Analyzing Quadratic Function K(x) = X^2 + 8x + 15 Vertex, Intercepts, And Graph

In the fascinating realm of mathematics, quadratic functions hold a special place. These functions, characterized by their elegant parabolic curves, find applications in diverse fields, from physics and engineering to economics and computer science. In this comprehensive guide, we embark on a journey to dissect and understand the quadratic function k(x) = x^2 + 8x + 15. We will transform it into its vertex form, pinpoint its vertex, unveil its x and y-intercepts, sketch its graph, and ultimately, delve into its domain and range.

(a) Transforming k(x) into Vertex Form: A Step-by-Step Approach

The vertex form of a quadratic function, expressed as k(x) = a(x - h)^2 + k, provides a treasure trove of information about the function's behavior. The vertex, a pivotal point on the parabola, is readily identified as (h, k), and the coefficient 'a' dictates the parabola's direction and concavity. To transform our given function, k(x) = x^2 + 8x + 15, into vertex form, we employ the powerful technique of completing the square. This method involves strategically manipulating the quadratic expression to create a perfect square trinomial, a trinomial that can be factored into the square of a binomial.

Our journey begins by isolating the x^2 and x terms: k(x) = (x^2 + 8x) + 15. Next, we embark on the quest to complete the square within the parentheses. To achieve this, we take half of the coefficient of our x term (which is 8), square it (resulting in 16), and add it inside the parentheses. However, to maintain the sanctity of the equation, we must also subtract this value outside the parentheses. This crucial step ensures that we haven't altered the fundamental nature of the function.

Our equation now takes the form: k(x) = (x^2 + 8x + 16) + 15 - 16. The expression within the parentheses is now a perfect square trinomial, which we can gracefully factor as (x + 4)^2. Simplifying the constants outside the parentheses, we arrive at the vertex form of our function: k(x) = (x + 4)^2 - 1. This elegant form unveils the secrets of our parabola, revealing its vertex and behavior.

(b) Pinpointing the Vertex: The Heart of the Parabola

The vertex, the cornerstone of a parabola, represents either the minimum or maximum point of the function. In the vertex form, k(x) = a(x - h)^2 + k, the vertex is effortlessly identified as the point (h, k). In our case, the vertex form of our function, k(x) = (x + 4)^2 - 1, reveals that h = -4 and k = -1. Therefore, the vertex of our parabola is located at the point (-4, -1). This point serves as a critical landmark in our understanding of the function's graph and behavior.

The vertex holds paramount importance as it signifies the parabola's axis of symmetry, an imaginary vertical line that gracefully divides the parabola into two congruent halves. The x-coordinate of the vertex, in our case -4, defines this axis of symmetry. Moreover, the vertex determines the minimum or maximum value of the function. Since the coefficient of our squared term (a) is positive (a = 1), our parabola opens upwards, implying that the vertex represents the minimum point of the function. The y-coordinate of the vertex, -1, signifies this minimum value.

(c) Unveiling the x-intercept(s): Where the Parabola Intersects the x-axis

The x-intercepts, the points where the parabola gracefully intersects the x-axis, hold significant geometric and algebraic meaning. At these points, the function's value, k(x), is zero. To unearth the x-intercepts, we set k(x) = 0 and embark on the quest to solve for x. Our equation transforms into: 0 = x^2 + 8x + 15. This quadratic equation can be solved using a variety of techniques, including factoring, employing the quadratic formula, or completing the square.

In this instance, the equation readily lends itself to factoring. We seek two numbers that gracefully multiply to 15 and sum to 8. These numbers, 3 and 5, emerge as the key to unlocking our solution. We rewrite the equation as: 0 = (x + 3)(x + 5). Applying the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero, we set each factor equal to zero and solve for x.

This yields two solutions: x + 3 = 0, which gives us x = -3, and x + 5 = 0, which leads us to x = -5. Therefore, our parabola gracefully intersects the x-axis at two distinct points: (-3, 0) and (-5, 0). These x-intercepts provide valuable insights into the function's behavior and its relationship with the x-axis.

(d) Determining the y-intercept: Where the Parabola Embraces the y-axis

The y-intercept, the point where the parabola elegantly intersects the y-axis, holds a special significance. At this point, the x-coordinate is zero. To unveil the y-intercept, we set x = 0 in our function, k(x) = x^2 + 8x + 15, and evaluate. This yields: k(0) = 0^2 + 8(0) + 15, which simplifies to k(0) = 15. Therefore, our parabola gracefully intersects the y-axis at the point (0, 15). This y-intercept provides a crucial data point for sketching the graph of the function.

The y-intercept, in conjunction with the x-intercepts and the vertex, paints a comprehensive picture of the parabola's position and orientation in the coordinate plane. It serves as another landmark in our quest to understand the function's behavior and its graphical representation.

(e) Sketching the Function: A Visual Representation of k(x)

Armed with the vertex, x-intercepts, and y-intercept, we are now equipped to sketch the graph of our quadratic function, k(x) = x^2 + 8x + 15. The vertex, located at (-4, -1), serves as our anchor point. The x-intercepts, (-3, 0) and (-5, 0), guide the parabola's path as it crosses the x-axis. The y-intercept, (0, 15), marks the point where the parabola embraces the y-axis.

Since the coefficient of our squared term (a) is positive (a = 1), we know that the parabola opens upwards, forming a U-shaped curve. Starting from the vertex, the parabola gracefully curves upwards, passing through the x-intercepts and the y-intercept. The axis of symmetry, a vertical line passing through the vertex (x = -4), ensures that the parabola is symmetrical about this line.

The sketch of the graph provides a visual representation of the function's behavior, allowing us to readily grasp its key features, including its minimum value (at the vertex), its roots (x-intercepts), and its overall shape. The graph serves as a powerful tool for understanding and communicating the properties of the quadratic function.

(f) Delving into the Domain and Range: Unveiling the Function's Boundaries

The domain and range of a function define the set of all possible input values (x-values) and output values (k(x)-values), respectively. For quadratic functions, the domain is typically all real numbers, as we can input any real number into the function and obtain a valid output. However, the range is constrained by the parabola's vertex and its direction of opening.

In our case, the parabola opens upwards, and its vertex, (-4, -1), represents the minimum point of the function. This implies that the function's output values (k(x)) will always be greater than or equal to -1. Therefore, the range of our function is all real numbers greater than or equal to -1, which can be expressed in interval notation as [-1, ∞).

The domain and range provide valuable insights into the function's behavior and its limitations. Understanding these boundaries allows us to accurately interpret and apply the function in various contexts.

In this comprehensive exploration, we have dissected the quadratic function k(x) = x^2 + 8x + 15, transforming it into its vertex form, pinpointing its vertex, unveiling its x and y-intercepts, sketching its graph, and delving into its domain and range. This journey has illuminated the intricate beauty and versatility of quadratic functions, showcasing their importance in mathematics and beyond. By mastering these fundamental concepts, we empower ourselves to tackle a wide range of mathematical challenges and appreciate the elegance of the parabolic curve.