Problem 12 - Algebra for College Students

Question

For Problems \(1-30\), find the vertex, focus, and directrix of the given parabola and sketch its graph. $$ y^{2}=4(x+1) $$

Step-by-Step Solution

Answer

Vertex: \((-1, 0)\), Focus: \((0, 0)\), Directrix: \(x = -2\).

Step 1: Identify the given equation type

The given equation is \( y^2 = 4(x+1) \). This is the equation of a parabola that opens horizontally. Parabolas with equations of the form \( y^2 = 4px \) open to the right if \( p > 0 \), and to the left if \( p < 0 \).

Step 2: Rewrite the equation in standard form

The standard form for a horizontally oriented parabola is \( (y-k)^2 = 4p(x-h) \), where \((h, k)\) is the vertex of the parabola. Rewrite the equation: \( (y - 0)^2 = 4(x + 1) \).

Step 3: Identify the vertex

From the equation \( (y-0)^2 = 4(x+1) \), identify the vertex as the point \((h, k)\). Here, \( h = -1 \) and \( k = 0 \), so the vertex is \((-1, 0)\).

Step 4: Determine the value of p and interpret it

The equation \( (y-0)^2 = 4(x+1) \) is in the form \( (y-k)^2 = 4p(x-h) \), where \( 4p = 4 \). Thus, \( p = 1 \). Since \( p > 0 \), the parabola opens to the right.

Step 5: Find the focus

The focus of the parabola is \( p \) units from the vertex in the direction the parabola opens. Since the vertex is \((-1, 0)\) and \( p = 1 \), the focus is at \((-1 + 1, 0) = (0, 0)\).

Step 6: Find the directrix

The directrix is a vertical line \( p \) units from the vertex in the direction opposite to the direction of the parabola opening. Hence, the directrix is \( x = h - p = -1 - 1 = -2 \).

Step 7: Sketch the graph

Plot the vertex at \((-1, 0)\) and the focus at \((0, 0)\). Draw a vertical line for the directrix at \( x = -2 \). Sketch the parabola opening to the right, passing through the vertex and curving around the focus.

Key Concepts

Vertex

The vertex of a parabola is a significant point that serves as the "tip" or the "turning point" of the curve. For a horizontally oriented parabola, like the one in the equation \( y^2 = 4(x+1) \), the vertex acts as the central point from which the parabola opens left or right.

In the provided equation, we can rewrite it in standard form: \( (y-0)^2 = 4(x+1) \). This allows us to identify the vertex coordinates directly. Here, the vertex is located at the point \((-1, 0)\).

Understanding the vertex helps in sketching the parabola as it tells us where the curve will turn. When plotting, place the vertex accurately to maintain the correct shape and size of the parabola. Every parabola has only one vertex, which is crucial for graphing and solving related problems.

Focus

The focus of a parabola is a unique point that defines its properties, including its shape and direction. It's an essential part of understanding how a parabola behaves and is positioned.

For the given parabola \( y^2 = 4(x+1) \), we need to determine the parameter \( p \). In this equation, \( 4p = 4 \), meaning \( p = 1 \). The vertex is located at \((-1, 0)\), and since the parabola opens to the right, the focus is found by moving \( p \) units horizontally from the vertex.

• To find the focus, add \( p \) to the x-coordinate of the vertex: \( (-1 + 1, 0) = (0, 0) \).

This focus tells the path of the parabola. A ray of light passing through the vertex would be directed towards the focus, exhibiting the reflective properties of a parabola.

Directrix

The directrix of a parabola is a line that, together with the focus, defines the curve. It provides balance in the geometric construction of a parabola and helps describe its "width" relative to the vertex and the focus.

For our specific parabola \( y^2 = 4(x+1) \), we found the vertex at \((-1, 0)\) and the focus at \((0, 0)\). The directrix is a vertical line defined as being \( p \) units away from the vertex, in the opposite direction from which the parabola opens. In this case, the directrix line is:

• \(x = h - p = -1 - 1 = -2\)

By understanding the directrix, we gain insight into the span and boundaries of the parabola. It acts as a guideline for sketching the curve and ensuring it maintains its correct proportions relative to the focus.