To analyze the behavior of the graph of the polynomial function \( f(x) = 2x^4 - 6x^3 - 8x^2 \) at its zeros, we first need to find its zeros.
Step 1: Factor the polynomial
Let's factor the polynomial:
\[ f(x) = 2x^2(x^2 - 3x - 4) \]
Next, we can further factor the quadratic \( x^2 - 3x - 4 \): To factor \( x^2 - 3x - 4 \), we look for two numbers that multiply to -4 and add to -3. These numbers are -4 and 1:
\[ x^2 - 3x - 4 = (x - 4)(x + 1) \]
Thus, we can factor \( f(x) \) completely as:
\[ f(x) = 2x^2(x - 4)(x + 1) \]
Step 2: Identify the zeros
Setting \( f(x) = 0 \) gives us the zeros of the polynomial:
- From \( 2x^2 = 0 \), we get \( x = 0 \) (with multiplicity 2).
- From \( x - 4 = 0 \), we get \( x = 4 \) (with multiplicity 1).
- From \( x + 1 = 0 \), we get \( x = -1 \) (with multiplicity 1).
Step 3: Analyze the behavior at the zeros
- At \( x = 0 \): Since the zero has even multiplicity (2), the graph touches the x-axis at this point and turns around.
- At \( x = 4 \): The zero has odd multiplicity (1), so the graph crosses the x-axis at this point.
- At \( x = -1 \): Similarly, since this zero also has odd multiplicity (1), the graph crosses the x-axis at this point.
Conclusion
Putting this analysis together, we see that:
- The graph touches the x-axis and turns around at \( x = 0 \).
- The graph crosses the x-axis at \( x = 4 \) and \( x = -1 \).
Among the options provided, the correct response is:
The graph crosses at the x-axis at 4 and -1, and touches the x-axis and turns around at 0.
1 answer