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Sagot :
Sure, let's graph and analyze the function [tex]\( s(x) = x^2 - 2x + 4 \)[/tex].
1. Identify the Vertex:
The vertex form of a parabola [tex]\( ax^2 + bx + c \)[/tex] can be found using the formula for the vertex [tex]\( x = -\frac{b}{2a} \)[/tex].
For [tex]\( s(x) = x^2 - 2x + 4 \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -2 \)[/tex]
So the x-coordinate of the vertex is:
[tex]\[ x = -\frac{-2}{2 \cdot 1} = 1 \][/tex]
Substitute [tex]\( x = 1 \)[/tex] back into the function to find the y-coordinate of the vertex:
[tex]\[ s(1) = (1)^2 - 2(1) + 4 = 1 - 2 + 4 = 3 \][/tex]
Thus, the vertex is [tex]\( (1, 3) \)[/tex].
2. Plot the Vertex:
On the Cartesian plane, plot the point [tex]\( (1, 3) \)[/tex].
3. Find Additional Points:
For better understanding, let's choose another x-value, for instance, [tex]\( x = 0 \)[/tex].
Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ s(0) = (0)^2 - 2(0) + 4 = 4 \][/tex]
So another point on the graph is [tex]\( (0, 4) \)[/tex].
Additionally, choose [tex]\( x = 2 \)[/tex]:
[tex]\[ s(2) = (2)^2 - 2(2) + 4 = 4 - 4 + 4 = 4 \][/tex]
So another point on the graph is [tex]\( (2, 4) \)[/tex].
4. Sketch the Parabola:
Using the vertex [tex]\( (1, 3) \)[/tex] and points [tex]\( (0, 4) \)[/tex] and [tex]\( (2, 4) \)[/tex], sketch the parabola opening upwards.
5. Analyze the Function:
- Intervals of Increasing and Decreasing:
- The function [tex]\( s(x) \)[/tex] is a parabola that opens upwards.
- It is decreasing on [tex]\( (-\infty, 1) \)[/tex] (as [tex]\( x \)[/tex] moves towards 1 from the left).
- It is increasing on [tex]\( (1, \infty) \)[/tex] (as [tex]\( x \)[/tex] moves away from 1 to the right).
- Maximum or Minimum:
- Since the parabola opens upwards, [tex]\( s(x) \)[/tex] has no maximum.
- It has a minimum value at the vertex.
- The minimum value is 3 at [tex]\( x = 1 \)[/tex].
- Domain and Range:
- The domain of [tex]\( s(x) \)[/tex], since it is a quadratic function, is [tex]\( (-\infty, \infty) \)[/tex].
- The range, given that the parabola opens upwards and the minimum value is 3, is [tex]\( [3, \infty) \)[/tex].
To summarize:
- Vertex: [tex]\( (1, 3) \)[/tex]
- Increasing on: [tex]\( (1, \infty) \)[/tex]
- Decreasing on: [tex]\( (-\infty, 1) \)[/tex]
- Maximum: Does not exist (DNE)
- Minimum: 3 at [tex]\( x = 1 \)[/tex]
- Domain: [tex]\( (-\infty, \infty) \)[/tex]
- Range: [tex]\( [3, \infty) \)[/tex]
1. Identify the Vertex:
The vertex form of a parabola [tex]\( ax^2 + bx + c \)[/tex] can be found using the formula for the vertex [tex]\( x = -\frac{b}{2a} \)[/tex].
For [tex]\( s(x) = x^2 - 2x + 4 \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -2 \)[/tex]
So the x-coordinate of the vertex is:
[tex]\[ x = -\frac{-2}{2 \cdot 1} = 1 \][/tex]
Substitute [tex]\( x = 1 \)[/tex] back into the function to find the y-coordinate of the vertex:
[tex]\[ s(1) = (1)^2 - 2(1) + 4 = 1 - 2 + 4 = 3 \][/tex]
Thus, the vertex is [tex]\( (1, 3) \)[/tex].
2. Plot the Vertex:
On the Cartesian plane, plot the point [tex]\( (1, 3) \)[/tex].
3. Find Additional Points:
For better understanding, let's choose another x-value, for instance, [tex]\( x = 0 \)[/tex].
Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ s(0) = (0)^2 - 2(0) + 4 = 4 \][/tex]
So another point on the graph is [tex]\( (0, 4) \)[/tex].
Additionally, choose [tex]\( x = 2 \)[/tex]:
[tex]\[ s(2) = (2)^2 - 2(2) + 4 = 4 - 4 + 4 = 4 \][/tex]
So another point on the graph is [tex]\( (2, 4) \)[/tex].
4. Sketch the Parabola:
Using the vertex [tex]\( (1, 3) \)[/tex] and points [tex]\( (0, 4) \)[/tex] and [tex]\( (2, 4) \)[/tex], sketch the parabola opening upwards.
5. Analyze the Function:
- Intervals of Increasing and Decreasing:
- The function [tex]\( s(x) \)[/tex] is a parabola that opens upwards.
- It is decreasing on [tex]\( (-\infty, 1) \)[/tex] (as [tex]\( x \)[/tex] moves towards 1 from the left).
- It is increasing on [tex]\( (1, \infty) \)[/tex] (as [tex]\( x \)[/tex] moves away from 1 to the right).
- Maximum or Minimum:
- Since the parabola opens upwards, [tex]\( s(x) \)[/tex] has no maximum.
- It has a minimum value at the vertex.
- The minimum value is 3 at [tex]\( x = 1 \)[/tex].
- Domain and Range:
- The domain of [tex]\( s(x) \)[/tex], since it is a quadratic function, is [tex]\( (-\infty, \infty) \)[/tex].
- The range, given that the parabola opens upwards and the minimum value is 3, is [tex]\( [3, \infty) \)[/tex].
To summarize:
- Vertex: [tex]\( (1, 3) \)[/tex]
- Increasing on: [tex]\( (1, \infty) \)[/tex]
- Decreasing on: [tex]\( (-\infty, 1) \)[/tex]
- Maximum: Does not exist (DNE)
- Minimum: 3 at [tex]\( x = 1 \)[/tex]
- Domain: [tex]\( (-\infty, \infty) \)[/tex]
- Range: [tex]\( [3, \infty) \)[/tex]
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