Problem 30 - Algebra 2

Question

Identify the vertex, the focus, and the directrix of each graph. Then sketch the graph. $$ (x-2)^{2}=4 y $$

Step-by-Step Solution

Answer

The vertex is at (2,0), the focus is at (2,1), and the directrix is the line \(y = -1\).

Step 1: Identify the Standard Form of a Parabola

Firstly, recognize that the given equation is in vertex form for a parabola that opens either up or down. The standard form for a parabola that opens up is \((x - h)^2 = 4a(y - k)\), where \((h, k)\) is the vertex of the parabola, and \(a\) is the distance from the vertex to the focus, and the distance from the vertex to the directrix.

Step 2: Identify the Vertex

In our equation \((x-2)^2 = 4y\), we have \(h = 2\) and \(k = 0\), so the vertex of the parabola is \((2, 0)\).

Step 3: Identify the Focus

We have \(4a = 4\), so \(a = 1\). Since our parabola opens up, the focus of the parabola will be \(a\) units above the vertex, at \((2, 1)\).

Step 4: Identify the Directrix

The directrix of the parabola will be \(a\) units below the vertex, at \(y = -1\).

Step 5: Sketch the Graph

Now that we have the vertex, focus, and directrix, we can sketch the graph. The parabola opens upward, vertex is at \((2, 0)\), it passes through the focus at \((2, 1)\) and the directrix is the line \(y = -1\).

Key Concepts

Vertex

When dealing with parabolas, the vertex is an essential point that describes its shape and position. It acts as the parabolic peak or valley. For the parabola expressed by \[(x-2)^2 = 4y\] the vertex is located at the point \((2, 0)\).

• The vertex in this form, \((h, k)\), serves as a helpful reference for graphing.
• Changing \(h\) and \(k\) shifts the parabola horizontally and vertically.

To visualize, imagine sliding the graph left or right by varying \(h\), or up and down by altering \(k\). This manipulation allows easy positioning of the parabola on the graph. Think of it as the anchoring point from which the parabola extends.

Focus

The focus of a parabola is a special point that helps to determine the "direction" in which the parabola opens. It is always located inside the parabola. For our current equation, \((x-2)^2 = 4y\), we have calculated the focus to be at \((2, 1)\).

• The focus is \(a\) units away from the vertex, \(a\) being a positive distance measured along the axis of symmetry.
• For this parabola opening upwards, the focus lies directly above the vertex.

The distance \(a\) is derived from the equation's component \(4a = 4\), thus \(a = 1\). This means the focus is exactly 1 unit above the vertex, enhancing our understanding of this parabolic curve's layout.

Directrix

The directrix of a parabola is a critical line used to define and understand the parabola's structure. It is perpendicular to the parabola's axis of symmetry and located away from the side that opens. For the equation \((x-2)^2 = 4y\), the directrix is positioned at \(y = -1\).

• The directrix is an imaginary boundary that shapes the path of the parabola.
• This line is equal in distance from the vertex to the focus, but in the opposite direction.

By considering the line \(y = -1\), we can visualize how the parabola balances between the focus above and the directrix below. This equilibrium helps in sketching accurate graphs and understanding how parabolas behave in a coordinate plane.