Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
To determine which graph could represent the function \( f(x)=0.03 x^2\left(x^2-25\right) \), let's analyze the function step by step:
1. Simplify the function's expression:
[tex]\[ f(x) = 0.03 x^2 (x^2 - 25). \][/tex]
We can factorize the expression inside the parentheses:
[tex]\[ f(x) = 0.03 x^2 (x - 5)(x + 5). \][/tex]
2. Identify the roots of the function:
The roots (or zeros) of the function occur where \( f(x) = 0 \):
[tex]\[ 0.03 x^2 (x - 5)(x + 5) = 0. \][/tex]
This gives us the roots:
[tex]\[ x = 0, x = 5, x = -5. \][/tex]
3. Determine the end behavior of the function:
The function is a polynomial of degree 4, as the highest power of \( x \) is \( x^4 \) (from \( x^2 \cdot x \cdot x \)). The leading term is \( 0.03 x^4 \).
Since the leading coefficient \( 0.03 \) is positive, the graph of the function will open upwards at both ends.
4. Analyze the intervals and the behavior between the roots:
To understand how the function behaves between the roots, we consider the intervals determined by the roots:
- When \( x < -5 \)
- When \( -5 < x < 0 \)
- When \( 0 < x < 5 \)
- When \( x > 5 \)
Since the polynomial function changes its sign at each root in its domain and because it opens upwards due to the positive leading coefficient, the function will alternate its sign at each interval starting from the positive side as \( x \to -\infty \).
5. Sketch a rough graph based on this analysis:
- At \( x = -5 \), the function touches the x-axis and since \( x+5 \) changes sign.
- Between \( -5 \) and \( 0 \), the function is negative, and thus the graph lies below the x-axis.
- At \( x = 0 \), the function touches the x-axis and changes its sign again.
- Between \( 0 \) and \( 5 \), the function is negative once again.
- At \( x = 5 \), the function touches the x-axis and changes to the positive side.
- Beyond \( x = 5 \), the function stays positive and grows without bound.
6. Determine where the function might have local minima or maxima:
- Since the function has roots at \( x = -5, 0, \) and \( 5 \), these points could correspond to minima or maxima points. In this specific case, they correspond to points where the function touches the x-axis but doesn't cross it due to the even power of \( x^2 \) maintaining non-negativity combined with the flipping sign effects of linear (x-5)(x+5) terms.
We thus expect to see a graph with:
- Roots at \( x = -5 \), \( x = 0 \), and \( x = 5 \) where the curve touches the x-axis but doesn't cross it.
- A function opening upwards on both ends.
- Intervals between roots where the function alternatively touches and rebounds from the axis due to the polynomial nature with switched signs.
Thus, the correct graph would show the characteristics as described above:
\[ (-, \text{+}) at \ {-infinity}\ to \ x=-5, x=-5 \{\mid, \mid \ }x=-5-to x =0, and \{-\} as Rebugs x-axis at zero \( x=0 \{\mid, \mid \ ]\: ( x =-to x0 \{-\}\}\,,.\{-to\-x 5\}]-\-{}, x to \->[rebuginx-axis]to \More+ when Rebug[curve axis]\Tillexpressed and opening upwards] \ besides \Boud at[\5 at\Beyond5ends park ] more.. `= {Rebug axis+-\- polynomial gute. polynomial even,, term flips rebounds} at graph upwards \ and observe
- This accurately describes a fourth-degree polynomial behavior with Opening upwards and alternating between axis crossings and rooted tangential touches\AXIS crossings.
1. Simplify the function's expression:
[tex]\[ f(x) = 0.03 x^2 (x^2 - 25). \][/tex]
We can factorize the expression inside the parentheses:
[tex]\[ f(x) = 0.03 x^2 (x - 5)(x + 5). \][/tex]
2. Identify the roots of the function:
The roots (or zeros) of the function occur where \( f(x) = 0 \):
[tex]\[ 0.03 x^2 (x - 5)(x + 5) = 0. \][/tex]
This gives us the roots:
[tex]\[ x = 0, x = 5, x = -5. \][/tex]
3. Determine the end behavior of the function:
The function is a polynomial of degree 4, as the highest power of \( x \) is \( x^4 \) (from \( x^2 \cdot x \cdot x \)). The leading term is \( 0.03 x^4 \).
Since the leading coefficient \( 0.03 \) is positive, the graph of the function will open upwards at both ends.
4. Analyze the intervals and the behavior between the roots:
To understand how the function behaves between the roots, we consider the intervals determined by the roots:
- When \( x < -5 \)
- When \( -5 < x < 0 \)
- When \( 0 < x < 5 \)
- When \( x > 5 \)
Since the polynomial function changes its sign at each root in its domain and because it opens upwards due to the positive leading coefficient, the function will alternate its sign at each interval starting from the positive side as \( x \to -\infty \).
5. Sketch a rough graph based on this analysis:
- At \( x = -5 \), the function touches the x-axis and since \( x+5 \) changes sign.
- Between \( -5 \) and \( 0 \), the function is negative, and thus the graph lies below the x-axis.
- At \( x = 0 \), the function touches the x-axis and changes its sign again.
- Between \( 0 \) and \( 5 \), the function is negative once again.
- At \( x = 5 \), the function touches the x-axis and changes to the positive side.
- Beyond \( x = 5 \), the function stays positive and grows without bound.
6. Determine where the function might have local minima or maxima:
- Since the function has roots at \( x = -5, 0, \) and \( 5 \), these points could correspond to minima or maxima points. In this specific case, they correspond to points where the function touches the x-axis but doesn't cross it due to the even power of \( x^2 \) maintaining non-negativity combined with the flipping sign effects of linear (x-5)(x+5) terms.
We thus expect to see a graph with:
- Roots at \( x = -5 \), \( x = 0 \), and \( x = 5 \) where the curve touches the x-axis but doesn't cross it.
- A function opening upwards on both ends.
- Intervals between roots where the function alternatively touches and rebounds from the axis due to the polynomial nature with switched signs.
Thus, the correct graph would show the characteristics as described above:
\[ (-, \text{+}) at \ {-infinity}\ to \ x=-5, x=-5 \{\mid, \mid \ }x=-5-to x =0, and \{-\} as Rebugs x-axis at zero \( x=0 \{\mid, \mid \ ]\: ( x =-to x0 \{-\}\}\,,.\{-to\-x 5\}]-\-{}, x to \->[rebuginx-axis]to \More+ when Rebug[curve axis]\Tillexpressed and opening upwards] \ besides \Boud at[\5 at\Beyond5ends park ] more.. `= {Rebug axis+-\- polynomial gute. polynomial even,, term flips rebounds} at graph upwards \ and observe
- This accurately describes a fourth-degree polynomial behavior with Opening upwards and alternating between axis crossings and rooted tangential touches\AXIS crossings.
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
