Problem 61 - College Algebra and Calculus: An Applied Approach

Question

Sketch the graph of the equation. Identify any intercepts and test for symmetry. \(y=1-x^{2}\)

Step-by-Step Solution

Answer

The graph of the equation \(y=1-x^2\) is a downward opening parabola with its vertex at (0, 1). The intercepts are at points (1, 0), (-1, 0), and (0, 1). The graph is symmetrical about the y-axis.

Step 1: Identify the Vertex

The equation given is \(y = 1 - x^2\), which is already in the vertex form, comparable to \(y=a(x-h)^2 + k\), where a is -1, h is 0 and k is 1. Thus, the vertex of the parabola is at the point (0, 1).

Step 2: Find the Intercepts

For x-intercepts, set \(y=0\) and solve for \(x\). The equation will be \(0 = 1 - x^2\), which gives \(x^2 = 1\) and therefore \(x = ±1\). The x-intercepts are (1, 0) and (-1, 0). The y-intercept is found by setting \(x=0\) in the equation. This provides \(y = 1 - 0 = 1\), so the y-intercept is at (0, 1).

Step 3: Check for Symmetry

Identify if the graph is symmetrical by checking what changing \(x\) to \(-x\) in the equation does. \(y = 1 - (-x)^2 = 1 - x^2\), which is symmetrical around the y-axis (since \(x = -x\)). The graph of this equation is symmetrical.

Step 4: Sketch the Graph

Begin by marking the vertex (0, 1) and the intercepts (1, 0), (-1, 0), and (0, 1) on the graph. The parabola opens downward since the coefficient of \(x^2\) is negative. Draw a smooth, curved line through these points to complete the parabolic graph.

Key Concepts

Vertex Identification

Understanding the vertex of a parabola is crucial because it represents the highest or lowest point of the graph. In the equation \(y = 1 - x^2\), we can relate it to the vertex form \(y = a(x-h)^2 + k\), where \(h\) and \(k\) are the coordinates of the vertex.
Here, the equation is already simplified such that \(h = 0\) and \(k = 1\), resulting in a vertex at the point \((0, 1)\).
This vertex tells us that the graph will reach its peak at this point, as the coefficient of \(x^2\) (\(a = -1\)) is negative, indicating the parabola opens downward.

Intercepts Calculation

Calculating intercepts helps in determining where the graph crosses the axes.

• X-Intercepts: To find the x-intercepts, set \(y = 0\) and solve the equation \(0 = 1 - x^2\).
  This results in \(x^2 = 1\), giving solutions \(x = 1\) and \(x = -1\).
  Thus, the x-intercepts are at the points \((1, 0)\) and \((-1, 0)\).
• Y-Intercept: Set \(x = 0\) to find the y-intercept. Substituting gives \(y = 1 - 0 = 1\). Thus, the y-intercept is \((0, 1)\).

These intercepts are critical points that help provide a skeletal framework for graphing the equation.

Symmetry Testing

Testing a graph for symmetry can reveal if it mirrors on a particular axis. For the equation \(y = 1 - x^2\), replace \(x\) with \(-x\) to test for y-axis symmetry.
When substituted, \(y = 1 - (-x)^2\) simplifies to \(y = 1 - x^2\).
Since the equation remains unchanged, the graph is symmetrical about the y-axis.
This form of symmetry is typical of parabolas that open upward or downward, as they naturally mirror evenly across the y-axis.

Graph Sketching

With all the properties identified, you can now sketch the graph effectively.

• Start by plotting the vertex \((0, 1)\), which is the top point here since the parabola opens downward.
• Plot the intercepts: \((1, 0)\) and \((-1, 0)\) on the x-axis, and \((0, 1)\) on the y-axis.
• Notice the downward opening direction, confirmed by the negative \(x^2\) coefficient.

Finally, draw a smooth curve through these points, ensuring the curve is symmetrical across the y-axis. This belly-down parabola will extend infinitely downward, growing wider but never touching any other point on the axes besides those identified.