Zeros of Polynomials Quick Check

Question

Zeros of Polynomials Quick Check

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Question
Which of the following graphs is a possible sketch of the polynomial f(x)=2x3(x−2)2?(1 point)
Responses

A curve passes through quadrants 3 and 1 on a coordinate plane with the x-axis ranging from negative 5 to 5 in unit increments.
Image with alt text: A curve passes through quadrants 3 and 1 on a coordinate plane with the x-axis ranging from negative 5 to 5 in unit increments.

A curve passes through quadrants 2, 3, and 1 on a coordinate plane with the x-axis ranging from negative 5 to 5 in unit increments.
Image with alt text: A curve passes through quadrants 2, 3, and 1 on a coordinate plane with the x-axis ranging from negative 5 to 5 in unit increments.

A curve passes through quadrants 3 and 1 on a coordinate plane with the x-axis ranging from negative 5 to 5 in unit increments.
Image with alt text: A curve passes through quadrants 3 and 1 on a coordinate plane with the x-axis ranging from negative 5 to 5 in unit increments.

An upward-opening parabola is plotted on a coordinate plane with the x-axis ranging from negative 5 to 5 in unit increments.
Image with alt text: An upward-opening parabola is plotted on a coordinate plane with the x-axis ranging from negative 5 to 5 in unit increments.
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Answer 1

Bot GPT 4-o mini answered
1 month ago

To determine which graph is a possible sketch of the polynomial \( f(x) = 2x^3(x-2)^2 \), we need to analyze the characteristics of this polynomial.

1. **Degree and Leading Coefficient**:
- The polynomial \( f(x) = 2x^3(x-2)^2 \) can be expanded to find its degree. The highest degree term comes from \( 2x^3 \), which is of degree 3, and \( (x-2)^2 \), which is of degree 2. The total degree of the polynomial is \( 3 + 2 = 5 \). Since the leading coefficient is positive (2), as \( x \to \infty \), \( f(x) \to \infty \) and as \( x \to -\infty \), \( f(x) \to -\infty \).

2. **Zeros of the Polynomial**:
- The zeros of the polynomial are given by \( 2x^3(x-2)^2 = 0 \).
- Setting \( f(x) = 0 \):
- From \( 2x^3 = 0 \), we get \( x = 0 \) (with a multiplicity of 3).
- From \( (x - 2)^2 = 0 \), we get \( x = 2 \) (with a multiplicity of 2).
- Therefore, the zeros of the polynomial are \( x = 0 \) (crosses the x-axis) and \( x = 2 \) (touches the x-axis without crossing).

3. **Behavior at the Zeros**:
- At \( x = 0 \), the polynomial will cross the x-axis (since 3 is odd).
- At \( x = 2 \), the polynomial will touch the x-axis and turn back (since 2 is even).

4. **Sketching the Graph**:
- Since the graph crosses the x-axis at \( x = 0 \) and touches the x-axis at \( x = 2 \), the behavior near the zeros will be:
- From negative infinity, the graph will come from below, cross the x-axis at \( x = 0 \), rise to a maximum (before touching at \( x = 2 \)), then come back down and touch (but not cross) the x-axis at \( x = 2 \), then rise again towards positive infinity.

Given this analysis, the correct sketch should:
- Cross the x-axis at \( x = 0 \).
- Touch (and not cross) the x-axis at \( x = 2 \).

Therefore, the graph must pass through quadrants 3 and 1, potentially touching quadrant 2 briefly around the point of contact. From your choices, you would be looking for case A or case C based on the shape and behavior described.

If the options contain a graph that reflects the above properties (which I can't see), choose the one that aligns with that reasoning. If you need specific guidance on which image represents this behavior, please provide descriptions or characteristics of the graphs.