Problem 26 - Algebra 2

Question

Identify the vertex and the \(y\) -intercept of the graph of each function. $$ y=(x-125)^{2}+125 $$

Step-by-Step Solution

Answer

The vertex of the graph is at (125, 125), and the y-intercept is at (0, 15750).

Step 1: Find the Vertex

The given quadratic equation is in vertex form, which is \(y = a(x-h)^{2} + k\) where \(h\) and \(k\) are the \(x\)-coordinate and \(y\)-coordinate of the vertex, respectively. Comparing the given equation \(y = (x-125)^{2} + 125\) to the vertex form, the vertex can be directly read off as \(h = 125\) and \(k = 125\). Therefore, the vertex is \(\text{(125, 125)}\).

Step 2: Find the y-intercept

To find the \(y\)-intercept, set \(x = 0\) in the equation \(y = (x-125)^{2} + 125\). After substituting, simplify the equation as follows: \[y = (0-125)^{2} + 125 = (-125)^{2} + 125 = 15625 + 125 = 15750\]. Therefore, the \(y\)-intercept is \(\text{(0, 15750)}\).

Key Concepts

Finding the Vertex

Understanding the vertex of a quadratic function is crucial for analyzing its graph. The vertex represents the highest or lowest point on the graph, depending on the direction the parabola opens. In the vertex form of a quadratic equation, which is expressed as y = a(x-h)^{2} + k, the coordinates (h, k) directly provide the vertex of the parabola.

In the example given, the quadratic function is presented as y = (x-125)^{2} + 125. By comparing it to the vertex form, we can see that h = 125 and k = 125. This means the vertex of the graph for our function is at the point (125, 125). This point can serve multiple purposes, including aiding in graphing the function and optimizing quadratic problems, such as finding maxima or minima in real-world situations.

Y-intercept of a Quadratic

The y-intercept is where the graph of a function crosses the y-axis. For any quadratic function, this occurs when x = 0. To find the y-intercept from the vertex form of the equation, we substitute 0 for x in the equation and solve for y.

In the provided example function y = (x-125)^{2} + 125, setting x to zero gives us y = (-125)^{2} + 125, which simplifies to 15750. Thus, the y-intercept of the function is the point (0, 15750). Knowing the y-intercept is fundamental for graphing and can be especially helpful in real-world contexts where the initial value of a scenario is modeled by a quadratic function.

Quadratic Function Analysis

Analyzing a quadratic function involves understanding its shape, direction, and key features such as the vertex, y-intercept, and x-intercepts. The graph of a quadratic function is a parabola which can open upwards (if a > 0) or downwards (if a < 0).

In addition to the vertex and y-intercept, it's also important to consider the axis of symmetry, which is a vertical line that passes through the vertex and divides the parabola into two mirror images. For our equation y = (x-125)^{2} + 125, the axis of symmetry is the line x = 125.

Moreover, by examining the coefficient of the squared term, we can determine the parabola's width and direction. A detailed analysis of a quadratic function leads to a deeper understanding of the function's behavior and is an essential skill in algebra and calculus.

Algebraic Transformations

Algebraic transformations refer to the different manipulations that can be applied to the standard form of a function, which alter its graph on the coordinate plane. In the context of quadratic functions, common transformations include vertical and horizontal shifts, reflections, and stretching or shrinking the graph.

By analyzing the vertex form of a quadratic equation y = a(x-h)^{2} + k, we can identify these transformations. The (h,k) values indicate a horizontal shift h units and a vertical shift k units. The coefficient a determines the direction (if a is negative, the parabola will open downwards) and the width (if |a| > 1, the parabola is narrower; if 0 < |a| < 1, it's wider) of the parabola.

Applying these concepts to the function y = (x-125)^{2} + 125, we observe that the parabola is shifted 125 units to the right and 125 units upwards. There is no reflection since a is positive, and the parabola is not narrower or wider than the standard one because a = 1. Learning these transformations helps students visualize and draw accurate graphs of quadratic functions and can be applied to other algebraic functions for analysis and sketching.