To find the [tex]\( x \)[/tex]-intercepts of the polynomial function [tex]\( f(x) = x^4 - 25x^2 \)[/tex], we need to solve the equation [tex]\( f(x) = 0 \)[/tex].
Step-by-step:
### Step 1: Factorize the polynomial
First, we can factorize [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = x^4 - 25x^2 \][/tex]
We notice that this can be written as a difference of squares:
[tex]\[ f(x) = (x^2)^2 - (5x)^2 \][/tex]
[tex]\[ f(x) = (x^2 - 5)(x^2 + 5) \][/tex]
### Step 2: Solve for [tex]\( x \)[/tex]
Set each factor equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 - 5 = 0 \quad \text{and} \quad x^2 + 5 = 0 \][/tex]
Solve [tex]\( x^2 - 5 = 0 \)[/tex]:
[tex]\[ x^2 = 5 \][/tex]
[tex]\[ x = \pm \sqrt{5} \][/tex]
[tex]\[ x = \pm 5 \][/tex]
Solve [tex]\( x^2 + 5 = 0 \)[/tex]:
[tex]\[ x^2 = -5 \][/tex]
Since [tex]\( x^2 = -5 \)[/tex] has no real solutions, we ignore this part.
Thus, the [tex]\( x \)[/tex]-intercepts are:
[tex]\[ x = -5, 0, 5 \][/tex]
### Step 3: Determine the behavior at each intercept
We need to determine whether the graph crosses the [tex]\( x \)[/tex]-axis or touches the [tex]\( x \)[/tex]-axis and turns around at each intercept.
1. At [tex]\( x = -5 \)[/tex]:
- To determine the behavior, we use the second derivative test. If the second derivative at [tex]\( x = -5 \)[/tex] is positive, the graph touches the [tex]\( x \)[/tex]-axis and turns around; if negative, the graph crosses the [tex]\( x \)[/tex]-axis.
- The second derivative of [tex]\( f(x) \)[/tex] is [tex]\( 12x^2 - 50 \)[/tex].
- Evaluating the second derivative at [tex]\( x = -5 \)[/tex]:
[tex]\[ f''(-5) = 12 \times (-5)^2 - 50 = 12 \times 25 - 50 = 300 - 50 = 250 \][/tex]
- Since [tex]\( f''(-5) > 0 \)[/tex], the graph touches the [tex]\( x \)[/tex]-axis and turns around.
2. At [tex]\( x = 0 \)[/tex]:
- Evaluate the second derivative at [tex]\( x = 0 \)[/tex]:
[tex]\[ f''(0) = 12 \times (0)^2 - 50 = -50 \][/tex]
- Since [tex]\( f''(0) < 0 \)[/tex], the graph crosses the [tex]\( x \)[/tex]-axis.
3. At [tex]\( x = 5 \)[/tex]:
- Evaluate the second derivative at [tex]\( x = 5 \)[/tex]:
[tex]\[ f''(5) = 12 \times (5)^2 - 50 = 12 \times 25 - 50 = 300 - 50 = 250 \][/tex]
- Since [tex]\( f''(5) > 0 \)[/tex], the graph touches the [tex]\( x \)[/tex]-axis and turns around.
### Step 4: Conclusion
Putting it all together, the results are:
- At [tex]\( x = -5 \)[/tex]: touches the [tex]\( x \)[/tex]-axis and turns around.
- At [tex]\( x = 0 \)[/tex]: crosses the [tex]\( x \)[/tex]-axis.
- At [tex]\( x = 5 \)[/tex]: touches the [tex]\( x \)[/tex]-axis and turns around.
Based on the given choices, the correct answer is:
[tex]\[
\textbf{None of the provided options are correct, but the closest would be: }
\][/tex]
[tex]\[
\text{0 , touches the } x \text{-axis and turns around; } 5 \text{ , touches the } x \text{-axis and turns around; } -5 \text{ , touches the } x \text{-axis and turns around}
\][/tex]
