Answer :
Sure! Let's analyze and graph the function [tex]\( y = -x^2 + 3 \)[/tex].
### Step-by-Step Solution
#### 1. Identify the Type of Function
The given function [tex]\( y = -x^2 + 3 \)[/tex] is a quadratic equation. Quadratic equations generally produce parabolas when graphed. The standard form of a quadratic equation is [tex]\( y = ax^2 + bx + c \)[/tex]. In this case, [tex]\( a = -1 \)[/tex], [tex]\( b = 0 \)[/tex], and [tex]\( c = 3 \)[/tex].
#### 2. Vertex of the Parabola
The vertex of a parabola given by [tex]\( y = ax^2 + bx + c \)[/tex] can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Plugging in the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ x = -\frac{0}{2(-1)} = 0 \][/tex]
To find the [tex]\( y \)[/tex]-coordinate of the vertex, substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = -0^2 + 3 = 3 \][/tex]
Thus, the vertex of the parabola is at [tex]\( (0, 3) \)[/tex].
#### 3. Plotting Points Around the Vertex
To give a clear idea of how the parabola looks, let's calculate some [tex]\( y \)[/tex]-values for a range of [tex]\( x \)[/tex]-values.
Selected [tex]\( x \)[/tex]-values and their corresponding [tex]\( y \)[/tex]-values are:
[tex]\[ \begin{aligned} x = -10, & \quad y = -(-10)^2 + 3 = -100 + 3 = -97 \\ x = -9.8, & \quad y = -(-9.8)^2 + 3 \approx -96 + 3 = -93 \\ x = -9.6, & \quad y = -(-9.6)^2 + 3 \approx -92 + 3 = -89 \\ & \quad \vdots \\ x = 0, & \quad y = -0^2 + 3 = 3 \quad (\text{vertex}) \\ x = 9.6, & \quad y = -(9.6)^2 + 3 \approx -92 + 3 = -89 \\ x = 9.8, & \quad y = -(9.8)^2 + 3 \approx -96 + 3 = -93 \\ x = 10, & \quad y = -(10)^2 + 3 = -100 + 3 = -97 \\ \end{aligned} \][/tex]
The values illustrate the symmetry of the function about the vertex.
#### 4. Characteristics of the Parabola
- Opens Downwards: Since [tex]\( a = -1 \)[/tex] is negative, the parabola opens downwards.
- Axis of Symmetry: The vertical line [tex]\( x = 0 \)[/tex] (the line [tex]\( x = -\frac{b}{2a} \)[/tex]) is the axis of symmetry.
#### 5. Graphing the Function
To graph the function, plot the vertex and the points calculated. Draw a smooth curve through these points, ensuring the parabola opens downwards.
Visually, the function will appear as a downward-facing parabola with its highest point (vertex) at [tex]\( (0, 3) \)[/tex], and extending symmetrically on both sides of the y-axis.
### Summary
The function [tex]\( y = -x^2 + 3 \)[/tex] is a downward-facing parabola with the vertex at [tex]\( (0, 3) \)[/tex]. The parabola is symmetric about the y-axis, with its values becoming more negative as the absolute value of [tex]\( x \)[/tex] increases.
Here are some of the key points and their [tex]\( y \)[/tex]-values for reference:
- [tex]\((-10, -97)\)[/tex]
- [tex]\((-9.8, -93)\)[/tex]
- [tex]\((-9.6, -89)\)[/tex]
- [tex]\(\ldots\)[/tex]
- [tex]\((0, 3)\)[/tex] (vertex)
- [tex]\(\ldots\)[/tex]
- [tex]\((9.6, -89)\)[/tex]
- [tex]\((9.8, -93)\)[/tex]
- [tex]\((10, -97)\)[/tex]
This completes the analysis and graphing of the quadratic function [tex]\( y = -x^2 + 3 \)[/tex].
### Step-by-Step Solution
#### 1. Identify the Type of Function
The given function [tex]\( y = -x^2 + 3 \)[/tex] is a quadratic equation. Quadratic equations generally produce parabolas when graphed. The standard form of a quadratic equation is [tex]\( y = ax^2 + bx + c \)[/tex]. In this case, [tex]\( a = -1 \)[/tex], [tex]\( b = 0 \)[/tex], and [tex]\( c = 3 \)[/tex].
#### 2. Vertex of the Parabola
The vertex of a parabola given by [tex]\( y = ax^2 + bx + c \)[/tex] can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Plugging in the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ x = -\frac{0}{2(-1)} = 0 \][/tex]
To find the [tex]\( y \)[/tex]-coordinate of the vertex, substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = -0^2 + 3 = 3 \][/tex]
Thus, the vertex of the parabola is at [tex]\( (0, 3) \)[/tex].
#### 3. Plotting Points Around the Vertex
To give a clear idea of how the parabola looks, let's calculate some [tex]\( y \)[/tex]-values for a range of [tex]\( x \)[/tex]-values.
Selected [tex]\( x \)[/tex]-values and their corresponding [tex]\( y \)[/tex]-values are:
[tex]\[ \begin{aligned} x = -10, & \quad y = -(-10)^2 + 3 = -100 + 3 = -97 \\ x = -9.8, & \quad y = -(-9.8)^2 + 3 \approx -96 + 3 = -93 \\ x = -9.6, & \quad y = -(-9.6)^2 + 3 \approx -92 + 3 = -89 \\ & \quad \vdots \\ x = 0, & \quad y = -0^2 + 3 = 3 \quad (\text{vertex}) \\ x = 9.6, & \quad y = -(9.6)^2 + 3 \approx -92 + 3 = -89 \\ x = 9.8, & \quad y = -(9.8)^2 + 3 \approx -96 + 3 = -93 \\ x = 10, & \quad y = -(10)^2 + 3 = -100 + 3 = -97 \\ \end{aligned} \][/tex]
The values illustrate the symmetry of the function about the vertex.
#### 4. Characteristics of the Parabola
- Opens Downwards: Since [tex]\( a = -1 \)[/tex] is negative, the parabola opens downwards.
- Axis of Symmetry: The vertical line [tex]\( x = 0 \)[/tex] (the line [tex]\( x = -\frac{b}{2a} \)[/tex]) is the axis of symmetry.
#### 5. Graphing the Function
To graph the function, plot the vertex and the points calculated. Draw a smooth curve through these points, ensuring the parabola opens downwards.
Visually, the function will appear as a downward-facing parabola with its highest point (vertex) at [tex]\( (0, 3) \)[/tex], and extending symmetrically on both sides of the y-axis.
### Summary
The function [tex]\( y = -x^2 + 3 \)[/tex] is a downward-facing parabola with the vertex at [tex]\( (0, 3) \)[/tex]. The parabola is symmetric about the y-axis, with its values becoming more negative as the absolute value of [tex]\( x \)[/tex] increases.
Here are some of the key points and their [tex]\( y \)[/tex]-values for reference:
- [tex]\((-10, -97)\)[/tex]
- [tex]\((-9.8, -93)\)[/tex]
- [tex]\((-9.6, -89)\)[/tex]
- [tex]\(\ldots\)[/tex]
- [tex]\((0, 3)\)[/tex] (vertex)
- [tex]\(\ldots\)[/tex]
- [tex]\((9.6, -89)\)[/tex]
- [tex]\((9.8, -93)\)[/tex]
- [tex]\((10, -97)\)[/tex]
This completes the analysis and graphing of the quadratic function [tex]\( y = -x^2 + 3 \)[/tex].
