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Magazine Chapter 10: Problem 67
Graph the equation. $$x=-y^{2}+2 y+5$$
Short Answer
Expert verified
Plot the vertex at (6, 1) and draw a leftward-opening parabola.
Step by step solution
01
Understanding the Equation Form
The given equation is \(x = -y^{2} + 2y + 5\). This equation is in the form of a quadratic equation in terms of \(y\). To graph this equation, it's useful to rewrite it in the standard form \(x = a(y - h)^2 + k\), which resembles the vertex form of a parabola.
02
Completing the Square
To rewrite \(x = -y^{2} + 2y + 5\) in vertex form, we complete the square for the quadratic in \(y\). - Start by factoring out -1 from the \(y^2\) terms: \(x = -(y^{2} - 2y) + 5\).- Take half of the coefficient of \(y\), square it, and add/subtract it inside the parentheses: \(x = -(y^{2} - 2y + 1 - 1) + 5\).- Simplify inside the parentheses: \(x = -((y - 1)^{2} - 1) + 5\).- Distribute the -1: \(x = -(y - 1)^{2} + 1 + 5\).- Final vertex form: \(x = -(y - 1)^{2} + 6\).
03
Identifying Key Features of the Parabola
In the equation \(x = -(y - 1)^{2} + 6\), we can identify the vertex of the parabola. The vertex is at \((h, k) = (6, 1)\). The \(-\) sign of the coefficient in front of \((y - 1)^2\) indicates the parabola opens leftwards.
04
Plotting the Vertex and Axis of Symmetry
Plot the vertex on the coordinate plane at the point (6, 1). The axis of symmetry is a horizontal line, \(y = 1\). This line helps in plotting the parabola symmetrically around this horizontal line.
05
Choosing Additional Points
Choose some \(y\) values around the vertex to calculate corresponding \(x\) values: - For \(y = 0\), \(x = -(0 - 1)^2 + 6 = 5\) - For \(y = 2\), \(x = -(2 - 1)^2 + 6 = 5\) - Plot these points: (5, 0) and (5, 2).
06
Drawing the Parabola
Use the vertex and additional points to draw a smooth, symmetrical leftward-opening parabola. The parabola passes through points (6, 1), (5, 0), and (5, 2). Ensure it maintains its shape consistent with the completed square form.
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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 a crucial concept in algebra, often represented in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. In the context of graphing, we're focused on equations like \(x = a(y^2) + by + c\), where the roles of \(x\) and \(y\) are interchangeable depending on which variable the quadratic is expressed.
- The quadratic equation inherently represents a parabolic graph.
- The highest power of the variable is squared, hence the term quadratic.
- This graph is a smooth, continuous curve.
- Vertex: The turning point of the parabola, either the maximum or minimum.
- Axis of symmetry: A line that divides the parabola into two symmetrical halves.
- Direction: Determined by the sign of the leading coefficient \(a\); if positive, the parabola opens upwards or right, and negative, it opens downwards or left.
Completing the Square
Completing the square is a method used to transform a quadratic equation into vertex form. For the equation \(x = -y^{2} + 2y + 5\), completing the square involves:
Completing the square not only helps in converting the standard quadratic equation but also provides insight into:
- Isolating the quadratic and linear terms: \(x = -(y^2 - 2y) + 5\).
- Finding half of the linear coefficient, squaring it, and adjusting the equation: Take half of \(-2\) squared, which results in \(1\), and manipulate: \(x = -(y^2 - 2y + 1 - 1) + 5\).
- This is rewritten as \(x = -((y - 1)^2 - 1) + 5\).
- Finally, expand and simplify: \(x = -(y - 1)^2 + 1 + 5 = -(y - 1)^2 + 6\).
Completing the square not only helps in converting the standard quadratic equation but also provides insight into:
- Determining the vertex directly from the equation transformed into vertex form.
- Understanding the adjustments required to shift and reshape the graph.
Vertex Form
The vertex form of a quadratic equation is expressed as \(x = a(y-h)^2 + k\), where
Benefits of expressing a quadratic equation in vertex form include:
- \((h, k)\) is the vertex.
- \(a\) determines the width and direction of the parabola's opening.
- The vertex is at \( (6, 1) \).
- The negative sign indicates the parabola opens leftward.
Benefits of expressing a quadratic equation in vertex form include:
- Simplifies identifying the vertex, which is crucial for graphing.
- Makes it easier to analyze changes in the parabola's position and shape by simply adjusting \((h, k)\) and \(a\).
Parabola
A parabola is a symmetric curve represented graphically by a quadratic equation. In this exercise:
Key features of graphing a parabola include:
To draw a parabola accurately:
- The equation \(x = -(y - 1)^2 + 6\) describes a parabola opening to the left.
- Its vertex, \((h, k) = (6, 1)\), indicates the maximum point in the case of a leftward opening parabola.
Key features of graphing a parabola include:
- Vertex: The primary point around which the parabola is shaped.
- Axis of symmetry: For the equation presented, it is the line \(y = 1\), ensuring the parabola mirrors across this line.
- Additional points: These are calculated to plot and understand the parabola's extent and exact shape.
To draw a parabola accurately:
- Identify and plot the vertex.
- Determine the direction of opening from the coefficient of the squared term.
- Assess the orientation i.e., horizontal in this case due to the expression \(x = \ldots\).
