Problem 10 - Precalculus

Question

Graph the equation after determining the \(x\) - and \(y\) -intercepts and whether the graph possesses any of the three types of symmetry described on page 58 $$x=y^{2}-1$$

Step-by-Step Solution

Answer

The graph has x-intercept \((-1, 0)\), y-intercepts \((0, 1)\) and \((0, -1)\), with symmetry about the x-axis.

Step 1: Identify the equation format

The given equation is in the form of \( x = y^2 - 1 \), which represents a parabola opening horizontally.

Step 2: Find the y-intercept

The y-intercept is found by setting \( x = 0 \) and solving for \( y \). The equation becomes \( 0 = y^2 - 1 \), which leads to \( y^2 = 1 \). Solving for \( y \), we get \( y = \pm 1 \). Thus, the y-intercepts are \((0, 1)\) and \((0, -1)\).

Step 3: Find the x-intercept

The x-intercept is found by setting \( y = 0 \) and solving for \( x \). Substituting \( y = 0 \) into the equation, we get \( x = 0^2 - 1 = -1 \). Thus, the x-intercept is \((-1, 0)\).

Step 4: Determine symmetry

To check symmetry about the x-axis, replace \( y \) with \( -y \), resulting in \( x = (-y)^2 - 1 = y^2 - 1 \), which is the same equation, indicating symmetry about the x-axis. To check symmetry about the y-axis, replace \( x \) with \( -x \), resulting in \( -x = y^2 - 1 \), which is not the same equation, indicating no symmetry about the y-axis. For symmetry about the origin, replace \( x \) with \( -x \) and \( y \) with \( -y \), leading to \( -x = (-y)^2 - 1 = y^2 - 1 \), which is also not the same as the original, indicating no symmetry about the origin.

Step 5: Draw the graph

Using the intercepts, plot the points \((0, 1), (0, -1)\), and \((-1, 0)\) on the coordinate system. Since the graph is a parabola opening sideways, sketch the parabola opening to the right from the vertex at \((-1, 0)\). The parabolic shape should pass through the y-intercepts and exhibit symmetry about the x-axis.

Key Concepts

x-intercepts

The x-intercepts of a graph are the points where the graph crosses the x-axis. This means that at these points, the y-value is zero. To find the x-intercepts:

• Set the y variable to zero in the equation.
• Solve the resulting equation for the x values.

In the provided exercise, we set the y value to zero in the equation: \[ x = y^2 - 1 \] This simplifies to: \[ x = 0^2 - 1, \] resulting in \[ x = -1. \] Therefore, the x-intercept for this parabola is located at the point \((-1, 0)\). This point indicates where the curve intersects the x-axis, crucial for sketching the graph.

y-intercepts

The y-intercepts of a graph are where the graph crosses the y-axis. At these points, the x-value is always zero. To find the y-intercepts:

• Set the x variable to zero.
• Solve the equation for y-values.

In our equation, \[ x = y^2 - 1, \] setting \( x = 0 \) gives:\[ 0 = y^2 - 1, \]which simplifies to \[ y^2 = 1. \] Solving for y, we find \( y = 1 \) and \( y = -1 \). Thus, the y-intercepts are the points \((0,1)\) and \((0,-1)\). These points highlight where the graph intersects the y-axis, and you can imagine them as the starting and stopping points when plotting.

symmetry

Symmetry in a graph refers to how a shape can be mirrored or replicated across an axis or a point. The three main types of symmetry in graphing are:

• Symmetry about the x-axis
• Symmetry about the y-axis
• Symmetry about the origin

To check for x-axis symmetry, replace y with \(-y\) in the equation. If the equation remains unchanged, the graph is symmetric about the x-axis. In our case: \[ x = (-y)^2 - 1 = y^2 - 1, \] indicating x-axis symmetry. For y-axis symmetry, replace x with \(-x\). If the equation still holds, then there is y-axis symmetry. Attempting this, the equation\(-x = y^2 - 1 \) differs from the original, so no y-axis symmetry exists here. Finally, origin symmetry requires both variables to be replaced: \(-x = (-y)^2 - 1 = y^2 - 1, \) which isn't equivalent to the original. Thus, the graph is not symmetric about the origin.

parabolas

Parabolas are U-shaped curves on a graph, typically represented by quadratic equations like \( y = ax^2 + bx + c \). However, the exercise given is formatted differently: \( x = y^2 - 1 \), indicating a parabola that opens sideways rather than upward or downward. This is an important distinction. For sideways-opening parabolas like this one, the vertex forms a turning point of the curve. Here, the vertex is at \((-1, 0)\). This vertex allows the parabola to extend rightward from it, creating a mirror image along the x-axis. Understanding the orientation of your parabola is vital when graphing; for instance, with this equation, the sideways opening results in the graph appearing horizontally symmetric. So, remember:

• The direction of the opening depends on the variable with the squared term.
• The vertex acts as the focal point from which the graph extends.
• This distinctive formation defines the parabola's shape.

These characteristics help us accurately draw and interpret the graph.