Problem 21 - Algebra and Trigonometry

Question

Use a graphing device to graph the parabola. $$y^{2}=-\frac{1}{3} x$$

Step-by-Step Solution

Answer

Graph the parabola opening to the left with vertex at (0,0).

Step 1: Recognize the Equation Type

The given equation is \( y^2 = -\frac{1}{3}x \), which is in the form of a parabola. In general, if you have \( y^2 = kx \) or \( x^2 = ky \), it represents a parabola.

Step 2: Determine the Orientation

Since the equation is of the form \( y^2 = -\frac{1}{3}x \), this indicates a horizontal parabola which opens to the left because of the negative sign before \( \frac{1}{3}x \).

Step 3: Identify Key Features

Identify the vertex of the parabola and any other key points. Here, the vertex of the parabola is at the origin (0,0).

Step 4: Graph the Parabola

Using a graphing device, first plot the vertex at (0,0). Next, draw the parabola opening to the left. Select values of \( y \), plug them into the equation, and solve for \( x \) to find points to plot. For example, if \( y = 1 \), then \( x = -\frac{1}{3} \), giving the point (\(-\frac{1}{3}, 1\)) on the graph.

Step 5: Verify Graph Symmetry

Check that the graph is symmetric across the x-axis because the equation involves \( y^2 \). For every point \((x, y)\), the corresponding point \((x, -y)\) should also be on the graph.

Key Concepts

Parabola Orientation

Understanding the orientation of a parabola is crucial when graphing these curves. A parabola can open upwards, downwards, left, or right. The direction depends on the equation's terms. For the given equation, \( y^2 = -\frac{1}{3}x \), it's important to notice the squared term: \( y^2 \). This leads to a horizontal parabola.

• When \( y^2 = kx \), it opens horizontally.
• If \( k > 0 \), the parabola opens to the right.
• If \( k < 0 \), as in our equation, it opens to the left.

Always look at the sign of the coefficient before the non-squared term. Here, because it's negative, the parabola orients to the left.

Vertex of the Parabola

The vertex is a key point on a parabola and acts as its "peak" or "valley," depending on orientation. For our equation \( y^2 = -\frac{1}{3}x \), this special point is the origin (0,0).

• A parabola in the form \( y^2 = kx \) or \( x^2 = ky \) without additional terms is centered at the origin.
• If a parabola has added constants, such as \( y^2 = -\frac{1}{3}(x - h) + k \), the vertex would shift to \((h, k)\).

Understanding the location of the vertex helps in sketching the graph accurately by knowing the center of symmetry and plot initiation.

Equation Symmetry

Symmetry in a parabolic graph ensures that for any point on the parabola, there is a directly corresponding point equidistant from the axis of symmetry. Our equation \( y^2 = -\frac{1}{3}x \) indicates symmetry around the x-axis.

• The equation contains \( y^2 \), suggesting reflection across the x-axis.
• For example, if a point \((x, y)\) exists on the graph, then \((x, -y)\) must also be on it.

This characteristic is critical for predicting the shape and positioning other key points in the graph.

Graphing Techniques

Graphing a parabola involves several key steps to ensure accuracy. Leveraging these techniques can aid in achieving a precise sketch.

• First, identify and plot the vertex, such as (0, 0) in our case.
• Next, determine the orientation and open the parabola accordingly. Our example opens to the left due to the equation's negative sign with \( x \).
• Select values of \( y \), substitute them into the equation, and calculate corresponding \( x \)-values. For example, if \( y = 1 \), then \( x = -\frac{1}{3} \), giving the point \((-\frac{1}{3}, 1)\).
• Verify symmetry by checking opposite points, like \((-\frac{1}{3}, -1)\) for \((-\frac{1}{3}, 1)\).

Using these methods makes the process of graph drawing intuitive and systematic, ensuring all aspects of the parabola are accurately represented.