Answer :
Let's analyze the polynomial function \( f(x) = x^4 + x^3 - 2x^2 \) step-by-step:
1. Finding the roots of the polynomial:
We start by finding the values where \( f(x) = 0 \). These are the points where the graph intersects or touches the x-axis. Solving the equation \( x^4 + x^3 - 2x^2 = 0 \):
[tex]\[ x^2 (x^2 + x - 2) = 0 \][/tex]
This gives us:
[tex]\[ x^2 = 0 \quad \text{or} \quad x^2 + x - 2 = 0 \][/tex]
Solving \( x^2 = 0 \) gives:
[tex]\[ x = 0 \][/tex]
Solving \( x^2 + x - 2 = 0 \) using the quadratic formula \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \):
[tex]\[ x = \frac{{-1 \pm \sqrt{1 + 8}}}{2} \][/tex]
[tex]\[ x = \frac{{-1 \pm 3}}{2} \][/tex]
[tex]\[ x = 1 \quad \text{or} \quad x = -2 \][/tex]
Therefore, the roots are \( x = -2 \), \( x = 0 \), and \( x = 1 \).
2. Checking the multiplicity of each root:
- For \( x = 0 \): The polynomial can be factored as \( f(x) = x^2(x + 2)(x - 1) \). The root \( x = 0 \) has a multiplicity of 2, meaning the graph touches the x-axis at this point.
- For \( x = -2 \): The root \( x = -2 \) has a multiplicity of 1 since \( x + 2 \) is a simple linear factor.
- For \( x = 1 \): The root \( x = 1 \) has a multiplicity of 1 since \( x - 1 \) is a simple linear factor.
3. Describing the behavior at each root:
- Since \( x = 0 \) has a multiplicity of 2, the graph touches the x-axis at this point.
- Since \( x = -2 \) has a multiplicity of 1, the graph crosses the x-axis at this point.
- Since \( x = 1 \) has a multiplicity of 1, the graph crosses the x-axis at this point.
Hence, we describe the graph as follows:
The graph of the polynomial function \( f(x) = x^4 + x^3 - 2x^2 \) touches the x-axis at \( x = 0 \), crosses the x-axis at \( x = -2 \) and \( x = 1 \).
Thus, the correct statement is:
The graph touches the x-axis at [tex]\( x = -2 \)[/tex] and [tex]\( x = 1 \)[/tex] and crosses the x-axis at [tex]\( x = 0 \)[/tex].
1. Finding the roots of the polynomial:
We start by finding the values where \( f(x) = 0 \). These are the points where the graph intersects or touches the x-axis. Solving the equation \( x^4 + x^3 - 2x^2 = 0 \):
[tex]\[ x^2 (x^2 + x - 2) = 0 \][/tex]
This gives us:
[tex]\[ x^2 = 0 \quad \text{or} \quad x^2 + x - 2 = 0 \][/tex]
Solving \( x^2 = 0 \) gives:
[tex]\[ x = 0 \][/tex]
Solving \( x^2 + x - 2 = 0 \) using the quadratic formula \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \):
[tex]\[ x = \frac{{-1 \pm \sqrt{1 + 8}}}{2} \][/tex]
[tex]\[ x = \frac{{-1 \pm 3}}{2} \][/tex]
[tex]\[ x = 1 \quad \text{or} \quad x = -2 \][/tex]
Therefore, the roots are \( x = -2 \), \( x = 0 \), and \( x = 1 \).
2. Checking the multiplicity of each root:
- For \( x = 0 \): The polynomial can be factored as \( f(x) = x^2(x + 2)(x - 1) \). The root \( x = 0 \) has a multiplicity of 2, meaning the graph touches the x-axis at this point.
- For \( x = -2 \): The root \( x = -2 \) has a multiplicity of 1 since \( x + 2 \) is a simple linear factor.
- For \( x = 1 \): The root \( x = 1 \) has a multiplicity of 1 since \( x - 1 \) is a simple linear factor.
3. Describing the behavior at each root:
- Since \( x = 0 \) has a multiplicity of 2, the graph touches the x-axis at this point.
- Since \( x = -2 \) has a multiplicity of 1, the graph crosses the x-axis at this point.
- Since \( x = 1 \) has a multiplicity of 1, the graph crosses the x-axis at this point.
Hence, we describe the graph as follows:
The graph of the polynomial function \( f(x) = x^4 + x^3 - 2x^2 \) touches the x-axis at \( x = 0 \), crosses the x-axis at \( x = -2 \) and \( x = 1 \).
Thus, the correct statement is:
The graph touches the x-axis at [tex]\( x = -2 \)[/tex] and [tex]\( x = 1 \)[/tex] and crosses the x-axis at [tex]\( x = 0 \)[/tex].
