Introduction to Quadratic Functions and Their Graphs

Quadratic functions are a fundamental part of mathematics, and understanding their behavior and properties is crucial for many fields of study and applications. A quadratic function is generally represented in the form y ax^2 bx c, where a, b, and c are constants, and a ≠ 0. The graph of a quadratic function is a parabola, which opens either upwards if a 0 or downwards if a 0.

Determining Minimum and Maximum Values of a Quadratic Function

Given the quadratic function y x^2 - 6x 8, we need to determine whether this function has a minimum or a maximum value. The coefficient of x^2 is positive, indicating that the parabola opens upwards, and hence the function has a minimum value.

Algebraic Method: Completing the Square

Step 1: Start with the given equation and complete the square.

An equation of the form y x^2 - 6x 8 can be rewritten by completing the square:

First, we can add and subtract 9 within the equation to complete the square:

y x^2 - 6x 9 - 9 8 (x - 3)^2 - 1

Step 2: Analyze the expression (x - 3)^2 - 1.

The term (x - 3)^2 is always non-negative because a square of any real number is non-negative. Therefore, the minimum value of (x - 3)^2 - 1 occurs when (x - 3)^2 0, which happens when x 3.

Hence, the function y x^2 - 6x 8 has a minimum value of -1 at x 3.

graphical Method: Finding the Vertex

The vertex of a parabola is the point where the parabola turns. For a quadratic function in the form y ax^2 bx c, the vertex occurs at x -b/(2a).

From the given function y x^2 - 6x 8, we can identify a 1, b -6, and c 8.

The x-coordinate of the vertex is calculated as:

x -(-6) / (2*1) 6 / 2 3

Substituting x 3 into the equation:

y (3)^2 - 6(3) 8 9 - 18 8 -1

So, the vertex of the parabola is at (3, -1), indicating that the minimum value of the function is -1.

Calculus Method: Using Derivatives

The first derivative of a function gives the slope of the function at any point. For a quadratic function, the first derivative will be a linear function, and the second derivative will be a constant, which helps us determine the nature of the critical points.

The first derivative of y x^2 - 6x 8 is:

y' 2x - 6

Solving for the critical points where the first derivative equals zero:

2x - 6 0 implies x 3

To confirm that this critical point is a minimum, we check the second derivative:

The second derivative of y is:

y'' 2

Since the second derivative is positive, the function has a local minimum at x 3.

Substituting x 3 into the original function:

y (3)^2 - 6(3) 8 9 - 18 8 -1

Hence, the function has a minimum value of -1 at x 3.

Conclusion

Through algebraic and calculus methods, we have demonstrated that the function y x^2 - 6x 8 has a minimum value of -1 at x 3. This function does not have a maximum value because the parabola opens upwards, indicating that the minimum value is a global minimum.

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