Problem 62 - College Algebra and Calculus: An Applied Approach

Question

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

Step-by-Step Solution

Answer

The graph is an upward opening parabola centered at (0,-1) with x-intercepts at (-1,0) and (1,0). The graph is symmetrical around the y-axis.

Step 1: Plot the Equation

The first step will be plotting the equation \(y=x^{2}-1\). The graph this equation produces is a parabola with a vertex at the point (0, -1) and opens upwards.

Step 2: Identify Intercepts

To find the y-intercept, set x=0 and solve for y. \(y = (0)^2 - 1 = -1\), thus the y-intercept is at (0, -1). To find the x-intercepts, set y=0 and solve for x. The solutions to the equation \(0=x^{2}-1\) are \(x=-1\) and \(x=1\), hence the x-intercepts are at (-1, 0) and (1, 0).

Step 3: Test for Symmetry

To test for symmetry, replace x with -x in the equation and simplify. The equation becomes: \(y=(-x)^2 - 1 = x^2 - 1\), which is the same as the given equation. Hence, the graph has symmetrical characteristics around y-axis.

Key Concepts

Understanding the Parabola

When dealing with quadratic equations like \(y = x^2 - 1\), the graph takes the form of a curve known as a parabola. A parabola is a symmetric curve that typically has a "U" or an upside-down "U" shape. The equation \(y = x^2 - 1\) is in its simplest form, and this indicates that the parabola opens upwards.
The vertex, which is the highest or lowest point on the parabola, can be determined directly from the equation. Here, the vertex is at the point (0, -1). This vertex represents the minimum point of the curve because the parabola opens upwards.

• The vertex of a quadratic function in the form \(y = ax^2 + bx + c\) can be found using \( x = -\frac{b}{2a} \). In our case, since \(a = 1\) and \(b = 0\), the vertex is simply at \(x = 0\).
• The constant \(-1\) in the equation shifts the entire parabola one unit downwards along the y-axis, compared to \(y = x^2\).

Understanding the basics of a parabola helps in sketching it more accurately and predicting its behavior.

Exploring Intercepts

Intercepts are key points where the graph crosses the axes. For the equation \(y = x^2 - 1\), they help in providing a detailed sketch of the graph. Intercepts consist of both the y-intercept and any x-intercepts.

Y-intercept

• The y-intercept is the point where the graph crosses the y-axis. For our equation, this occurs when \(x = 0\). By substituting \(x = 0\) into the equation, we find \(y = -1\). Therefore, the y-intercept is at the point (0, -1).

X-intercepts

• The x-intercepts represent the points where the graph crosses the x-axis. These occur when \(y = 0\). Solving \(x^2 - 1 = 0\) gives two solutions: \(x = -1\) and \(x = 1\). Thus, the parabola touches the x-axis at points (-1, 0) and (1, 0).

These intercepts create a framework for placing the parabola correctly on the graph and understanding its intersection with the axes.

Axis of Symmetry

Symmetry is another important factor when graphing parabolas. It simplifies the process of understanding the entire curve. For the equation \(y = x^2 - 1\), the symmetry can be observed about the vertical line that passes through the vertex.
To test for symmetry, we replace \(x\) with \(-x\) in the equation. If the transformed equation is the same as the original, symmetry is confirmed.

• In our equation, replacing \(x\) with \(-x\) results in \(y = (-x)^2 - 1 = x^2 - 1\). As we can see, this remains unchanged, indicating that the parabola is symmetric about the y-axis.

Understanding symmetry aids not only in sketching the graph but also in solving quadratic equations, as any solutions or characteristics on one side will reflect on the other.