Comprehensive Analysis Of Quadratic Function D(x) = 3x² - 12x + 11
This article provides a comprehensive analysis of the quadratic function d(x) = 3x² - 12x + 11. We will explore various aspects of this function, including converting it to vertex form, identifying the vertex, determining the x and y-intercepts, sketching the graph, and delving into the mathematical principles behind these operations. Understanding quadratic functions is crucial in various fields, from physics and engineering to economics and computer science. This guide aims to provide a clear and detailed explanation of each step, making it accessible to students and anyone interested in learning more about quadratic functions.
(a) Writing the Function in Vertex Form
To express the given quadratic function d(x) = 3x² - 12x + 11 in vertex form, we need to complete the square. The vertex form of a quadratic function is given by d(x) = a(x - h)² + k, where (h, k) represents the vertex of the parabola. Vertex form is particularly useful as it allows us to immediately identify the vertex of the parabola, which is a critical point for understanding the function's behavior. The process of completing the square involves manipulating the quadratic expression to create a perfect square trinomial, which can then be factored into the form (x - h)². This transformation allows us to rewrite the function in a way that explicitly shows the vertex coordinates.
First, we factor out the coefficient of the x² term (which is 3) from the first two terms of the function:
d(x) = 3(x² - 4x) + 11
Next, we complete the square inside the parentheses. To do this, we take half of the coefficient of the x term (-4), square it ((-2)² = 4), and add it inside the parentheses. However, since we're adding it inside the parentheses, which are being multiplied by 3, we must also subtract 3 times this value outside the parentheses to maintain the equality of the function. This ensures that we are not changing the overall value of the function, only its form.
d(x) = 3(x² - 4x + 4) + 11 - 3(4)
Now we can rewrite the expression inside the parentheses as a squared term:
d(x) = 3(x - 2)² + 11 - 12
Finally, we simplify the constant term:
d(x) = 3(x - 2)² - 1
This is the vertex form of the function. By rewriting the function in vertex form, we gain immediate insight into the key characteristics of the parabola, particularly the location of its vertex. The vertex form makes it clear how the parabola is shifted horizontally and vertically from the basic parabola y = x², and the coefficient 'a' reveals whether the parabola opens upwards or downwards and the degree of its vertical stretch or compression. This form is essential for quickly understanding and sketching the graph of the quadratic function.
(b) Identifying the Vertex
Now that we have the function in vertex form, d(x) = 3(x - 2)² - 1, identifying the vertex is straightforward. As mentioned earlier, the vertex form of a quadratic function is given by d(x) = a(x - h)² + k, where (h, k) is the vertex. The vertex represents the point where the parabola changes direction; it is the minimum point if the parabola opens upwards (a > 0) and the maximum point if the parabola opens downwards (a < 0). Understanding the vertex is crucial for determining the range of the function and for solving optimization problems.
Comparing our function to the general vertex form, we can see that h = 2 and k = -1. Therefore, the vertex of the parabola is:
(2, -1)
The x-coordinate of the vertex, h, gives the axis of symmetry for the parabola, which is the vertical line x = h that divides the parabola into two symmetrical halves. The y-coordinate of the vertex, k, represents the minimum value of the function when the parabola opens upwards (as it does in this case since a = 3, which is positive). The vertex is a critical point in understanding the behavior of the quadratic function, as it helps us determine the function's minimum or maximum value and the symmetry of its graph. This information is vital for sketching the graph accurately and for solving various mathematical problems involving quadratic functions.
(c) Determining the x-intercept(s)
To find the x-intercept(s) of the function d(x) = 3x² - 12x + 11, we need to determine the values of x for which d(x) = 0. The x-intercepts are the points where the parabola intersects the x-axis, representing the real roots or solutions of the quadratic equation 3x² - 12x + 11 = 0. These intercepts provide valuable information about the behavior of the function and its relationship to the x-axis. There are several methods to find these intercepts, including factoring, using the quadratic formula, or completing the square.
In this case, the quadratic expression is not easily factorable, so we will use the quadratic formula. The quadratic formula is a general solution for any quadratic equation in the form ax² + bx + c = 0 and is given by:
x = (-b ± √(b² - 4ac)) / (2a)
For our function, a = 3, b = -12, and c = 11. Substituting these values into the quadratic formula, we get:
x = (12 ± √((-12)² - 4 * 3 * 11)) / (2 * 3)
x = (12 ± √(144 - 132)) / 6
x = (12 ± √12) / 6
We can simplify √12 as 2√3:
x = (12 ± 2√3) / 6
Now, we divide both terms in the numerator by 6:
x = 2 ± (√3) / 3
Therefore, the x-intercepts are:
x = 2 + (√3) / 3 and x = 2 - (√3) / 3
These are the two points where the parabola intersects the x-axis. Knowing the x-intercepts, along with the vertex, gives us a good understanding of the parabola's position and shape on the coordinate plane. The x-intercepts are crucial for understanding where the function's values change sign and for solving real-world problems that involve finding the zeros of a quadratic function.
(d) Determining the y-intercept(s)
To find the y-intercept of the function d(x) = 3x² - 12x + 11, we need to determine the value of d(x) when x = 0. The y-intercept is the point where the parabola intersects the y-axis, representing the value of the function when the input is zero. This intercept is easily found by substituting x = 0 into the function's equation. The y-intercept is a key point for sketching the graph, as it provides a clear indication of the function's value at the y-axis.
Substituting x = 0 into the function, we get:
d(0) = 3(0)² - 12(0) + 11
d(0) = 0 - 0 + 11
d(0) = 11
Therefore, the y-intercept is:
(0, 11)
The y-intercept, along with the vertex and x-intercepts, gives us a comprehensive view of the parabola's position and shape on the coordinate plane. The y-intercept is particularly useful in practical applications where the initial value of a quantity is of interest. For example, in modeling projectile motion, the y-intercept could represent the initial height of the projectile. In economic models, it might represent the fixed costs of production. Thus, finding the y-intercept is not only a mathematical exercise but also a valuable step in interpreting the function in real-world contexts.
(e) Sketching the Function
Now that we have determined the vertex, x-intercepts, and y-intercept, we can sketch the function d(x) = 3x² - 12x + 11. Sketching the graph of a quadratic function involves plotting these key points and understanding the overall shape of the parabola. The vertex gives us the minimum or maximum point, the x-intercepts tell us where the graph crosses the x-axis, and the y-intercept indicates where the graph crosses the y-axis. These points, along with the symmetry of the parabola, allow us to create an accurate representation of the function's behavior.
We know the following:
- Vertex: (2, -1)
- x-intercepts: x = 2 + (√3) / 3 ≈ 2.58 and x = 2 - (√3) / 3 ≈ 1.42
- y-intercept: (0, 11)
First, plot the vertex (2, -1). Since the coefficient of the x² term (a = 3) is positive, the parabola opens upwards, meaning the vertex is the minimum point. This tells us that the parabola will have a U-shape, opening upwards from the vertex.
Next, plot the x-intercepts, which are approximately 2.58 and 1.42. These points lie on the x-axis and show where the function's value is zero. Plot the y-intercept (0, 11), which is the point where the parabola intersects the y-axis.
Now, sketch the parabola. Draw a smooth curve that passes through the plotted points, ensuring the curve is symmetrical about the vertical line passing through the vertex (the axis of symmetry, x = 2). The parabola should open upwards from the vertex, passing through the x-intercepts and the y-intercept. The steeper the curve, the faster the function's values are changing. In this case, the coefficient a = 3 indicates a vertical stretch, making the parabola narrower than the standard parabola y = x².
The sketch should visually represent the key features of the quadratic function: its minimum value at the vertex, its intersections with the x-axis (x-intercepts), and its intersection with the y-axis (y-intercept). The sketch provides a valuable tool for visualizing the function's behavior and understanding its properties. It also helps in solving related problems, such as finding the range of the function or determining the intervals where the function is increasing or decreasing. A well-sketched graph is an essential aid in understanding quadratic functions and their applications.
(f) Discussion Category: Mathematics
This discussion falls squarely within the category of mathematics, specifically within the subfields of algebra and calculus. The analysis of the quadratic function d(x) = 3x² - 12x + 11 involves algebraic techniques such as completing the square, using the quadratic formula, and identifying key features of the function. These techniques are fundamental in algebra and are used extensively in higher-level mathematics and related fields. The process of sketching the function also involves graphical analysis, which is a key concept in both algebra and calculus. Understanding quadratic functions is a cornerstone of mathematical education, as they serve as a building block for more advanced topics such as polynomial functions, conic sections, and calculus concepts like derivatives and integrals. The ability to analyze and manipulate quadratic functions is essential for solving a wide range of mathematical problems and real-world applications.
Furthermore, the concepts explored in this analysis, such as finding intercepts, the vertex, and the range of a function, are crucial in calculus. For instance, the vertex of a parabola can be found using calculus techniques, and the x-intercepts represent the roots of the equation, which are often found using numerical methods in calculus. The ability to sketch functions and understand their graphical behavior is also fundamental in calculus for understanding concepts like limits, continuity, and differentiability. Therefore, this discussion firmly belongs within the domain of mathematics, encompassing algebraic manipulation, graphical analysis, and foundational concepts for calculus.
