Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the graph of the given equation. $$ (x+2)^{2}=4 $$

Step-by-Step Solution

Answer

The graph is a parabola with vertex (-2,0), touching the x-axis at x = 0 and x = -4.

Step 1: Identify the Type of Equation

The equation \((x+2)^2 = 4\) is a quadratic equation in the form of a perfect square. It implies that when expanded, the equation will have the form \(ax^2 + bx + c = 0\). However, it is presented as a completed square form.

Step 2: Solve for x

Take the square root of both sides to eliminate the square: \(\sqrt{(x+2)^2} = \pm \sqrt{4}\). This simplifies to \(x+2 = \pm 2\). Therefore, we have two solutions:1. \(x+2 = 2\) leading to \(x = 0\)2. \(x+2 = -2\) leading to \(x = -4\).

Step 3: Identify Key Points

The roots or solutions of the equation, \(x = 0\) and \(x = -4\), are key points where the graph touches or intersects the x-axis.

Step 4: Sketch the Graph

The given equation \((x+2)^2 = 4\) represents a parabola opening upwards with the vertex at \((x+2) = 0\), or \(x = -2\). The graph intersects the x-axis at \(x = 0\) and \(x = -4\). To sketch:- Plot the vertex at \((-2, 0)\).- Mark the points where the graph meets the x-axis: \((-4, 0)\) and \((0, 0)\).- Draw a parabola opening upwards passing through these points and vertex.

Key Concepts

Understanding the Parabola

A parabola is a U-shaped curve that is seen in graphs of quadratic equations. In its most basic form, a parabola is represented by the equation \(y = ax^2 + bx + c\), a quadratic equation. The term "quadratic" comes from "quad," meaning square, due to the exponent of 2.

Let's break down its structure:

• The highest degree of \(x\) is 2, identifying the equation as quadratic.
• If \(a > 0\), the parabola opens upwards ("happy") and if \(a < 0\), it opens downwards ("sad").

In our exercise, the parabola stems from the equation \((x+2)^2 = 4\) which can be thought of as \(x^2 + 4x + 4 = 4\) upon expansion and rearrangement. The vertex is a significant point here, determining the maximum or minimum of the parabola, situated at \((h, k)\) when in vertex form \(y = a(x-h)^2 + k\).

In the example, the vertex form is evident, allowing immediate identification of the vertex at \((-2, 0)\). This makes the process of sketching clearer and more intuitive.

Graphing Parabolic Equations

The process of graphing quadratic equations like a parabola revolves around identifying key features to accurately represent the equation visually.

Here’s a step-by-step guide:

• Start with the vertex, an essential anchor for the graph which gives the general shape.
• Identify the axis of symmetry, which is a vertical line passing through the vertex, here \(x = -2\).
• Locate the solutions to know where the parabola crosses the x-axis, that is, root(s) like \(x = 0\) and \(x = -4\) in our problem.
• Ensure the direction of the opening is correct, which is up in this scenario, as implied by the original vertex form.
• Finally, draw a smooth curve through these points ensuring it reflects these attributes and aligns with the opening direction.

By identifying and plotting these aspects, the graph effectively represents the quadratic equation, showing its intersection points and shape detailedly.

Finding Solutions of Quadratic Equations

Solutions or roots of quadratic equations are the values of \(x\) which make the equation true, often represented as points where the parabola crosses the x-axis.

In theory, for a standard quadratic equation \(ax^2 + bx + c = 0\), you can use several methods to find solutions:

• Factorization: Breaking down into simple expressions to find roots.
• Completing the square: Solution through equation rearrangement into vertex form.
• Quadratic formula: \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), a reliable method for direct computation.

For the exercise given, completing the square was already achieved, simplifying it to \((x+2)^2 = 4\). Taking the square root gives solutions almost instantaneously: \(x + 2 = \pm 2\) leads to an explicit answer set \(x = 0\) and \(x = -4\). Each solution shows where the parabola intersects the x-axis, crucial in constructing an accurate graph.