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Sagot :
Sure, let's go through the detailed solution step-by-step.
### Given Polynomial Function:
[tex]$ f(x) = (x + 2)^2 (x - 1)^2 $[/tex]
### (a) End Behavior:
We need to determine how the graph behaves as [tex]\(x \to \pm\infty\)[/tex].
- The polynomial function [tex]\(f(x) = (x+2)^2 (x-1)^2\)[/tex] is a product of two squared terms.
- Each squared term contributes to a positive value as [tex]\( x \to \pm\infty \)[/tex].
- Thus, the leading term will dominate the end behavior, and since the leading coefficient (from expanding the polynomial) is positive and of even degree (degree 4), the end behavior is that the polynomial will rise on both ends.
Answer:
The graph will rise to the left and rise to the right.
### (b) Real Zeros:
Next, we identify the real zeros and how the graph behaves at those points.
- To find the zeros of [tex]\(f(x)\)[/tex], set [tex]\(f(x) = 0\)[/tex]:
[tex]$ (x + 2)^2 (x - 1)^2 = 0 $[/tex]
- The solutions to this are [tex]\(x = -2\)[/tex] and [tex]\(x = 1\)[/tex].
1. At [tex]\(x = -2\)[/tex]:
- The zero [tex]\(x = -2\)[/tex] has a multiplicity of 2. Because the multiplicity is even, the graph touches the x-axis at this point but does not cross it.
2. At [tex]\(x = 1\)[/tex]:
- Similarly, the zero [tex]\(x = 1\)[/tex] has a multiplicity of 2. Thus, the graph also touches the x-axis at this point but does not cross it.
Answer:
- Zeros where the graph crosses the x-axis: None
- Zeros where the graph touches but does not cross the x-axis: [tex]\( -2, 1 \)[/tex]
### (c) Y-Intercept:
To find the y-intercept, we evaluate [tex]\(f(x)\)[/tex] at [tex]\(x = 0\)[/tex]:
[tex]$ f(0) = (0 + 2)^2 (0 - 1)^2 = 2^2 \cdot (-1)^2 = 4 $[/tex]
Answer:
The y-intercept is 4.
### (d) Graphing the Function:
To graph [tex]\(f(x) = (x + 2)^2 (x - 1)^2\)[/tex]:
1. Plot Points Where the Graph Intersects the Axes:
- The zeros at [tex]\(x = -2\)[/tex] and [tex]\(x = 1\)[/tex] must be plotted.
- The y-intercept is at [tex]\( (0, 4) \)[/tex].
2. Behavior at the Zeros:
- At [tex]\(x = -2\)[/tex] and [tex]\(x = 1\)[/tex], the graph touches the x-axis but does not cross it, due to the even multiplicities.
3. Drawing the Graph:
- Ensure the graph rises to the left and rises to the right.
- The graph should touch the x-axis at [tex]\( -2 \)[/tex] and [tex]\( 1 \)[/tex] and should smoothly bounce off the axis at these points.
- Make sure to pass through the y-intercept at [tex]\( (0, 4) \)[/tex].
This should give you a complete representation of the graph of the function [tex]\(f(x) = (x + 2)^2 (x - 1)^2\)[/tex].
### Given Polynomial Function:
[tex]$ f(x) = (x + 2)^2 (x - 1)^2 $[/tex]
### (a) End Behavior:
We need to determine how the graph behaves as [tex]\(x \to \pm\infty\)[/tex].
- The polynomial function [tex]\(f(x) = (x+2)^2 (x-1)^2\)[/tex] is a product of two squared terms.
- Each squared term contributes to a positive value as [tex]\( x \to \pm\infty \)[/tex].
- Thus, the leading term will dominate the end behavior, and since the leading coefficient (from expanding the polynomial) is positive and of even degree (degree 4), the end behavior is that the polynomial will rise on both ends.
Answer:
The graph will rise to the left and rise to the right.
### (b) Real Zeros:
Next, we identify the real zeros and how the graph behaves at those points.
- To find the zeros of [tex]\(f(x)\)[/tex], set [tex]\(f(x) = 0\)[/tex]:
[tex]$ (x + 2)^2 (x - 1)^2 = 0 $[/tex]
- The solutions to this are [tex]\(x = -2\)[/tex] and [tex]\(x = 1\)[/tex].
1. At [tex]\(x = -2\)[/tex]:
- The zero [tex]\(x = -2\)[/tex] has a multiplicity of 2. Because the multiplicity is even, the graph touches the x-axis at this point but does not cross it.
2. At [tex]\(x = 1\)[/tex]:
- Similarly, the zero [tex]\(x = 1\)[/tex] has a multiplicity of 2. Thus, the graph also touches the x-axis at this point but does not cross it.
Answer:
- Zeros where the graph crosses the x-axis: None
- Zeros where the graph touches but does not cross the x-axis: [tex]\( -2, 1 \)[/tex]
### (c) Y-Intercept:
To find the y-intercept, we evaluate [tex]\(f(x)\)[/tex] at [tex]\(x = 0\)[/tex]:
[tex]$ f(0) = (0 + 2)^2 (0 - 1)^2 = 2^2 \cdot (-1)^2 = 4 $[/tex]
Answer:
The y-intercept is 4.
### (d) Graphing the Function:
To graph [tex]\(f(x) = (x + 2)^2 (x - 1)^2\)[/tex]:
1. Plot Points Where the Graph Intersects the Axes:
- The zeros at [tex]\(x = -2\)[/tex] and [tex]\(x = 1\)[/tex] must be plotted.
- The y-intercept is at [tex]\( (0, 4) \)[/tex].
2. Behavior at the Zeros:
- At [tex]\(x = -2\)[/tex] and [tex]\(x = 1\)[/tex], the graph touches the x-axis but does not cross it, due to the even multiplicities.
3. Drawing the Graph:
- Ensure the graph rises to the left and rises to the right.
- The graph should touch the x-axis at [tex]\( -2 \)[/tex] and [tex]\( 1 \)[/tex] and should smoothly bounce off the axis at these points.
- Make sure to pass through the y-intercept at [tex]\( (0, 4) \)[/tex].
This should give you a complete representation of the graph of the function [tex]\(f(x) = (x + 2)^2 (x - 1)^2\)[/tex].
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