Problem 9 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-20: Sketch the graph of the equation, and label the \(x\) - and \(y\)-intercepts. $$ x=\frac{1}{4} y^{2} $$

Step-by-Step Solution

Answer

The graph is a parabola opening to the right with its vertex and intercepts at the origin (0,0).

Step 1: Identify Equation Type

The given equation is \( x = \frac{1}{4} y^{2} \). This is a quadratic equation in terms of \( y \), resembling the standard form of a parabola \( y = ax^2 + bx + c \), but solved for \( x \). Therefore, it represents a parabola that opens horizontally.

Step 2: Determine the Direction and Vertex of the Parabola

Since the parabola is given in the form \( x = \frac{1}{4} y^2 \), it will open to the right. The vertex, the point from which the parabola symmetrically expands, is at the origin (0,0) because there are no constant or linear terms for either \( x \) or \( y \).

Step 3: Find the x-intercept

For the x-intercept, set \( y = 0 \) in the equation \( x = \frac{1}{4} y^2 \). Solving, we get \( x = \frac{1}{4} (0)^2 = 0 \). Hence, the x-intercept is at \( (0,0) \).

Step 4: Find the y-intercepts

To find the y-intercepts, set \( x = 0 \). The equation becomes \( 0 = \frac{1}{4} y^2 \). Solving for \( y \) gives \( y = 0 \). Therefore, the y-intercept is also at \( (0,0) \), which coincidentally is the same as the x-intercept.

Step 5: Sketch the Graph

The graph is a parabola that opens to the right and has its vertex at the origin. Plot the intercept \( (0,0) \) on the coordinate plane. Since the parabola opens to the right, sketch a symmetrical U-shape originating from the origin extending to the right, with the vertical symmetry line being \( y = 0 \).

Key Concepts

Understanding the x-intercept for Parabolas

The x-intercept of a graph is where it crosses the x-axis. For parabolas, determining the x-intercept involves setting the value of the dependent variable, usually y, to zero in the equation. In the given exercise, the equation is given as \(x = \frac{1}{4}y^2\).

To find the x-intercept:

• Substitute \(y = 0\) into the equation. This results in \(x = \frac{1}{4}(0)^2 = 0\).
• Thus, the x-intercept is found at the point \((0, 0)\), where the parabola meets the x-axis.

It's important to note that parabolas might have multiple x-intercepts, one, or none, depending on their orientation and position. In this case, because the parabola opens horizontally and is positioned at the origin, the x-intercept is directly at \((0, 0)\). This means that the vertex and the x-intercept coincide in this scenario.

Identifying the y-intercept of a Parabola

The y-intercept is where the graph intersects the y-axis. For a parabola, the y-intercept can be found by setting the variable x to zero in the equation. With our equation \(x = \frac{1}{4}y^2\), we can determine the y-intercept:

Follow these steps:

• Set \(x = 0\) in the equation, resulting in \(0 = \frac{1}{4}y^2\).
• Solving for \(y\) gives \(y = 0\).

The y-intercept is also at \((0, 0)\). Interestingly, in this problem, the y-intercept and the x-intercept are the same, occurring at the origin. This coincides with the vertex of the parabola, a common scenario with parabolas that are centered at the origin.

Vertex of a Parabola Explained

The vertex of a parabola is a crucial feature that signifies the turning point of the graph. It acts as the point where the parabola changes direction. In the equation \(x = \frac{1}{4}y^2\), the vertex can be seen directly by its form.

Key points about the vertex:

• The vertex is the point \((h, k)\) on the parabolic graph, derived from the equation's general form. For our specific equation \(x = ay^2\), the absence of added constants implies \((h, k) = (0, 0)\).
• The vertex is also the point of symmetry, meaning the parabola is a mirror image on either side of the line passing through the vertex. In this exercise, the parabola is symmetric around the line \(y = 0\).
• The vertex indicates the minimum or maximum x-value for parabolas opening horizontally. Since this parabola opens to the right, the vertex represents the minimum x-value of the parabola.

At \((0, 0)\), the vertex is both the starting point and a critical feature coordinating the structure of the parabola.