Question
Use the zeros to determine which of the following graphs is a sketch of the polynomial f(x)=−x(x+2)2 .(1 point)ResponsesImage with alt text: A curve with arrows at both the ends is plotted on a coordinate plane. The x-axis ranges from negative 6 to 2 in unit increments. The y-axis does not show any values.Image with alt text: A downward-opening parabola is plotted on a coordinate plane with the x-axis ranging from negative 4 to 2 in unit increments. Image with alt text: A curve passes through quadrants 2 and 4 on a coordinate plane with the x-axis ranging from negative 6 to 2 in unit increments.Image with alt text: A curve passes through quadrants 2 and 4 on a coordinate plane with the x-axis ranging from negative 2 to 6 in unit increments.Skip to navigationpage 14 of 14
Answers
Answered by
GPT-4o mini
To determine which graph represents the polynomial \( f(x) = -x(x+2)^2 \), we first need to analyze the function.
1. **Identify the Zeros**:
The zeros of the polynomial are found by setting \( f(x) = 0 \):
\[
-x(x+2)^2 = 0
\]
This gives us:
\[
x = 0 \quad \text{and} \quad (x+2)^2 = 0 \implies x = -2
\]
Thus, the zeros of the function are \( x = 0 \) and \( x = -2 \). The zero at \( x = -2 \) has multiplicity 2 because of the squared term.
2. **Behavior at the Zeros**:
- At \( x = 0 \), the graph crosses the x-axis because this zero has odd multiplicity (1).
- At \( x = -2 \), the graph touches the x-axis but does not cross it because this zero has even multiplicity (2).
3. **End Behavior**:
Since the leading coefficient of the polynomial is negative (as indicated by the minus sign in front), the ends of the graph will both point downwards:
- As \( x \to -\infty \), \( f(x) \to +\infty \)
- As \( x \to +\infty \), \( f(x) \to -\infty \)
4. **Sketching the Graph**:
- The graph will start high on the left side (as \( x \to -\infty \)) and will approach the x-axis at \( x = -2 \), touching it there.
- Then, it will go up to cross the x-axis at \( x = 0 \) before heading downwards as \( x \) increases beyond 0.
Now, based on this analysis, the correct graph should:
- Touch at \( x = -2 \)
- Cross at \( x = 0 \)
- Start high on the left and end low on the right.
Among the options given, select the graph that matches these characteristics to conclude.
1. **Identify the Zeros**:
The zeros of the polynomial are found by setting \( f(x) = 0 \):
\[
-x(x+2)^2 = 0
\]
This gives us:
\[
x = 0 \quad \text{and} \quad (x+2)^2 = 0 \implies x = -2
\]
Thus, the zeros of the function are \( x = 0 \) and \( x = -2 \). The zero at \( x = -2 \) has multiplicity 2 because of the squared term.
2. **Behavior at the Zeros**:
- At \( x = 0 \), the graph crosses the x-axis because this zero has odd multiplicity (1).
- At \( x = -2 \), the graph touches the x-axis but does not cross it because this zero has even multiplicity (2).
3. **End Behavior**:
Since the leading coefficient of the polynomial is negative (as indicated by the minus sign in front), the ends of the graph will both point downwards:
- As \( x \to -\infty \), \( f(x) \to +\infty \)
- As \( x \to +\infty \), \( f(x) \to -\infty \)
4. **Sketching the Graph**:
- The graph will start high on the left side (as \( x \to -\infty \)) and will approach the x-axis at \( x = -2 \), touching it there.
- Then, it will go up to cross the x-axis at \( x = 0 \) before heading downwards as \( x \) increases beyond 0.
Now, based on this analysis, the correct graph should:
- Touch at \( x = -2 \)
- Cross at \( x = 0 \)
- Start high on the left and end low on the right.
Among the options given, select the graph that matches these characteristics to conclude.