-4 - 1 0 x -frac{4}{10} -0.4

Since we are interested in integer values, we again check the nearest integers. The function value at x 0 is, as we have seen, 3, which is the highest value among our evaluated points.

Conclusion

In conclusion, the maximum value of the function y 3 - 4x - 5x^2 for integer values of x is 3, which occurs at x 0. This value can be confirmed through both the vertex theorem and the evaluation of the function at the nearest integer points. Understanding these procedures is crucial for working with quadratic functions in a variety of applications.