Problem 41 - Algebra and Trigonometry with Analytic Geometry

Question

Identify the graph of the equation as a parabola (with vertical or horizontal axis), circle, ellipse, or hyperbola. $$x^{2}+3 x=3 y-6$$

Step-by-Step Solution

Answer

The graph is a parabola with a vertical axis.

Step 1: Rearrange the Equation

Start by rearranging the given equation to move all terms to one side: \[ x^2 + 3x - 3y + 6 = 0 \] Now, focus on the terms involving \(x\) and \(y\) to identify the type of conic.

Step 2: Complete the Square for x

To understand the graph better, complete the square for the \(x\)-terms. Take the equation:\[ x^2 + 3x = 3y - 6 \]The goal is to rewrite \(x^2 + 3x\) as a complete square. Take half of the coefficient of \(x\) (\(\frac{3}{2}\)), square it (\((\frac{3}{2})^2 = \frac{9}{4}\)), and add and subtract it inside the equation:\[ x^2 + 3x + \frac{9}{4} - \frac{9}{4} = 3y - 6 \]This transforms into:\[ (x + \frac{3}{2})^2 - \frac{9}{4} = 3y - 6 \]

Step 3: Simplify and Recognize the Conic

Solve the rearranged equation for \(y\):\[ (x + \frac{3}{2})^2 - \frac{9}{4} = 3y - 6 \].Add \(\frac{9}{4}\) to both sides:\[ (x + \frac{3}{2})^2 = 3y - 6 + \frac{9}{4} \].Simplify the right side:\[ (x + \frac{3}{2})^2 = 3y - \frac{15}{4} \].Divide through by 3 to make \(y\) the subject:\[ y = \frac{1}{3}(x + \frac{3}{2})^2 + \frac{5}{4} \].This is in the form \(y = a(x-h)^2 + k\), indicating a parabola with a vertical axis.

Key Concepts

Parabola

A parabola is a U-shaped curve that can open upwards, downwards, or sideways in a coordinate plane. It is a type of conic section, like circles, ellipses, and hyperbolas. Parabolas are defined by quadratic equations, with the standard form for parabolas with vertical axes being \( y = a(x-h)^2 + k \).

• The vertex of the parabola is at the point \((h, k)\).
• The value of \(a\) determines how wide or narrow the parabola is, as well as its direction. When \(a > 0\), the parabola opens upwards, while it opens downwards for \(a < 0\).
• The axis of symmetry is a vertical line that passes through the vertex, found at \(x = h\).

In our exercise, the derived equation \( y = \frac{1}{3}(x + \frac{3}{2})^2 + \frac{5}{4} \) identifies the shape as a parabola. It has a vertical axis because the equation is in the standard form, with \((x-h)^2\) on the right side.

Completing the Square

Completing the square is a technique used in algebra to transform a quadratic expression into a perfect square trinomial, making it easier to solve or graph. The method involves adding and subtracting a particular value, calculated from the coefficient of the linear term. This simplifies the quadratic into a form that reveals its vertex and essentially translates the parabola.

• First, identify the coefficient of the linear term in \( ax^2 + bx + c \). Here it is \(b\).
• Take half of \(b\), square it, and add or subtract this square to/from the equation.
• Finally, rewrite the expression as a binomial square.

In the given exercise, we completed the square for \(x^2 + 3x\) by adding and subtracting \(\left(\frac{3}{2}\right)^2\), simplifying to \((x + \frac{3}{2})^2\). This process is crucial for identifying the conic section and aligning it with a recognizable form for graphing.

Graph of Equation

Graphing an equation is all about visualizing the shape formed by plotting solutions to the equation in the coordinate plane. For conics like parabolas, this involves understanding the structure revealed by the equation's form.

• Start by identifying the type of conic section through the equation's format, as seen with quadratic equations indicating parabolas.
• Once you know it's a parabola, use information about the vertex, axis of symmetry, and direction to plot correctly.
• Points such as the vertex and direction can be drawn from the completed square form.

In this solution, the equation was transformed into \( y = \frac{1}{3}(x + \frac{3}{2})^2 + \frac{5}{4} \), showing a vertically oriented parabola. The vertex \((-\frac{3}{2},\frac{5}{4})\) serves as a starting point on the graph. The positive \(\frac{1}{3}\) in front of the squared term shows that it opens upwards, creating a U-shape on the grid.