Chapter 2: Problem 15
\(y=2(x-3)^2+1\)
Short Answer
Expert verified
The vertex of the parabola is located at the point (3, 1) and the standard form of the equation is \(y=2x^2 - 12x + 19\).
Step by step solution
01
Identify the vertex of the parabola
From the equation \(y = 2*(x-3)^2 +1\), we can see it is in the vertex form. Hence, the vertex of the parabola is at the point (h, k). Since our h is represented by '3' and our k is represented by '1', thus the vertex of the parabola from this equation is at the point (3, 1).
02
Derive the standard form
To expand this equation into standard form which is \(y=ax^2+bx+c\), we begin by multiplying out the binomial squared in the equation. \((x-3)^2\) becomes \(x^2 - 6x + 9\). Then multiply each of these terms by the 'a' value, which in this scenario, is 2. So: \((2)*x^2 - (2)*6x + (2)*9\). Simplify to get \(2x^2 -12x + 18\). Adding the 'k' value (1), the standard form becomes: \(y=2x^2 - 12x + 19\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Standard Form of a Quadratic Equation
The standard form of a quadratic equation is an essential tool for analyzing and graphing parabolas. It is written as \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are coefficients, and \(x\) and \(y\) are variables. The standard form makes it easy to identify key characteristics of the parabola such as its direction, width, and the y-intercept.
In the case of \(y = 2(x-3)^2 + 1\), by expanding and simplifying as shown in the step by step solution, we obtain the standard form \(y = 2x^2 - 12x + 19\). The coefficient \(a = 2\) indicates that the parabola opens upwards and will be narrower than the parabola of \(y = x^2\), due to it being stretched vertically. The coefficients \(b = -12\) and \(c = 19\) help us find the axis of symmetry and the y-intercept respectively.
It's important to be able to manipulate the quadratic equation to and from the standard form to utilize different methods of analyzing the function's graph. Mastery of this concept grants the ability to quickly decipher the behavior of quadratic functions, which is valuable for various applications in mathematics and science.
In the case of \(y = 2(x-3)^2 + 1\), by expanding and simplifying as shown in the step by step solution, we obtain the standard form \(y = 2x^2 - 12x + 19\). The coefficient \(a = 2\) indicates that the parabola opens upwards and will be narrower than the parabola of \(y = x^2\), due to it being stretched vertically. The coefficients \(b = -12\) and \(c = 19\) help us find the axis of symmetry and the y-intercept respectively.
It's important to be able to manipulate the quadratic equation to and from the standard form to utilize different methods of analyzing the function's graph. Mastery of this concept grants the ability to quickly decipher the behavior of quadratic functions, which is valuable for various applications in mathematics and science.
Graphing Parabolas
Graphing a parabola is a foundational skill in understanding quadratic functions. To graph a parabola, one must first identify its vertex, which is the highest or lowest point on the graph depending on whether the parabola opens upwards or downwards. The vertex form of a quadratic function, \(y = a(x-h)^2 + k\), provides this critical point directly, with \(h, k\) representing the vertex's coordinates.
In our exercise, the vertex of the parabola is at \(3, 1\), as determined in the solution's step 1. Once the vertex is plotted, we use the value of \(a\) to understand the parabola's shape and direction. For example, with \(a = 2\), each step away from the vertex horizontally means we go up by \(2^2 = 4\) units vertically, creating a steeper curve.
In our exercise, the vertex of the parabola is at \(3, 1\), as determined in the solution's step 1. Once the vertex is plotted, we use the value of \(a\) to understand the parabola's shape and direction. For example, with \(a = 2\), each step away from the vertex horizontally means we go up by \(2^2 = 4\) units vertically, creating a steeper curve.
Drawing the Parabola
The next step is to plot additional points by choosing x-values around the vertex and calculating their corresponding y-values. One can then sketch the parabola, ensuring it's symmetrical around its axis of symmetry (the line that runs through the vertex). The graph is a powerful visual representation of the quadratic function and is imperative for solving various real-world problems.Quadratic Functions
Quadratic functions represent a class of polynomial functions where the highest degree of \(x\) is two. These functions are crucial in algebra and are identified by their characteristic 'U' shaped curve known as a parabola. Their general form is \(y = ax^2 + bx + c\), which we often manipulate into different forms like the vertex or factored form depending on the context of the problem.
The properties of quadratic functions, such as the direction of opening, the vertex, the axis of symmetry, and intercepts, are determinable from its equation. For instance, the sign of the \(a\) coefficient indicates whether the parabola opens up (if positive) or down (if negative). Additionally, the vertex can be found directly from the vertex form \(y = a(x-h)^2 + k\) as shown in the exercise.
The properties of quadratic functions, such as the direction of opening, the vertex, the axis of symmetry, and intercepts, are determinable from its equation. For instance, the sign of the \(a\) coefficient indicates whether the parabola opens up (if positive) or down (if negative). Additionally, the vertex can be found directly from the vertex form \(y = a(x-h)^2 + k\) as shown in the exercise.