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Sagot :
Sure, let's work through the problem step-by-step.
### Step 1: Identifying the Function
The function given in the problem is:
[tex]\[ y = x^2 - 1 \][/tex]
### Step 2: Understanding the Function
This is a quadratic function in the standard form [tex]\( y = ax^2 + bx + c \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 0 \)[/tex], and [tex]\( c = -1 \)[/tex].
### Step 3: Graphing the Function
To graph [tex]\( y = x^2 - 1 \)[/tex], recognize that it is a parabola opening upwards since the coefficient of [tex]\( x^2 \)[/tex] is positive. The vertex form of a quadratic function is:
[tex]\[ y = a(x - h)^2 + k \][/tex]
For our function:
[tex]\[ y = 1(x - 0)^2 - 1 \][/tex]
This tells us that the vertex of the parabola is at the point [tex]\((0, -1)\)[/tex].
### Step 4: Finding Important Points
To get a better sense of the graph, let's identify a few key points by substituting various values of [tex]\( x \)[/tex]:
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 0^2 - 1 = -1 \][/tex]
Point: [tex]\((0, -1)\)[/tex]
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 1^2 - 1 = 0 \][/tex]
Point: [tex]\((1, 0)\)[/tex]
- When [tex]\( x = -1 \)[/tex]:
[tex]\[ y = (-1)^2 - 1 = 0 \][/tex]
Point: [tex]\((-1, 0)\)[/tex]
- When [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 2^2 - 1 = 3 \][/tex]
Point: [tex]\((2, 3)\)[/tex]
- When [tex]\( x = -2 \)[/tex]:
[tex]\[ y = (-2)^2 - 1 = 3 \][/tex]
Point: [tex]\((-2, 3)\)[/tex]
### Step 5: Plotting the Points
Plot these points on a coordinate grid:
- [tex]\((0, -1)\)[/tex] (vertex)
- [tex]\((1, 0)\)[/tex]
- [tex]\((-1, 0)\)[/tex]
- [tex]\((2, 3)\)[/tex]
- [tex]\((-2, 3)\)[/tex]
### Step 6: Drawing the Parabola
Connect these points smoothly to form the parabolic curve. Ensure that the parabola opens upwards and is symmetric around the y-axis.
### Step 7: Analysis of the Function
- Vertex: The lowest point on the graph is the vertex [tex]\((0, -1)\)[/tex].
- Axis of Symmetry: The vertical line [tex]\( x = 0 \)[/tex] (the y-axis).
- Y-Intercept: This occurs when [tex]\( x = 0 \)[/tex], so the y-intercept is [tex]\((0, -1)\)[/tex].
- X-Intercepts: These occur where [tex]\( y = 0 \)[/tex]. Solving the equation [tex]\( x^2 - 1 = 0 \)[/tex]:
[tex]\[ x^2 = 1 \][/tex]
[tex]\[ x = \pm 1 \][/tex]
Thus, the x-intercepts are [tex]\((1, 0)\)[/tex] and [tex]\((-1, 0)\)[/tex].
By following these steps, we've effectively analyzed and graphed the function [tex]\( y = x^2 - 1 \)[/tex].
### Step 1: Identifying the Function
The function given in the problem is:
[tex]\[ y = x^2 - 1 \][/tex]
### Step 2: Understanding the Function
This is a quadratic function in the standard form [tex]\( y = ax^2 + bx + c \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 0 \)[/tex], and [tex]\( c = -1 \)[/tex].
### Step 3: Graphing the Function
To graph [tex]\( y = x^2 - 1 \)[/tex], recognize that it is a parabola opening upwards since the coefficient of [tex]\( x^2 \)[/tex] is positive. The vertex form of a quadratic function is:
[tex]\[ y = a(x - h)^2 + k \][/tex]
For our function:
[tex]\[ y = 1(x - 0)^2 - 1 \][/tex]
This tells us that the vertex of the parabola is at the point [tex]\((0, -1)\)[/tex].
### Step 4: Finding Important Points
To get a better sense of the graph, let's identify a few key points by substituting various values of [tex]\( x \)[/tex]:
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 0^2 - 1 = -1 \][/tex]
Point: [tex]\((0, -1)\)[/tex]
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 1^2 - 1 = 0 \][/tex]
Point: [tex]\((1, 0)\)[/tex]
- When [tex]\( x = -1 \)[/tex]:
[tex]\[ y = (-1)^2 - 1 = 0 \][/tex]
Point: [tex]\((-1, 0)\)[/tex]
- When [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 2^2 - 1 = 3 \][/tex]
Point: [tex]\((2, 3)\)[/tex]
- When [tex]\( x = -2 \)[/tex]:
[tex]\[ y = (-2)^2 - 1 = 3 \][/tex]
Point: [tex]\((-2, 3)\)[/tex]
### Step 5: Plotting the Points
Plot these points on a coordinate grid:
- [tex]\((0, -1)\)[/tex] (vertex)
- [tex]\((1, 0)\)[/tex]
- [tex]\((-1, 0)\)[/tex]
- [tex]\((2, 3)\)[/tex]
- [tex]\((-2, 3)\)[/tex]
### Step 6: Drawing the Parabola
Connect these points smoothly to form the parabolic curve. Ensure that the parabola opens upwards and is symmetric around the y-axis.
### Step 7: Analysis of the Function
- Vertex: The lowest point on the graph is the vertex [tex]\((0, -1)\)[/tex].
- Axis of Symmetry: The vertical line [tex]\( x = 0 \)[/tex] (the y-axis).
- Y-Intercept: This occurs when [tex]\( x = 0 \)[/tex], so the y-intercept is [tex]\((0, -1)\)[/tex].
- X-Intercepts: These occur where [tex]\( y = 0 \)[/tex]. Solving the equation [tex]\( x^2 - 1 = 0 \)[/tex]:
[tex]\[ x^2 = 1 \][/tex]
[tex]\[ x = \pm 1 \][/tex]
Thus, the x-intercepts are [tex]\((1, 0)\)[/tex] and [tex]\((-1, 0)\)[/tex].
By following these steps, we've effectively analyzed and graphed the function [tex]\( y = x^2 - 1 \)[/tex].
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