Answer :
Let's analyze the given polynomial function [tex]\( f(x) = -x^2 (x-2)^2 (x+1) \)[/tex] step by step.
### (a) End behavior of the graph of [tex]\( f \)[/tex]
To determine the end behavior, we need to look at the leading term of the polynomial when it is fully expanded.
1. The powers of [tex]\( x \)[/tex] in the polynomial are:
- [tex]\( x^2 \)[/tex] from [tex]\(-x^2\)[/tex],
- [tex]\( (x-2)^2 \)[/tex] contributes [tex]\( x^2 \)[/tex],
- [tex]\( (x+1) \)[/tex] contributes [tex]\( x \)[/tex].
2. Multiplying these together, the highest power of [tex]\( x \)[/tex] is:
[tex]\[ -x^2 \cdot x^2 \cdot x = -x^5 \][/tex]
3. The leading term is thus [tex]\( -x^5 \)[/tex].
Given that the leading coefficient is negative ([tex]\( -1 \)[/tex]) and the highest degree is odd ([tex]\( x^5 \)[/tex]), the end behavior is:
- As [tex]\( x \to +\infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex]
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex]
Thus, the end behavior of the graph of [tex]\( f \)[/tex] is:
[tex]\[ \text{as } x \to \pm\infty, f(x) \to -\infty \][/tex]
### (b) Real zeros and their behavior at the x-axis
To find the real zeros of the polynomial, we set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ -x^2 (x-2)^2 (x+1) = 0 \][/tex]
The zeros of [tex]\( f(x) \)[/tex] are the solutions to [tex]\( -x^2 = 0 \)[/tex], [tex]\( (x-2)^2 = 0 \)[/tex], and [tex]\( x+1 = 0 \)[/tex].
1. [tex]\( -x^2 = 0 \Rightarrow x = 0 \)[/tex]
2. [tex]\( (x-2)^2 = 0 \Rightarrow x = 2 \)[/tex]
3. [tex]\( x+1 = 0 \Rightarrow x = -1 \)[/tex]
Next, determine whether each zero crosses or touches the x-axis:
- For [tex]\( x = 0 \)[/tex]: [tex]\( x^2 \)[/tex] has even multiplicity (2), so the graph touches the x-axis but does not cross it.
- For [tex]\( x = 2 \)[/tex]: [tex]\( (x-2)^2 \)[/tex] has even multiplicity (2), so the graph touches the x-axis but does not cross it.
- For [tex]\( x = -1 \)[/tex]: [tex]\( x + 1 \)[/tex] has odd multiplicity (1), so the graph crosses the x-axis.
Thus:
[tex]\[ \text{Zero(s) where the graph crosses the x-axis: } -1 \][/tex]
[tex]\[ \text{Zero(s) where the graph touches, but does not cross the x-axis: } 0, 2 \][/tex]
### (c) Finding the [tex]\( y \)[/tex]-intercept
The [tex]\( y \)[/tex]-intercept occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -0^2 (0-2)^2 (0+1) = 0 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept is:
[tex]\[ (0, 0) \][/tex]
### (d) Graph of [tex]\( f(x) \)[/tex]
Based on the above information, we can plot the points where the graph intersects the x-axis and y-axis:
- Intersects the x-axis at [tex]\( (-1, 0), (0, 0), \)[/tex] and [tex]\( (2, 0) \)[/tex].
- Intersects the y-axis at [tex]\( (0, 0) \)[/tex].
Behavior at zeros:
- At [tex]\( x = -1 \)[/tex], the graph crosses the x-axis.
- At [tex]\( x = 0 \)[/tex], the graph touches the x-axis but does not cross it.
- At [tex]\( x = 2 \)[/tex], the graph touches the x-axis but does not cross it.
To graph [tex]\( f(x) = -x^2 (x-2)^2 (x+1) \)[/tex]:
1. Plot the points [tex]\( (-1, 0) \)[/tex], [tex]\( (0, 0) \)[/tex], and [tex]\( (2, 0) \)[/tex] on the x-axis.
2. Note that at [tex]\( x = -1 \)[/tex] the graph will cross the x-axis.
3. Note that at [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex], the graph will touch the x-axis but not cross.
4. Ensure the end behavior as [tex]\( x \to \pm\infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex].
Use this information to sketch the graph.
### (a) End behavior of the graph of [tex]\( f \)[/tex]
To determine the end behavior, we need to look at the leading term of the polynomial when it is fully expanded.
1. The powers of [tex]\( x \)[/tex] in the polynomial are:
- [tex]\( x^2 \)[/tex] from [tex]\(-x^2\)[/tex],
- [tex]\( (x-2)^2 \)[/tex] contributes [tex]\( x^2 \)[/tex],
- [tex]\( (x+1) \)[/tex] contributes [tex]\( x \)[/tex].
2. Multiplying these together, the highest power of [tex]\( x \)[/tex] is:
[tex]\[ -x^2 \cdot x^2 \cdot x = -x^5 \][/tex]
3. The leading term is thus [tex]\( -x^5 \)[/tex].
Given that the leading coefficient is negative ([tex]\( -1 \)[/tex]) and the highest degree is odd ([tex]\( x^5 \)[/tex]), the end behavior is:
- As [tex]\( x \to +\infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex]
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex]
Thus, the end behavior of the graph of [tex]\( f \)[/tex] is:
[tex]\[ \text{as } x \to \pm\infty, f(x) \to -\infty \][/tex]
### (b) Real zeros and their behavior at the x-axis
To find the real zeros of the polynomial, we set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ -x^2 (x-2)^2 (x+1) = 0 \][/tex]
The zeros of [tex]\( f(x) \)[/tex] are the solutions to [tex]\( -x^2 = 0 \)[/tex], [tex]\( (x-2)^2 = 0 \)[/tex], and [tex]\( x+1 = 0 \)[/tex].
1. [tex]\( -x^2 = 0 \Rightarrow x = 0 \)[/tex]
2. [tex]\( (x-2)^2 = 0 \Rightarrow x = 2 \)[/tex]
3. [tex]\( x+1 = 0 \Rightarrow x = -1 \)[/tex]
Next, determine whether each zero crosses or touches the x-axis:
- For [tex]\( x = 0 \)[/tex]: [tex]\( x^2 \)[/tex] has even multiplicity (2), so the graph touches the x-axis but does not cross it.
- For [tex]\( x = 2 \)[/tex]: [tex]\( (x-2)^2 \)[/tex] has even multiplicity (2), so the graph touches the x-axis but does not cross it.
- For [tex]\( x = -1 \)[/tex]: [tex]\( x + 1 \)[/tex] has odd multiplicity (1), so the graph crosses the x-axis.
Thus:
[tex]\[ \text{Zero(s) where the graph crosses the x-axis: } -1 \][/tex]
[tex]\[ \text{Zero(s) where the graph touches, but does not cross the x-axis: } 0, 2 \][/tex]
### (c) Finding the [tex]\( y \)[/tex]-intercept
The [tex]\( y \)[/tex]-intercept occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -0^2 (0-2)^2 (0+1) = 0 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept is:
[tex]\[ (0, 0) \][/tex]
### (d) Graph of [tex]\( f(x) \)[/tex]
Based on the above information, we can plot the points where the graph intersects the x-axis and y-axis:
- Intersects the x-axis at [tex]\( (-1, 0), (0, 0), \)[/tex] and [tex]\( (2, 0) \)[/tex].
- Intersects the y-axis at [tex]\( (0, 0) \)[/tex].
Behavior at zeros:
- At [tex]\( x = -1 \)[/tex], the graph crosses the x-axis.
- At [tex]\( x = 0 \)[/tex], the graph touches the x-axis but does not cross it.
- At [tex]\( x = 2 \)[/tex], the graph touches the x-axis but does not cross it.
To graph [tex]\( f(x) = -x^2 (x-2)^2 (x+1) \)[/tex]:
1. Plot the points [tex]\( (-1, 0) \)[/tex], [tex]\( (0, 0) \)[/tex], and [tex]\( (2, 0) \)[/tex] on the x-axis.
2. Note that at [tex]\( x = -1 \)[/tex] the graph will cross the x-axis.
3. Note that at [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex], the graph will touch the x-axis but not cross.
4. Ensure the end behavior as [tex]\( x \to \pm\infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex].
Use this information to sketch the graph.
