Chapter 2: Problem 25
19–44 ? Make a table of values and sketch the graph of the equation. Find the x- and y-intercepts and test for symmetry. $$ y=1-x^{2} $$
Short Answer
Expert verified
The x-intercepts are (1, 0) and (-1, 0); the y-intercept is (0, 1); the graph is symmetric about the y-axis.
Step by step solution
01
Make a Table of Values
Choose several values for \( x \), and plug them into the equation \( y = 1 - x^2 \) to find corresponding \( y \) values. For example, for \( x = -2, -1, 0, 1, 2 \), calculate the following:- For \( x = -2 \), \( y = 1 - (-2)^2 = 1 - 4 = -3 \)- For \( x = -1 \), \( y = 1 - (-1)^2 = 1 - 1 = 0 \)- For \( x = 0 \), \( y = 1 - 0^2 = 1 \)- For \( x = 1 \), \( y = 1 - 1^2 = 0 \)- For \( x = 2 \), \( y = 1 - 2^2 = -3 \)Thus, the table of values is:\[ \begin{array}{c|c}x & y \\hline-2 & -3 \-1 & 0 \0 & 1 \1 & 0 \2 & -3 \\end{array} \]
02
Sketch the Graph
Using the table of values from Step 1, plot the points \((-2, -3), (-1, 0), (0, 1), (1, 0), (2, -3)\) on a graph. Connect these points with a smooth curve. You should see that the graph forms a downward-opening parabola centered around \((0, 1)\). This is because the equation is a quadratic function with a negative leading coefficient.
03
Find the x-intercepts
To find the x-intercepts, set \( y = 0 \) and solve for \( x \):\[ 1 - x^2 = 0 \]\[ x^2 = 1 \]\[ x = \pm 1 \]The x-intercepts are \((1, 0)\) and \((-1, 0)\).
04
Find the y-intercept
To find the y-intercept, set \( x = 0 \) and solve for \( y \):\[ y = 1 - 0^2 = 1 \]The y-intercept is \((0, 1)\).
05
Test for Symmetry
Test for symmetry by substituting \( -x \) for \( x \) in the equation:\[ y = 1 - (-x)^2 = 1 - x^2 \]The equation remains the same, indicating that the graph is symmetric about the y-axis. This is confirmed by the fact that the equation is even, as expected for a parabola.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parabolas
Parabolas are a key concept in understanding quadratic functions. They are the distinctive U-shaped curves formed on a graph, defined by a quadratic equation of the form \( ax^2 + bx + c \). When the leading coefficient (\( a \)) is negative, like in the equation \( y = 1 - x^2 \), the parabola opens downward. Conversely, if \( a \) is positive, the parabola opens upward.
This downward-opening parabola reflects that as \( x \) moves away from the vertex, the \( y \)-values decrease. The vertex of a parabola is its highest or lowest point, depending on the orientation. For the equation \( y = 1 - x^2 \), the vertex is at the point \( (0, 1) \), where \( 0 \) is along the x-axis and \( 1 \) is the maximum height on the y-axis.
This downward-opening parabola reflects that as \( x \) moves away from the vertex, the \( y \)-values decrease. The vertex of a parabola is its highest or lowest point, depending on the orientation. For the equation \( y = 1 - x^2 \), the vertex is at the point \( (0, 1) \), where \( 0 \) is along the x-axis and \( 1 \) is the maximum height on the y-axis.
- The vertex is a crucial point as it provides symmetry and serves as a reference to determine the direction of the graph.
- A parabola is symmetric about its axis of symmetry, which in this case is the y-axis because the vertex is aligned along x = 0.
Intercepts
Finding intercepts is an essential part of graphing and understanding equations. Intercepts are the points where the graph crosses the x-axis and y-axis.
**X-Intercepts:** To find x-intercepts, set \( y = 0 \) in the equation \( y = 1 - x^2 \) and solve for \( x \). The equation becomes \( 1 - x^2 = 0 \). Solving it, we find \( x = \pm 1 \), giving us x-intercepts at the points \( (-1, 0) \) and \( (1, 0) \). These intercepts show where the parabola crosses the x-axis and are crucial for understanding the basic shape of the graph.
**Y-Intercept:** To find the y-intercept, set \( x = 0 \) and solve for \( y \). In this equation, that gives \( y = 1 \). Therefore, the y-intercept is at the point \( (0, 1) \). The y-intercept indicates where the parabola crosses the y-axis, providing a base point for sketching the curve.
Intercepts provide critical touchpoints that frame the graph's position and are invaluable for accurately plotting the function's representation.
**X-Intercepts:** To find x-intercepts, set \( y = 0 \) in the equation \( y = 1 - x^2 \) and solve for \( x \). The equation becomes \( 1 - x^2 = 0 \). Solving it, we find \( x = \pm 1 \), giving us x-intercepts at the points \( (-1, 0) \) and \( (1, 0) \). These intercepts show where the parabola crosses the x-axis and are crucial for understanding the basic shape of the graph.
**Y-Intercept:** To find the y-intercept, set \( x = 0 \) and solve for \( y \). In this equation, that gives \( y = 1 \). Therefore, the y-intercept is at the point \( (0, 1) \). The y-intercept indicates where the parabola crosses the y-axis, providing a base point for sketching the curve.
Intercepts provide critical touchpoints that frame the graph's position and are invaluable for accurately plotting the function's representation.
Graph Symmetry
Symmetry is a fundamental property of parabolas, making them predictable and easier to graph. For a function to be symmetric about a line, substituting \( -x \) for \( x \) in the function should yield an equivalent expression.
For the equation \( y = 1 - x^2 \), replace \( x \) with \( -x \) and notice the equation remains unchanged: \( y = 1 - (-x)^2 = 1 - x^2 \). This confirms the function's symmetry about the y-axis, indicating the parabola is an even function.
For the equation \( y = 1 - x^2 \), replace \( x \) with \( -x \) and notice the equation remains unchanged: \( y = 1 - (-x)^2 = 1 - x^2 \). This confirms the function's symmetry about the y-axis, indicating the parabola is an even function.
- The symmetry about the y-axis ensures that each point on one side of the axis has a corresponding point directly opposite on the other side.
- This reflectional symmetry helps in constructing the complete graph by mirroring points across the y-axis.