Chapter 10: Problem 20
Sketch the graph of the given equation. \((x+2)^{2}=4\)
Short Answer
Expert verified
Graph a parabola opening upwards with vertex at (-2, 0) and x-intercepts at 0 and -4.
Step by step solution
01
Identify the Type of Equation
The given equation \((x+2)^2 = 4\) is a quadratic equation in disguise. It's a simple quadratic equation having a squared term.
02
Solve for x
To sketch the graph, we first solve the equation. Take the square root of both sides: \[\sqrt{(x+2)^2} = \sqrt{4}\]which results in \[|x+2| = 2\]This gives two solutions: \[x + 2 = 2\] leading to \[x = 0\] and \[x + 2 = -2\] leading to \[x = -4\].
03
Analyze the Solutions
The solutions \(x = 0\) and \(x = -4\) are the points where the graph of the equation intersects the x-axis. These are the roots of the equation.
04
Determine the Vertex
The function \((x+2)^2\) is a perfect square trinomial with the vertex at \((-2, 0)\), which is the lowest point of the parabola since the coefficient of the squared term is positive, indicating it opens upwards.
05
Sketch the Graph
Plot the vertex at \((-2, 0)\) and the intercepts at \((0, 0)\) and \((-4, 0)\). The graph is a parabola opening upwards, centered at \(x = -2\). Draw the parabola ensuring it passes through these points.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
A quadratic equation is an algebraic expression of the form \(ax^2 + bx + c = 0\), where \(a, b,\) and \(c\) are constants, and \(a\) is not equal to zero. This type of equation results in a parabolic graph.
The characteristic feature of a quadratic equation is the presence of the squared term \(x^2\). In our exercise, the equation is \((x+2)^2=4\). To see it as a standard quadratic equation, you can expand and rearrange it to form \(x^2 + 4x + 4 = 4\), which simplifies to \(x^2 + 4x = 0\) after subtracting 4 from both sides.
When graphed, quadratic equations often create 'U' shaped curves known as parabolas, which makes solving them visually fascinating. Identifying these features helps to understand and graph the equations efficiently.
The characteristic feature of a quadratic equation is the presence of the squared term \(x^2\). In our exercise, the equation is \((x+2)^2=4\). To see it as a standard quadratic equation, you can expand and rearrange it to form \(x^2 + 4x + 4 = 4\), which simplifies to \(x^2 + 4x = 0\) after subtracting 4 from both sides.
When graphed, quadratic equations often create 'U' shaped curves known as parabolas, which makes solving them visually fascinating. Identifying these features helps to understand and graph the equations efficiently.
Solving Equations
Solving quadratic equations involves finding the values of \(x\) that make the equation true. In our original problem, solving \((x+2)^2=4\) means finding the value or values of \(x\) that satisfy the equation.
We start by taking the square root of both sides, resulting in \(|x+2| = 2\). Since the absolute value represents distance in both directions on a number line, we consider both possibilities. This gives us two linear equations:
We start by taking the square root of both sides, resulting in \(|x+2| = 2\). Since the absolute value represents distance in both directions on a number line, we consider both possibilities. This gives us two linear equations:
- \(x + 2 = 2\)
- \(x + 2 = -2\)
- \(x = 0\)
- \(x = -4\)
Parabola
A parabola is a symmetrical, curved shape that is mimicked by the graph of a quadratic equation. It can open either upwards or downwards, depending on the coefficient of the squared term in the equation. If positive, the parabola opens upwards like a smile; if negative, it opens downwards like a frown.
In our case, the expanded form of the quadratic equation is \(x^2 + 4x = 0\). The coefficient of \(x^2\) is positive \((1)\), indicating that our parabola opens upwards. This provides immense insight when graphing, as it dictates the general orientation and shape the graph should take.
The parabola’s property of symmetry means it reflects identically on both sides of its vertex, creating a mirror image about the line of symmetry.
In our case, the expanded form of the quadratic equation is \(x^2 + 4x = 0\). The coefficient of \(x^2\) is positive \((1)\), indicating that our parabola opens upwards. This provides immense insight when graphing, as it dictates the general orientation and shape the graph should take.
The parabola’s property of symmetry means it reflects identically on both sides of its vertex, creating a mirror image about the line of symmetry.
Vertex of a Parabola
The vertex is a critical point on the parabola. It represents the peak or the lowest point, depending on the orientation of the parabola. For an upward-opening parabola, the vertex is the lowest point.
In a vertex form quadratic equation \(a(x-h)^2 + k\), the vertex is given by the point \((h, k)\).Our specific equation, \((x+2)^2=4\), can be rewritten to reveal its vertex at \((-2, 0)\).
This vertex serves as the parabola's lowest point and its line of symmetry. Hence, the parabola is symmetric around the vertical line \(x = -2\). Identifying the vertex helps in sketching the accurate shape of a parabola and determining its basic position relative to the coordinate plane.
In a vertex form quadratic equation \(a(x-h)^2 + k\), the vertex is given by the point \((h, k)\).Our specific equation, \((x+2)^2=4\), can be rewritten to reveal its vertex at \((-2, 0)\).
This vertex serves as the parabola's lowest point and its line of symmetry. Hence, the parabola is symmetric around the vertical line \(x = -2\). Identifying the vertex helps in sketching the accurate shape of a parabola and determining its basic position relative to the coordinate plane.
