Problem 25 - Algebra and Trigonometry

Question

19–44 ? Make a table of values and sketch the graph of the equation. Find the x- and y-intercepts and test for symmetry. $$ y=1-x^{2} $$

Step-by-Step Solution

Answer

The x-intercepts are (1, 0) and (-1, 0); the y-intercept is (0, 1); the graph is symmetric about the y-axis.

Step 1: 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:\[ \begin{array}{c|c}x & y \\hline-2 & -3 \-1 & 0 \0 & 1 \1 & 0 \2 & -3 \\end{array} \]

Step 2: 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.

Step 3: 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 \((-1, 0)\).

Step 4: 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 \((0, 1)\).

Step 5: 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.

Key Concepts

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.

• 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.

Understanding parabolas helps in analyzing the behavior and properties of quadratic functions, such as determining the direction and width of the curve.

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.

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.

• 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.

Recognizing symmetry in graphs is a helpful tool to simplify graphing and analysis, ensuring models are both accurate and efficient in depicting functional relationships.