Answer :
Given the equation of the parabola:
[tex]\[ y = \frac{1}{4}(x - 5)^2 - 4 \][/tex]
We want to identify the vertex, focus, directrix, and the direction in which the parabola opens. Let's break this down step by step:
### 1. Vertex Form and Vertex
The equation given is already in vertex form:
[tex]\[ y = a(x - h)^2 + k \][/tex]
Here, \(a = \frac{1}{4}\), \(h = 5\), and \(k = -4\). The vertex form tells us that the vertex (h, k) is:
[tex]\[ \text{Vertex} = (5, -4) \][/tex]
### 2. Focus and Directrix
To find the focus and directrix, we need to understand the parameter \( p \) which determines the distance from the vertex to the focus and directrix.
For a parabola in the form:
[tex]\[ y = \frac{1}{4p}(x - h)^2 + k \][/tex]
We have the distance \( p \) such that:
[tex]\[ \frac{1}{4p} = a \][/tex]
Given \( a = \frac{1}{4} \):
[tex]\[ \frac{1}{4p} = \frac{1}{4} \][/tex]
Solving for \( p \):
[tex]\[ 4p = 4 \][/tex]
[tex]\[ p = 1 \][/tex]
The focus is found by moving \( p \) units from the vertex in the direction the parabola opens. Since \( y \) is in terms of \((x - h)^2\), the parabola opens vertically.
For a parabola that opens upward (since \( a > 0 \)):
[tex]\[ \text{Focus} = (h, k + p) \][/tex]
So, substituting \( h = 5 \), \( k = -4 \), and \( p = 1 \):
[tex]\[ \text{Focus} = (5, -3) \][/tex]
The directrix is a horizontal line \( p \) units away from the vertex in the opposite direction of the focus:
[tex]\[ \text{Directrix} = y = k - p \][/tex]
So, substituting \( k = -4 \) and \( p = 1 \):
[tex]\[ \text{Directrix} = y = -5 \][/tex]
### 3. Direction of Opening
The value of \( a \) tells us that the parabola opens upwards since \( a = \frac{1}{4} > 0 \).
### Summary
- Vertex: (5, -4)
- Focus: (5, -3)
- Directrix: \( y = -5 \)
- Direction of Opening: Upwards
### Sketching the Graph
To sketch the graph:
1. Plot the vertex at (5, -4).
2. Since the parabola opens upwards, plot the focus at (5, -3) above the vertex.
3. Draw the directrix as the line \( y = -5 \) below the vertex.
4. Sketch the parabola opening upwards, ensuring the vertex is the lowest point, and it passes through the focus point.
This will give you a clear visual representation of how the parabola behaves based on the given equation.
[tex]\[ y = \frac{1}{4}(x - 5)^2 - 4 \][/tex]
We want to identify the vertex, focus, directrix, and the direction in which the parabola opens. Let's break this down step by step:
### 1. Vertex Form and Vertex
The equation given is already in vertex form:
[tex]\[ y = a(x - h)^2 + k \][/tex]
Here, \(a = \frac{1}{4}\), \(h = 5\), and \(k = -4\). The vertex form tells us that the vertex (h, k) is:
[tex]\[ \text{Vertex} = (5, -4) \][/tex]
### 2. Focus and Directrix
To find the focus and directrix, we need to understand the parameter \( p \) which determines the distance from the vertex to the focus and directrix.
For a parabola in the form:
[tex]\[ y = \frac{1}{4p}(x - h)^2 + k \][/tex]
We have the distance \( p \) such that:
[tex]\[ \frac{1}{4p} = a \][/tex]
Given \( a = \frac{1}{4} \):
[tex]\[ \frac{1}{4p} = \frac{1}{4} \][/tex]
Solving for \( p \):
[tex]\[ 4p = 4 \][/tex]
[tex]\[ p = 1 \][/tex]
The focus is found by moving \( p \) units from the vertex in the direction the parabola opens. Since \( y \) is in terms of \((x - h)^2\), the parabola opens vertically.
For a parabola that opens upward (since \( a > 0 \)):
[tex]\[ \text{Focus} = (h, k + p) \][/tex]
So, substituting \( h = 5 \), \( k = -4 \), and \( p = 1 \):
[tex]\[ \text{Focus} = (5, -3) \][/tex]
The directrix is a horizontal line \( p \) units away from the vertex in the opposite direction of the focus:
[tex]\[ \text{Directrix} = y = k - p \][/tex]
So, substituting \( k = -4 \) and \( p = 1 \):
[tex]\[ \text{Directrix} = y = -5 \][/tex]
### 3. Direction of Opening
The value of \( a \) tells us that the parabola opens upwards since \( a = \frac{1}{4} > 0 \).
### Summary
- Vertex: (5, -4)
- Focus: (5, -3)
- Directrix: \( y = -5 \)
- Direction of Opening: Upwards
### Sketching the Graph
To sketch the graph:
1. Plot the vertex at (5, -4).
2. Since the parabola opens upwards, plot the focus at (5, -3) above the vertex.
3. Draw the directrix as the line \( y = -5 \) below the vertex.
4. Sketch the parabola opening upwards, ensuring the vertex is the lowest point, and it passes through the focus point.
This will give you a clear visual representation of how the parabola behaves based on the given equation.
