Analyzing The Quadratic Function F(x) = 2(x + 2)² + 1 Vertex Intercepts And Properties
In this article, we will delve into the properties of the quadratic function $f(x) = 2(x + 2)^2 + 1$. We will determine its vertex, y-intercept(s), and x-intercept(s). Understanding these key features allows us to sketch the graph of the function and gain insights into its behavior. Quadratic functions are fundamental in mathematics and have wide-ranging applications in various fields, including physics, engineering, and economics. This exploration will provide a solid foundation for working with quadratic functions and their applications.
a) Determining the Vertex of the Graph
To state the vertex of the graph of the function $f(x) = 2(x + 2)^2 + 1$, we need to recognize the vertex form of a quadratic equation. The vertex form is given by $f(x) = a(x - h)^2 + k$, where $(h, k)$ represents the vertex of the parabola. This form is particularly useful because it directly reveals the vertex coordinates, making it easy to identify the highest or lowest point on the graph.
In our given function, $f(x) = 2(x + 2)^2 + 1$, we can see a clear resemblance to the vertex form. By comparing the given function with the general vertex form, we can identify the values of $a$, $h$, and $k$. Here, $a = 2$, and we can rewrite the expression $(x + 2)$ as $(x - (-2))$, which means $h = -2$. The value of $k$ is simply $1$. Therefore, the vertex of the parabola is at the point $(-2, 1)$.
The coefficient $a$ plays a crucial role in determining the shape and direction of the parabola. Since $a = 2$ in our function, and $2 > 0$, the parabola opens upwards. This means that the vertex represents the minimum point of the function. If $a$ were negative, the parabola would open downwards, and the vertex would represent the maximum point.
Understanding the vertex form not only helps in identifying the vertex but also provides insights into the transformations applied to the basic parabola $y = x^2$. The term $(x - h)$ represents a horizontal shift of the parabola by $h$ units, and the term $k$ represents a vertical shift by $k$ units. In our case, the parabola is shifted 2 units to the left (due to $h = -2$) and 1 unit upwards (due to $k = 1$).
Therefore, by carefully analyzing the given function and comparing it with the vertex form, we can confidently state that the vertex of the graph of the function $f(x) = 2(x + 2)^2 + 1$ is (-2, 1). This point is the minimum value of the function, and the parabola opens upwards, indicating that the function increases as we move away from the vertex in either direction. This understanding is crucial for sketching the graph and analyzing the behavior of the quadratic function.
b) Determining the y-intercept(s) of the Graph
To state the y-intercept(s) of the graph of the function $f(x) = 2(x + 2)^2 + 1$, we need to understand what the y-intercept represents. The y-intercept is the point where the graph of the function intersects the y-axis. At this point, the x-coordinate is always 0. Therefore, to find the y-intercept, we need to evaluate the function at $x = 0$.
Substituting $x = 0$ into the function $f(x) = 2(x + 2)^2 + 1$, we get:
This calculation shows that when $x = 0$, the value of the function is $f(0) = 9$. This means that the graph of the function intersects the y-axis at the point $(0, 9)$. Therefore, the y-intercept of the graph is 9.
The y-intercept is an important feature of the graph as it provides a point of reference on the y-axis. It helps in visualizing the vertical position of the parabola. In this case, since the y-intercept is 9, we know that the parabola crosses the y-axis at a relatively high point. This information, combined with the vertex, gives us a clearer picture of the parabola's shape and position on the coordinate plane.
In the context of real-world applications, the y-intercept can represent the initial value of a quantity being modeled by the quadratic function. For example, if the function represents the height of a projectile over time, the y-intercept would represent the initial height of the projectile. Understanding how to calculate and interpret the y-intercept is crucial for applying quadratic functions in various practical scenarios.
Therefore, by substituting $x = 0$ into the function and evaluating, we have determined that the y-intercept of the graph of the function $f(x) = 2(x + 2)^2 + 1$ is 9. This point, $(0, 9)$, provides valuable information about the graph's intersection with the y-axis and contributes to a comprehensive understanding of the function's behavior.
c) Determining the x-intercept(s) of the Graph
To state the x-intercept(s) of the graph of the function $f(x) = 2(x + 2)^2 + 1$, we need to find the points where the graph intersects the x-axis. The x-intercepts are the values of $x$ for which the function $f(x)$ is equal to 0. Therefore, we need to solve the equation $2(x + 2)^2 + 1 = 0$.
Let's solve the equation step by step:
First, we subtract 1 from both sides of the equation:
Next, we divide both sides by 2:
(x + 2)^2 = - rac{1}{2}
Now, we need to take the square root of both sides. However, we encounter a problem: the square root of a negative number is not a real number. In other words, there is no real number that, when squared, will result in a negative value. This means that the equation $(x + 2)^2 = - rac{1}{2}$ has no real solutions.
What does this imply for the graph of the function? It means that the parabola does not intersect the x-axis at any point. The x-intercepts are the real solutions to the equation $f(x) = 0$. Since we found that there are no real solutions, we can conclude that the graph has no x-intercepts.
This result is consistent with the fact that the vertex of the parabola is at $(-2, 1)$ and the parabola opens upwards. Since the vertex is above the x-axis and the parabola opens upwards, it will never cross the x-axis. This further confirms our conclusion that there are no x-intercepts.
Understanding the absence of x-intercepts can provide valuable insights into the behavior of the function. It tells us that the function's values are always positive, meaning it never reaches or goes below zero. This can be important in various applications where the function represents a quantity that cannot be negative.
Therefore, by attempting to solve the equation $2(x + 2)^2 + 1 = 0$ and encountering a situation where we need to take the square root of a negative number, we have determined that the graph of the function $f(x) = 2(x + 2)^2 + 1$ has no x-intercepts. This conclusion is supported by the fact that the parabola opens upwards and its vertex is above the x-axis.
In summary, we have thoroughly analyzed the quadratic function $f(x) = 2(x + 2)^2 + 1$ and determined its key properties. We found that the vertex of the graph is at (-2, 1), the y-intercept is 9, and there are no x-intercepts. These features provide a comprehensive understanding of the function's behavior and its graphical representation.
Understanding the vertex, intercepts, and the direction of opening of a parabola is crucial for solving a variety of mathematical problems and for applying quadratic functions in real-world scenarios. This analysis serves as a strong foundation for further exploration of quadratic functions and their applications in various fields.