Find The Vertex Of The Quadratic Function Y = X^2 - 4x + 1

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In mathematics, particularly in algebra, understanding quadratic functions is crucial. A key feature of a quadratic function is its vertex, which represents the minimum or maximum point of the parabola. This article provides a comprehensive guide on how to find the vertex of a quadratic function, using the example function y=x2−4x+1y = x^2 - 4x + 1 as a practical illustration. We will explore different methods, ensuring a clear understanding for readers of all levels. Let's dive into the world of quadratic functions and master the art of finding the vertex.

Understanding Quadratic Functions

At its core, a quadratic function is a polynomial function of degree two. The general form of a quadratic function is expressed as:

f(x)=ax2+bx+cf(x) = ax^2 + bx + c

where a, b, and c are constants, and a is not equal to zero. The graph of a quadratic function is a parabola, a U-shaped curve that opens either upwards or downwards, depending on the sign of the coefficient a. If a is positive, the parabola opens upwards, and the vertex represents the minimum point of the function. Conversely, if a is negative, the parabola opens downwards, and the vertex represents the maximum point. Understanding these basic properties is the first step in finding the vertex of a given quadratic function.

Key Concepts and Terminology

Before we delve into the methods for finding the vertex, let's define some key concepts and terminologies:

  • Parabola: The U-shaped curve that represents the graph of a quadratic function.
  • Vertex: The point where the parabola changes direction. It is either the minimum point (if the parabola opens upwards) or the maximum point (if the parabola opens downwards).
  • Axis of Symmetry: A vertical line that passes through the vertex and divides the parabola into two symmetrical halves. The equation of the axis of symmetry is given by x=−b/2ax = -b/2a.
  • Coefficients: The constants a, b, and c in the general form of the quadratic function. These coefficients play a crucial role in determining the shape and position of the parabola.

These concepts are essential for a solid understanding of quadratic functions and will be instrumental in our quest to find the vertex. Remember, the vertex is not just a point on the graph; it's a critical feature that provides valuable information about the function's behavior.

Methods to Find the Vertex

There are several methods to find the vertex of a quadratic function. We will explore three primary methods:

  1. Using the Vertex Formula: This is a direct and efficient method that utilizes a specific formula to calculate the coordinates of the vertex.
  2. Completing the Square: This method involves transforming the quadratic function into vertex form, which directly reveals the vertex coordinates.
  3. Using Calculus (Differentiation): This method employs calculus principles to find the critical point of the function, which corresponds to the vertex.

Each method offers a unique approach to finding the vertex, and the choice of method often depends on the specific form of the quadratic function and personal preference. Let's examine each method in detail.

1. Using the Vertex Formula

The vertex formula is a straightforward method for finding the vertex of a quadratic function. Given the quadratic function in the standard form y=ax2+bx+cy = ax^2 + bx + c, the vertex (h,k)(h, k) can be found using the following formulas:

  • h=−b/2ah = -b / 2a
  • k=f(h)k = f(h)

Where h represents the x-coordinate of the vertex, and k represents the y-coordinate of the vertex. The formula for h is derived from the axis of symmetry, and k is obtained by substituting h back into the original quadratic function.

Applying the Vertex Formula to the Example

Let's apply the vertex formula to our example function, y=x2−4x+1y = x^2 - 4x + 1. First, identify the coefficients a, b, and c:

  • a=1a = 1
  • b=−4b = -4
  • c=1c = 1

Now, calculate the x-coordinate of the vertex, h:

h=−(−4)/(2∗1)=4/2=2h = -(-4) / (2 * 1) = 4 / 2 = 2

Next, calculate the y-coordinate of the vertex, k, by substituting h = 2 back into the original function:

k=(2)2−4(2)+1=4−8+1=−3k = (2)^2 - 4(2) + 1 = 4 - 8 + 1 = -3

Therefore, the vertex of the quadratic function y=x2−4x+1y = x^2 - 4x + 1 is (2,−3)(2, -3). This method provides a quick and efficient way to determine the vertex, especially when the quadratic function is given in standard form. The vertex formula is a powerful tool in analyzing quadratic functions and understanding their graphical representation.

2. Completing the Square

Completing the square is another effective method for finding the vertex of a quadratic function. This method involves transforming the quadratic function from its standard form (y=ax2+bx+cy = ax^2 + bx + c) into vertex form, which is given by:

y=a(x−h)2+ky = a(x - h)^2 + k

where (h,k)(h, k) represents the vertex of the parabola. The vertex form directly reveals the vertex coordinates, making it a valuable tool for analysis and graphing.

Steps for Completing the Square

To complete the square, follow these steps:

  1. Factor out the coefficient a from the x2x^2 and x terms. In our example, y=x2−4x+1y = x^2 - 4x + 1, the coefficient a is 1, so we don't need to factor anything out in this case.
  2. Take half of the coefficient of the x term, square it, and add and subtract it inside the parentheses. The coefficient of the x term is -4. Half of -4 is -2, and (-2)^2 is 4. So, we add and subtract 4 inside the expression.
  3. Rewrite the expression as a perfect square trinomial plus a constant. The perfect square trinomial can be factored into the form (x−h)2(x - h)^2.
  4. Simplify the expression to obtain the vertex form, y=a(x−h)2+ky = a(x - h)^2 + k.

Applying Completing the Square to the Example

Let's apply the completing the square method to our example function, y=x2−4x+1y = x^2 - 4x + 1:

  1. Since a = 1, we can skip the factoring step.

  2. Take half of the coefficient of the x term (-4), which is -2, and square it: (-2)^2 = 4. Add and subtract 4 inside the expression:

    y=x2−4x+4−4+1y = x^2 - 4x + 4 - 4 + 1

  3. Rewrite the expression as a perfect square trinomial plus a constant:

    y=(x−2)2−4+1y = (x - 2)^2 - 4 + 1

  4. Simplify the expression:

    y=(x−2)2−3y = (x - 2)^2 - 3

Now, the function is in vertex form, y=(x−2)2−3y = (x - 2)^2 - 3. By comparing this to the vertex form y=a(x−h)2+ky = a(x - h)^2 + k, we can identify the vertex as (h,k)=(2,−3)(h, k) = (2, -3). Completing the square provides a systematic way to transform the quadratic function into a form that directly reveals the vertex coordinates. This method is particularly useful when you need to understand the transformations applied to the basic parabola y=x2y = x^2.

3. Using Calculus (Differentiation)

Calculus provides another powerful method for finding the vertex of a quadratic function. This method involves using differentiation to find the critical points of the function, which correspond to the vertex. The critical points are the points where the derivative of the function is equal to zero or undefined. For a quadratic function, the critical point will always be the vertex.

Steps for Using Calculus

To find the vertex using calculus, follow these steps:

  1. Find the derivative of the quadratic function with respect to x. The derivative represents the slope of the tangent line to the curve at any given point.
  2. Set the derivative equal to zero and solve for x. This will give you the x-coordinate of the vertex, h.
  3. Substitute the value of h back into the original function to find the y-coordinate of the vertex, k.

Applying Calculus to the Example

Let's apply the calculus method to our example function, y=x2−4x+1y = x^2 - 4x + 1:

  1. Find the derivative of the function:

    dy/dx=2x−4dy/dx = 2x - 4

  2. Set the derivative equal to zero and solve for x:

    2x−4=02x - 4 = 0

    2x=42x = 4

    x=2x = 2

    This gives us the x-coordinate of the vertex, h = 2.

  3. Substitute h = 2 back into the original function to find the y-coordinate, k:

    k=(2)2−4(2)+1=4−8+1=−3k = (2)^2 - 4(2) + 1 = 4 - 8 + 1 = -3

Therefore, the vertex of the quadratic function y=x2−4x+1y = x^2 - 4x + 1 is (2,−3)(2, -3). The calculus method provides a more general approach to finding the vertex, as it can be applied to a wide range of functions, not just quadratic functions. This method is particularly useful when dealing with more complex functions where algebraic methods may be more challenging.

Analyzing the Results

We have now explored three different methods for finding the vertex of the quadratic function y=x2−4x+1y = x^2 - 4x + 1: the vertex formula, completing the square, and calculus (differentiation). Regardless of the method used, we consistently arrived at the same vertex coordinates: (2,−3)(2, -3). This consistency reinforces the accuracy and reliability of these methods.

The vertex (2,−3)(2, -3) provides valuable information about the parabola represented by the function y=x2−4x+1y = x^2 - 4x + 1. Since the coefficient of the x2x^2 term is positive (a = 1), the parabola opens upwards, and the vertex represents the minimum point of the function. This means that the function reaches its lowest value at x=2x = 2, and the minimum value is y=−3y = -3.

Implications of the Vertex

The vertex also helps us understand the axis of symmetry of the parabola. The axis of symmetry is a vertical line that passes through the vertex, dividing the parabola into two symmetrical halves. In this case, the axis of symmetry is the line x=2x = 2. This means that the parabola is symmetrical about the vertical line passing through x=2x = 2.

Furthermore, the vertex can be used to determine the range of the quadratic function. Since the parabola opens upwards and the vertex is the minimum point, the range of the function is all y-values greater than or equal to the y-coordinate of the vertex. Therefore, the range of y=x2−4x+1y = x^2 - 4x + 1 is y≥−3y ≥ -3.

In summary, finding the vertex of a quadratic function is not just about locating a point on the graph; it's about gaining a deeper understanding of the function's behavior, its symmetry, and its range. The vertex serves as a crucial anchor point for analyzing and interpreting quadratic functions.

Conclusion

In conclusion, finding the vertex of a quadratic function is a fundamental skill in mathematics with significant applications. We have explored three distinct methods for determining the vertex: using the vertex formula, completing the square, and employing calculus (differentiation). Each method offers a unique perspective and approach, yet they all converge on the same result, reinforcing the robustness of mathematical principles.

In our example, the quadratic function y=x2−4x+1y = x^2 - 4x + 1 yielded a vertex of (2,−3)(2, -3). This vertex not only pinpoints the minimum point of the parabola but also provides insights into the function's axis of symmetry and range. The ability to accurately and efficiently find the vertex empowers us to analyze and interpret quadratic functions effectively.

Whether you prefer the directness of the vertex formula, the transformative nature of completing the square, or the generality of calculus, mastering these methods will enhance your understanding of quadratic functions and their graphical representations. The vertex serves as a cornerstone in the study of parabolas, and its determination is a key step in unlocking the full potential of quadratic functions in mathematical modeling and problem-solving.

Therefore, the correct answer is D. (2, -3).