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Sagot :
To graph the parabola given by the equation [tex]\( y = -\frac{1}{2} x^2 \)[/tex], follow these steps:
### 1. Understanding the Parabola
The parabola described by the equation [tex]\( y = -\frac{1}{2} x^2 \)[/tex] opens downwards because the coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\(-\frac{1}{2}\)[/tex]) is negative.
### 2. Identify Key Points
To graph the parabola, it's useful to identify key points such as the vertex and several additional points on both sides of the vertex.
#### Vertex:
The vertex of the parabola [tex]\( y = ax^2 + bx + c \)[/tex] in this case occurs where [tex]\( b = 0 \)[/tex] and [tex]\( c = 0 \)[/tex], so:
[tex]\[ x = 0 \][/tex]
[tex]\[ y = -\frac{1}{2} (0)^2 = 0 \][/tex]
Thus, the vertex is at [tex]\( (0, 0) \)[/tex].
#### Additional Points:
Choose a few values of [tex]\( x \)[/tex] around the vertex, calculate the corresponding values of [tex]\( y \)[/tex], and plot these points. Here are a few:
For [tex]\( x = -2 \)[/tex]:
[tex]\[ y = -\frac{1}{2} (-2)^2 = -\frac{1}{2} \times 4 = -2 \][/tex]
For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = -\frac{1}{2} (-1)^2 = -\frac{1}{2} \times 1 = -\frac{1}{2} = -0.5 \][/tex]
For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = -\frac{1}{2} (1)^2 = -\frac{1}{2} \times 1 = -0.5 \][/tex]
For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = -\frac{1}{2} (2)^2 = -\frac{1}{2} \times 4 = -2 \][/tex]
### 3. Plot the Points
Now plot the points:
- Vertex: [tex]\( (0, 0) \)[/tex]
- [tex]\( (-2, -2) \)[/tex]
- [tex]\( (-1, -0.5) \)[/tex]
- [tex]\( (1, -0.5) \)[/tex]
- [tex]\( (2, -2) \)[/tex]
### 4. Sketch the Parabola
1. Draw the Cartesian plane with the x-axis (horizontal) and the y-axis (vertical).
2. Mark the plotted points on the Cartesian plane.
3. Draw a smooth curve through the points, ensuring the curve is symmetrical about the y-axis.
### Important Note
The parabola is symmetric around the y-axis because the parabolic equation [tex]\( y = -\frac{1}{2} x^2 \)[/tex] lacks a linear term ([tex]\( bx \)[/tex]) and a constant term ([tex]\( c \)[/tex]). Every point [tex]\( (x, y) \)[/tex] on one side of the y-axis has a corresponding point [tex]\( (-x, y) \)[/tex] on the other side.
### Summary of the Plotted Points
[tex]\[ \begin{array}{c|c} x & y \\ \hline -2 & -2 \\ -1 & -0.5 \\ 0 & 0 \\ 1 & -0.5 \\ 2 & -2 \\ \end{array} \][/tex]
By connecting these points with a smooth curve, you will be able to see the parabolic shape that opens downward.
### Graph Output
```
y-axis
6 |
5 |
4 |
3 |
2 |
1 |
0 | . (0,0)
-1 | . .
-2 | X X
-3 |
-4 |
-5 |
-6 |
______________________________________ x-axis
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
```
This graph should give you a clear view of the parabolic curve described by the equation [tex]\( y = -\frac{1}{2} x^2 \)[/tex].
### 1. Understanding the Parabola
The parabola described by the equation [tex]\( y = -\frac{1}{2} x^2 \)[/tex] opens downwards because the coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\(-\frac{1}{2}\)[/tex]) is negative.
### 2. Identify Key Points
To graph the parabola, it's useful to identify key points such as the vertex and several additional points on both sides of the vertex.
#### Vertex:
The vertex of the parabola [tex]\( y = ax^2 + bx + c \)[/tex] in this case occurs where [tex]\( b = 0 \)[/tex] and [tex]\( c = 0 \)[/tex], so:
[tex]\[ x = 0 \][/tex]
[tex]\[ y = -\frac{1}{2} (0)^2 = 0 \][/tex]
Thus, the vertex is at [tex]\( (0, 0) \)[/tex].
#### Additional Points:
Choose a few values of [tex]\( x \)[/tex] around the vertex, calculate the corresponding values of [tex]\( y \)[/tex], and plot these points. Here are a few:
For [tex]\( x = -2 \)[/tex]:
[tex]\[ y = -\frac{1}{2} (-2)^2 = -\frac{1}{2} \times 4 = -2 \][/tex]
For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = -\frac{1}{2} (-1)^2 = -\frac{1}{2} \times 1 = -\frac{1}{2} = -0.5 \][/tex]
For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = -\frac{1}{2} (1)^2 = -\frac{1}{2} \times 1 = -0.5 \][/tex]
For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = -\frac{1}{2} (2)^2 = -\frac{1}{2} \times 4 = -2 \][/tex]
### 3. Plot the Points
Now plot the points:
- Vertex: [tex]\( (0, 0) \)[/tex]
- [tex]\( (-2, -2) \)[/tex]
- [tex]\( (-1, -0.5) \)[/tex]
- [tex]\( (1, -0.5) \)[/tex]
- [tex]\( (2, -2) \)[/tex]
### 4. Sketch the Parabola
1. Draw the Cartesian plane with the x-axis (horizontal) and the y-axis (vertical).
2. Mark the plotted points on the Cartesian plane.
3. Draw a smooth curve through the points, ensuring the curve is symmetrical about the y-axis.
### Important Note
The parabola is symmetric around the y-axis because the parabolic equation [tex]\( y = -\frac{1}{2} x^2 \)[/tex] lacks a linear term ([tex]\( bx \)[/tex]) and a constant term ([tex]\( c \)[/tex]). Every point [tex]\( (x, y) \)[/tex] on one side of the y-axis has a corresponding point [tex]\( (-x, y) \)[/tex] on the other side.
### Summary of the Plotted Points
[tex]\[ \begin{array}{c|c} x & y \\ \hline -2 & -2 \\ -1 & -0.5 \\ 0 & 0 \\ 1 & -0.5 \\ 2 & -2 \\ \end{array} \][/tex]
By connecting these points with a smooth curve, you will be able to see the parabolic shape that opens downward.
### Graph Output
```
y-axis
6 |
5 |
4 |
3 |
2 |
1 |
0 | . (0,0)
-1 | . .
-2 | X X
-3 |
-4 |
-5 |
-6 |
______________________________________ x-axis
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
```
This graph should give you a clear view of the parabolic curve described by the equation [tex]\( y = -\frac{1}{2} x^2 \)[/tex].
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