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Sagot :
Certainly! Let's go through the problem step-by-step to understand the solution for the given equation [tex]\( y = -\frac{1}{2} x^2 \)[/tex].
### Step 1: Understanding the Equation
The equation [tex]\( y = -\frac{1}{2} x^2 \)[/tex] is a quadratic equation in which [tex]\( y \)[/tex] is expressed as a function of [tex]\( x \)[/tex].
### Step 2: Identify the Components of the Quadratic Equation
This quadratic equation is in the form of [tex]\( y = ax^2 + bx + c \)[/tex], where:
- [tex]\( a = -\frac{1}{2} \)[/tex]
- [tex]\( b = 0 \)[/tex]
- [tex]\( c = 0 \)[/tex]
### Step 3: Analyzing the Parabola
Since this is a quadratic equation, its graph will be a parabola. Let's identify some key characteristics of this parabola:
#### Vertex
The vertex of a parabola [tex]\( y = ax^2 + bx + c \)[/tex] is given by the point [tex]\(\left( -\frac{b}{2a}, f\left(-\frac{b}{2a}\right) \right) \)[/tex]. Here, [tex]\( b = 0 \)[/tex], so the vertex is at [tex]\( x = 0 \)[/tex].
Plug [tex]\( x = 0 \)[/tex] back into the equation to find the corresponding [tex]\( y \)[/tex]-value:
[tex]\[ y = -\frac{1}{2} (0)^2 = 0 \][/tex]
Therefore, the vertex of the parabola is at [tex]\( (0, 0) \)[/tex].
#### Direction of the Parabola
The coefficient of [tex]\( x^2 \)[/tex] (in this case, [tex]\( a \)[/tex]) determines the direction the parabola opens. Since [tex]\( a = -\frac{1}{2} \)[/tex] is negative, the parabola opens downwards.
### Step 4: Plotting Key Points
To get a sense of the shape, let’s calculate the value of [tex]\( y \)[/tex] for a few values of [tex]\( x \)[/tex]:
When [tex]\( x = 1 \)[/tex]:
[tex]\[ y = -\frac{1}{2} (1)^2 = -\frac{1}{2} \][/tex]
When [tex]\( x = -1 \)[/tex]:
[tex]\[ y = -\frac{1}{2} (-1)^2 = -\frac{1}{2} \][/tex]
When [tex]\( x = 2 \)[/tex]:
[tex]\[ y = -\frac{1}{2} (2)^2 = -2 \][/tex]
When [tex]\( x = -2 \)[/tex]:
[tex]\[ y = -\frac{1}{2} (-2)^2 = -2 \][/tex]
### Step 5: Sketching the Graph
Putting it all together:
- The vertex is at (0, 0).
- The parabola opens downwards.
- Key points we calculated are (1, -0.5), (-1, -0.5), (2, -2), and (-2, -2).
Using these points and the direction, we can sketch the parabola.
### Final Result
The function [tex]\( y = -\frac{1}{2} x^2 \)[/tex] describes a downward-opening parabola with its vertex at the origin (0, 0) and a symmetric shape about the y-axis. The key points are symmetric around the y-axis: as you move away from the vertex, the y-values become more negative, indicating the downward opening.
Thus, the detailed solution reveals the characteristics and shape of the quadratic function [tex]\( y = -\frac{1}{2} x^2 \)[/tex].
### Step 1: Understanding the Equation
The equation [tex]\( y = -\frac{1}{2} x^2 \)[/tex] is a quadratic equation in which [tex]\( y \)[/tex] is expressed as a function of [tex]\( x \)[/tex].
### Step 2: Identify the Components of the Quadratic Equation
This quadratic equation is in the form of [tex]\( y = ax^2 + bx + c \)[/tex], where:
- [tex]\( a = -\frac{1}{2} \)[/tex]
- [tex]\( b = 0 \)[/tex]
- [tex]\( c = 0 \)[/tex]
### Step 3: Analyzing the Parabola
Since this is a quadratic equation, its graph will be a parabola. Let's identify some key characteristics of this parabola:
#### Vertex
The vertex of a parabola [tex]\( y = ax^2 + bx + c \)[/tex] is given by the point [tex]\(\left( -\frac{b}{2a}, f\left(-\frac{b}{2a}\right) \right) \)[/tex]. Here, [tex]\( b = 0 \)[/tex], so the vertex is at [tex]\( x = 0 \)[/tex].
Plug [tex]\( x = 0 \)[/tex] back into the equation to find the corresponding [tex]\( y \)[/tex]-value:
[tex]\[ y = -\frac{1}{2} (0)^2 = 0 \][/tex]
Therefore, the vertex of the parabola is at [tex]\( (0, 0) \)[/tex].
#### Direction of the Parabola
The coefficient of [tex]\( x^2 \)[/tex] (in this case, [tex]\( a \)[/tex]) determines the direction the parabola opens. Since [tex]\( a = -\frac{1}{2} \)[/tex] is negative, the parabola opens downwards.
### Step 4: Plotting Key Points
To get a sense of the shape, let’s calculate the value of [tex]\( y \)[/tex] for a few values of [tex]\( x \)[/tex]:
When [tex]\( x = 1 \)[/tex]:
[tex]\[ y = -\frac{1}{2} (1)^2 = -\frac{1}{2} \][/tex]
When [tex]\( x = -1 \)[/tex]:
[tex]\[ y = -\frac{1}{2} (-1)^2 = -\frac{1}{2} \][/tex]
When [tex]\( x = 2 \)[/tex]:
[tex]\[ y = -\frac{1}{2} (2)^2 = -2 \][/tex]
When [tex]\( x = -2 \)[/tex]:
[tex]\[ y = -\frac{1}{2} (-2)^2 = -2 \][/tex]
### Step 5: Sketching the Graph
Putting it all together:
- The vertex is at (0, 0).
- The parabola opens downwards.
- Key points we calculated are (1, -0.5), (-1, -0.5), (2, -2), and (-2, -2).
Using these points and the direction, we can sketch the parabola.
### Final Result
The function [tex]\( y = -\frac{1}{2} x^2 \)[/tex] describes a downward-opening parabola with its vertex at the origin (0, 0) and a symmetric shape about the y-axis. The key points are symmetric around the y-axis: as you move away from the vertex, the y-values become more negative, indicating the downward opening.
Thus, the detailed solution reveals the characteristics and shape of the quadratic function [tex]\( y = -\frac{1}{2} x^2 \)[/tex].
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