Question

Sketch the graph of f(x)=2x^4−6x^3−8x^2 using Desmos. How does the graph behave at the zeros of the polynomial?(1 point) The graph touches the the x-axis and turns around at 4 and -1, and crosses the x-axis at 0. The graph touches the the x-axis and turns around at 4 and -1, and crosses the x-axis at 0. The graph crosses the x-axis at 4 and -1, and touches the x-axis and turns around at 0. The graph crosses the x-axis at 4 and -1, and touches the x-axis and turns around at 0. The graph crosses the x-axis at 4 and -1. The graph crosses the x-axis at 4 and -1. The graph touches the x-axis and turns around at 4 and -1.

Answer 1

To analyze the behavior of the polynomial function \( f(x) = 2x^4 - 6x^3 - 8x^2 \), we need to find its zeros and determine the nature of those zeros (whether the graph touches or crosses the x-axis).

1. **Finding the zeros**:
We can start by factoring the polynomial:

\[
f(x) = 2x^2(x^2 - 3x - 4)
\]

Now we need to factor \( x^2 - 3x - 4 \):

\[
x^2 - 3x - 4 = (x - 4)(x + 1)
\]

Thus, we can rewrite \( f(x) \) as:

\[
f(x) = 2x^2 (x - 4)(x + 1)
\]

The zeros of the function are:
- From \( 2x^2 = 0 \) => \( x = 0 \) (with multiplicity 2)
- From \( x - 4 = 0 \) => \( x = 4 \) (with multiplicity 1)
- From \( x + 1 = 0 \) => \( x = -1 \) (with multiplicity 1)

2. **Determining the behavior at the zeros**:
- At \( x = 0 \) (multiplicity 2): The graph **touches** the x-axis and turns around.
- At \( x = 4 \) (multiplicity 1): The graph **crosses** the x-axis.
- At \( x = -1 \) (multiplicity 1): The graph **crosses** the x-axis.

From this analysis, we can conclude:
- The graph touches the x-axis and reverses direction at \( x = 0 \).
- The graph crosses the x-axis at \( x = 4 \) and \( x = -1 \).

Given the multiple-choice options you've provided, the correct option is:

- The graph touches the x-axis and turns around at 0, and crosses the x-axis at 4 and -1.

Feel free to sketch this function on Desmos or any graphing tool to visualize these behaviors accurately!