Problem 30 - Algebra 2

Question

Rewrite each equation in vertex form. $$ y=-2 x^{2}+6 x+1 $$

Step-by-Step Solution

Answer

\(y = -2(x - \frac{3}{2})^2 + \frac{11}{2}\)

Step 1: Understand Vertex Form

Understand that the vertex form of a quadratic equation is given by \(y = a(x-h)^2 + k\), where \((h, k)\) is the vertex of the parabola, and \(a\) is the coefficient that determines the width and direction of the parabola.

Step 2: Factor the Coefficient of \(x^2\)

Factor out the coefficient of \(x^2\) from the first two terms. Since the coefficient is -2, we get \(y = -2(x^2 - 3x) + 1\).

Step 3: Complete the Square

Rewrite the equation in a way that allows us to complete the square for the expression in the parentheses. We achieve this by adding and subtracting the square of half of the coefficient of \(x\) inside the parentheses, which in this case is \((\frac{3}{2})^2 = \frac{9}{4}\). So, \(y = -2(x^2 - 3x + \frac{9}{4} - \frac{9}{4}) + 1\).

Step 4: Simplify the Equation

Simplify the expression by combining terms within the parentheses and adjusting the constant term outside. The equation now becomes \(y = -2((x - \frac{3}{2})^2 - \frac{9}{4}) + 1\).

Step 5: Expand and Combine Like Terms

Expand the constant terms to get \(y = -2(x - \frac{3}{2})^2 + \frac{9}{2} + 1\). Combine the constant terms to obtain the final vertex form: \(y = -2(x - \frac{3}{2})^2 + \frac{11}{2}\).

Key Concepts

Completing the Square

Understanding the process of completing the square is essential to converting quadratic equations into vertex form. This method involves creating a perfect square trinomial from a quadratic expression, which can then be written as the square of a binomial. To start, consider a generic quadratic equation in the format of \(ax^2 + bx + c\). The goal is to manipulate this into a form where \(x\) only appears once, in a squared term.

Here's a step-by-step guide on how to complete the square:

• Rewrite the equation separating the \(x^2\) and \(x\)-terms from the constant \(c\), if necessary.
• If \(a eq 1\), factor it out from the \(x^2\) and \(x\) terms.
• Take half of the coefficient of \(x\) from the binomial, square it, and add and subtract this square within the parentheses.
• Rewrite the equation as the binomial square and adjust the constant terms accordingly.

For our exercise, after factoring \(-2\) from the first two terms, we added and subtracted \(\frac{9}{4}\) to complete the square, leading to the creation of a perfect square trinomial \((x - \frac{3}{2})^2\).

Parabolas in Algebra

In algebra, parabolas are the graphs formed by quadratic equations and are U-shaped curves that can either open upwards or downwards. The direction in which a parabola opens is determined by the sign of the coefficient \(a\) in the quadratic equation \(y = ax^2 + bx + c\). If \(a\) is positive, the parabola opens upwards, and if \(a\) is negative, as in our sample exercise, it opens downwards.

A key feature of parabolas is their vertex, the highest or lowest point on the graph, depending on the direction the parabola opens. The vertex form of the quadratic equation, \(y = a(x-h)^2 + k\), highlights the coordinates of the vertex \((h, k)\). Using the vertex form allows you to easily determine important characteristics of the parabola, such as its vertex, axis of symmetry, and the direction it opens.

Quadratic Equations

Quadratic equations are fundamental in algebra and can be identified by their standard form, \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, and \(a\) is nonzero. These equations describe parabolas when graphed and have widespread applications in various fields, including physics, engineering, and finance.

Solving quadratic equations can be approached in multiple ways: factoring, using the quadratic formula, graphing, or completing the square. Each method has its own context of use, and understanding how to manipulate these equations into different forms, such as vertex form or factored form, can provide different insights into the properties of the parabola, such as its roots or vertex.

By rewriting the equation in vertex form, as illustrated in the exercise, we can easily determine the vertex of the parabola and gain a clearer view of its graphical representation. This form is particularly useful for solving real-world problems where the maximum or minimum point of a parabolic trajectory must be identified.